4.2 right triangle trigonometry homework answers

Section 4.2: Right Triangle Trigonometry

Learning objectives.

In this section you will:

• Write proportions using similar triangles
• Use right triangles to evaluate trigonometric functions
• Use trigonometric functions to solve triangles
• Find function values for common angles
• Use right-triangle trigonometry to solve applied problems

Similar Triangles

Two triangles are said to be similar if their corresponding angles have the same measure. In the interactive activity from Geogebra below, triangle ABC is similar to triangle DEF because angles A and D have the same measure, angles B and E have the same measure, and angles C and F have the same measure.

Made with Geogebra

On the blue triangle on the left, drag around points A, B, and/or C. Observe the ratios described beneath the triangles (like c/b and f/e). What do you notice about the ratios between the two similar triangles?

This is the other important property of similar triangles: corresponding sides are proportional . So, as long as we have a pair of similar triangles, ABC and DEF, their corresponding sides are proportional: $\frac{c}{b}=\frac{f}{e}$, $\frac{c}{a}=\frac{f}{d}$, and $\frac{b}{a}=\frac{e}{d}$, to name a few such proportions.

Similar Right Triangles

Now, we're building up to something magical. Recall (from somewhere in the depths of your geometry education) that the angles in a triangle should always add up to $180^\circ$. You can check with the triangles in the applet above. We're going to use this fact to create classes of right triangles. In other words, if we know we have a right triangle (a triangle with a $90^\circ$ angle), we can use one of the other angles in as a reference and create a whole class of similar triangles.

Let's start with a right triangle with another angle that is $28^\circ$. This is enough to describe a class of similar triangles, since the third angle in the triangle must be $180-28-90 = 62^\circ$; therefore, we have defined a group of similar triangles with angle measures of $90^\circ, 28^\circ$, and $62^\circ$. Now, if we create ratios using the sides of those triangles, we know that those ratios will always produce the same value. Check it out in the demo applet below:

If we resize the triangle, but don't change the angles, the ratios of the sides stay the same. This is such a special feature of right triangles that mathematicians decided to name those ratios!

Right Triangle Trigonometry

We can define the trigonometric functions in terms a reference angle $t$ and the lengths of the sides of the triangle. The adjacent side is the leg closest to the reference angle. (Adjacent means “next to.”) The opposite side is the leg across from the angle. The hypotenuse is the side of the triangle opposite the right angle. These sides are labeled in the figure below:

Given a class of right triangles with reference angle $t$, we know that the ratio of, say, the opposite side and the hypotenuse will always be a particular number given a particular value of $t$. So, we're going to name each of those ratios.

The Trigonometric Ratios

Given a right triangle with an acute angle of $t$, we define the following trigonometric ratios:

Sine: $\sin(t)=\frac{\text{opposite}}{\text{hypotenuse}}$

Cosine: $\cos(t)=\frac{\text{adjacent}}{\text{hypotenuse}}$

Tangent: $\tan(t) = \frac{\text{opposite}}{\text{adjacent}}$

A common mnemonic for remembering these relationships is SohCahToa (or as some students like to write it $S\frac{o}{h}C\frac{a}{h}T\frac{o}{a}$), formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”

Given the triangle shown below, find the value of the three main trigonometric ratios:

First, we want to label the sides. Since $\alpha$ is our reference, the side adjacent to the angle is 15, and the hypotenuse of the triangle is 17. The opposite side currently has an unknown length. Good news, though, we can use our old friend the Pythagorean Theorem to find the length of the opposite side: $$15^2+x^2 = 17^2$$ $$x^2 = 17^2 - 15^2$$ $$x^2 = 64$$ $$x = 8$$ Now, we find the values of the trigonometric ratios: $$\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}}=\frac{8}{17}$$ $$\cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}}=\frac{15}{17}$$ $$\tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}}=\frac{8}{15}$$

The Reciprocal Trigonometric Functions

In addition to sine, cosine, and tangent, there are three more functions. They can be defined either in terms of the sides of the right triangle, or by their relationship to the three main trigonometric functions.

The Reciprocal Trigonometric Ratios

Cosecant: $\csc(t)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{1}{\sin(t)}$

Secant: $\sec(t)=\frac{\text{hypotenuse}}{\text{adjacent}} = \frac{1}{\cos(t)}$

Cotangent: $\cot(t) = \frac{\text{adjacent}}{\text{opposite}}=\frac{1}{\tan(t)}$

Note the confusing terminology: cosecant is the reciprocal of sine , while secant is the reciprocal of cosine . This is one of those inconvenient math definitions that we have to accept and make our piece with.

Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals.

Using the triangle shown below, evaluate $\sin(\alpha), \cos(\alpha), \tan(\alpha), \csc(\alpha), \sec(\alpha)$, and $\cot(\alpha)$.

Again, we don't know the length of one of the legs, but we can figure that out using the Pythagorean Theorem:

Now, name the sides of the triangle according to the reference angle:

Next, evaluate the main trig ratios:

Now, to find the rest, we use the fact that cosecant is the reciprocal of sine: $$\csc(\alpha) = \frac{1}{\sin(\alpha)} = \frac{5}{4}$$ secant is the reciprocal of cosine: $$\sec(\alpha) = \frac{1}{\cos(\alpha)} = \frac{5}{3}$$ and cotangent is the reciprocal of tangent: $$\cot(\alpha) = \frac{1}{\tan(\alpha)} = \frac{3}{4}$$.

Using Trigonometric Functions to Solve Triangles

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

Find the unknown sides of the triangle:

Let's start by naming all of the sides of the triangle. Using the 30 degree angle as a reference, the side labelled a is adjacent, the side labelled 7 is opposite, and the side labelled c is the hypotenuse. If we want to solve for side a, then we are looking for the adjacent side, and we know the lenght of the opposite side. The common trigonometric ratio involving the opposite and adjacent sides is tangent.

Now, we use the tangent ratio to create an equation: $$\tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}}$$ $$\tan(30^\circ) = \frac{7}{a}$$ We want to solve for a, but it's in the denominator. So, we can multiply on both sides of the equation by a, then divide to isolate a: $$a\cdot \tan(30^\circ) = 7$$ $$a = \frac{7}{\tan(30^\circ)}$$ Because of the magical properties of similar right triangles, we know $\tan(30^\circ)$ is always a particular value, and we can use our calculator to get an approximation of this value. Be careful not to round too soon though; it's more accurate to enter $\frac{7}{\tan(30^\circ)}$ all at once in your calculator. If you do so, the approximate answer rounded to four decimal places is $$a \approx 12.1244$$.

Next, let's solve for $c$. To get the most accurate answer, we want to use as much of the original, given information as possible, and avoid using our rounded answer for a if at all possible. So, c is the hypotenuse, and we were given the length of the opposite side. The basic trig ratio involving the opposite side and the hypotenuse is sine, so we create the equation: $$\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\sin(30^\circ) = \frac{7}{c}$$ $$c = \frac{7}{\sin(30^\circ)}$$ Luckily, this time our answer happens to be exact (no need to round): $$c = 14$$

We can also use trigonometric ratios to solve for missing angles in a right triangle. First, we must gently introduce inverse trigonometric functions. We will study these inverse functions more deeply in a later section. For now, we only need to know the following relationship:

Inverse Trigonometric Functions

There are a couple of extremely important caveats to notice and understand. First of all:

$\sin^{-1}(x)$ is not the same as $\frac{1}{\sin(x)}$

$\cos^{-1}(x)$ is not the same as $\frac{1}{\cos(x)}$, $\tan^{-1}(x)$ is not the same as $\frac{1}{\tan(x)}$.

In other words, that exponent is not really an exponent. It's rather a tag that means "we want to use the inverse function here." Confusing, we know. If someone would make us monarchs of math, we'd get rid of this confusing notation for good. Alas. If we did want, for example, $\frac{1}{\sin(x)}$, we'd simply use $\csc(x)$.

Also notice that the inverse trigonometric functions swap the inputs and outputs of the trig functions. For example, with $\sin(a) = b$, the input (a) is an angle and the output, b, is the ratio of the opposite side over the hypotenuse of a triangle with reference angle a. With $\sin^{-1}(b) = a$, the input is that ratio of opposite side over hypotenuse, and the output tells us the reference angle we'd need to produce that ratio. Therefore, we'll be able to use these inverse functions to solve for angles.

Finally, notice that these inverse trigonometric functions are defined only for particular intervals of angles. That will be nothing to worry about in this section, since we're dealing in right triangles. Later on, we'll have to be a little bit more careful when interpreting the results after applying an inverse trigonometric function.

Given two sides of a right triangle, solve for an angle.

• Determine which sides of the triangle are involved and the corresponding "main" trigonometric ratio (i.e. sine, cosine, or tangent).
• Write an equation using that trigonometric ratio and the sides involved.
• Rewrite the equation using the corresponding inverse trigonometric function.
• Use a calculator to evaluate approximately.

In the picture, the side labeled 9 is adjacent to the reference angle, and the side labeled 12 is the hypotenuse. The common ratio involving adjacent and hypotenuse is cosine, so $$\cos(\theta) = \frac{9}{12}$$ Rewrite as an inverse function: $$\theta = \cos^{-1}\left(\frac{9}{12}\right)$$ Use a calculator to evaluate approximately: $$\theta \approx 41.4096^\circ$$

In the picture, the side labeled 6 is opposite to the reference angle, and the side labeled 10 is the hypotenuse. The common ratio involving opposite and hypotenuse is sine, so $$\sin(\theta) = \frac{6}{10}$$ Rewrite as an inverse function: $$\theta = \sin^{-1}\left(\frac{6}{10}\right)$$ Use a calculator to evaluate approximately: $$\theta \approx 36.8699^\circ$$

Special Right Triangles

$30^\circ-60^\circ-90^\circ$ triangles.

There are some triangles that are so common, and so special, that it's worth knowing the values of sine and cosine that they produce. To build the first type of special right triangle, we'll start with an equilateral triangle. In an equilateral triangle, all the angles are the same measure ($60^\circ$) and all the sides are the same length (we'll call that length $2x$ for now):

To create a right triangle that we can work with, we'll draw an altitude in this triangle. That will split the base exactly in half, and it will split the top angle exactly in half:

Now we can use the Pythagorean Theorem to solve for the height of this triangle: $$x^2+h^2=(2x)^2$$ $$x^2+h^2=4x^2$$ $$h^2= 3x^2$$ $$h = x\sqrt{3}$$

That gives us this reference triangle:

Now we can figure out the sine and cosine of $30^\circ$: $$\sin(30^\circ) = \frac{x}{2x} = \frac{1}{2}$$ $$\cos(30^\circ) = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}$$ and the sine and cosine of $60^\circ$: $$\sin(60^\circ) = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}$$ $$\cos(60^\circ) = \frac{x}{2x} = \frac{1}{2}$$

$45^\circ-45^\circ-45^\circ$ Triangles

There's another set of special sine and cosine values that we want to explore. Our door into these will be an isoscoles right triangle:

Again, we'll use the Pythagorean Theorem to solve, but this time for the hypotenuse: $$x^2 + x^2 = c^2$$ $$2x^2 = c^2$$ $$c = x\sqrt{2}$$

And then we can find the sine and cosine of $45^\circ$: $$\sin(45^\circ)=\frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}}$$ $$\cos(45^\circ)=\frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Commonly, we'll rationalize the denominator of this fraction: $$\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}$$

The sines and cosines of these angles are going to become our particular friends in this chapter. We'll also want to get to know them well when we're thinking in terms of radians. We recommend that you commit to memorizing the following (like back in the days when you learned your times tables!).

Sines and Cosines of Special Angles

$$\sin(30^\circ) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$ $$\cos(30^\circ) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

$$\sin(60^\circ) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$\cos(60^\circ) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

$$\sin(45^\circ) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ $$\cos(45^\circ) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height.

Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of $75^\circ$ with the ground? Round to the nearest foot.

First, draw the situation:

We know the side opposite the reference angle is 50 feet, and we want to solve for the hypotenuse (which represents the length of the ladder). Therefore, we'll use a sine function to solve the problem: $$\sin(75^\circ)=\frac{50}{x}$$ $$x = \frac{50}{\sin(75^\circ}$$ $$x \approx 51.76 \text{ ft}$$

A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 43°, and that the angle of depression to the bottom of the tower is 31°. How tall is the tower?

Please forgive the author's hand-drawn illustration:

As pictured above, this situation creates two right triangles. We want to solve in each of the opposite side to our reference angles, and we know that the adjacent side is of length 325 ft. Hence, we can use the tangent function: $$\tan(43^\circ) = \frac{x_1}{325}$$ $$x_1 = 325\tan(43^\circ)$$ and $$\tan(31^\circ) = \frac{x_2}{325}$$ $$x_2 = 325\tan(31^\circ)$$ If we add up these values, we have the total height of the tower (remember not to round until the very end!): $$x_1 + x_2 = 325\tan(43^\circ) + 325\tan(31^\circ) \approx 498.35 \text{ ft}$$

Practice Exercises:

In the following exercises, find the requested values.

1. Find $\theta, a,$ and $c$.

2. Find $\alpha, b$, and $c$.

3. Find $\theta, a$, and $c$.

4. Find $\beta, b$, and $c$.

1. $\theta = 30^{\circ}$,

$a = 3\sqrt{3}$,

$c = \sqrt{108} = 6\sqrt{3}$

2. $\alpha = 56^{\circ}$,

$b = 12 \tan(34^{\circ}) = 8.094$,

$c = 12\sec(34^{\circ}) = \dfrac{12}{\cos(34^{\circ})} \approx 14.475$

3. $\theta = 43^{\circ}$,

$a = 6\cot(47^{\circ}) = \dfrac{6}{\tan(47^{\circ})} \approx 5.595$,

$c = 6\csc(47^{\circ}) = \dfrac{6}{\sin(47^{\circ})} \approx 8.204$

4. $\beta = 40^{\circ}$,

$b = 2.5 \tan(50^{\circ}) \approx 2.979$,

$c = 2.5\sec(50^{\circ}) = \dfrac{2.5}{\cos(50^{\circ})} \approx 3.889$

In the following exercises, answer assuming $\theta$ is an angle in a right triangle.

5. If $\theta = 30^{\circ}$ and the side opposite $\theta$ has length $4$, how long is the side adjacent to $\theta$?

6. If $\theta = 15^{\circ}$ and the hypotenuse has length $10$, how long is the side opposite $\theta$?

7. If $\theta = 87^{\circ}$ and the side adjacent to $\theta$ has length $2$, how long is the side opposite $\theta$?

8. If $\theta = 38.2^{\circ}$ and the side opposite $\theta$ has lengh $14$, how long is the hypoteneuse?

9. If $\theta = 2.05^{\circ}$ and the hypotenuse has length $3.98$, how long is the side adjacent to $\theta$?

10. If $\theta = 42^{\circ}$ and the side adjacent to $\theta$ has length $31$, how long is the side opposite $\theta$?

5. The side adjacent to $\theta$ has length $4\sqrt{3} \approx 6.928$

6. The side opposite $\theta$ has length $10 \sin(15^{\circ}) \approx 2.588$

7. The side opposite $\theta$ is $2\tan(87^{\circ}) \approx 38.162$

8. The hypoteneuse has length $14 \csc(38.2^{\circ}) = \dfrac{14}{\sin(38.2^{\circ})} \approx 22.639$

9. The side adjacent to $\theta$ has length $3.98 \cos(2.05^{\circ}) \approx 3.977$

10. The side opposite $\theta$ has length $31\tan(42^{\circ}) \approx 27.912$

In the following exercises, find the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places.

11. 3, 4, and 5

12. 5, 12, and 13

13. 336, 527, and 625

11. $36.87^{\circ}$ and $53.13^{\circ}$

12. $22.62^{\circ}$ and $67.38^{\circ}$

13. $32.52^{\circ}$ and $57.48^{\circ}$

14. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is $21.4^{\circ}$. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem.

15. The broadcast tower for radio station WSAZ (Home of ``Algebra in the Morning with Carl and Jeff'') has two enormous flashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is $7.970^{\circ}$ and to the second light is $7.125^{\circ}$. Find the distance between the lights to the nearest foot.

16. From a firetower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is $2.5^{\circ}$. How far away from the base of the tower is the fire?

17. The ranger in question 16. sees a Sasquatch running directly from the fire towards the firetower. The ranger takes two sightings. At the first sighting, the angle of depression from the tower to the Sasquatch is $6^{\circ}$. The second sighting, taken just 10 seconds later, gives the the angle of depression as $6.5^{\circ}$. How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the firetower from his location at the second sighting? Round your answer to the nearest minute.

18. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is $50^{\circ}$ and the angle of depression to the base of the tree is $10^{\circ}$. What is the height of the tree? Round your answer to the nearest foot.

19. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The first sighting had an angle of depression of $8.2^{\circ}$ and the second sighting had an angle of depression of $25.9^{\circ}$. How far had the boat traveled between the sightings?

20. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a $43^{\circ}$ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground?

21. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place.

22. At Cliffs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a fire is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the fire? Express your answer using degree measure rounded to one decimal place.

23. Shelving is being built at the Utility Muffin Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall?

24. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place.

25. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let $\theta$ denote the angle of depression from the top of the tower to a point on the ground. If the range of the rifle with a tranquilizer dart is 300 feet, find the smallest value of $\theta$ for which the corresponding point on the ground is in range of the rifle. Round your answer to the nearest hundreth of a degree.

26. The rule of thumb for safe ladder use states that the length of the ladder should be at least four times as long as the distance from the base of the ladder to the wall. Assuming the ladder is resting against a wall which is `plumb' (that is, makes a $90^{\circ}$ angle with the ground), determine the acute angle the ladder makes with the ground, rounded to the nearest tenth of a degree.

14. The tree is about 47 feet tall.

15. The lights are about 75 feet apart.

16. The fire is about 4581 feet from the base of the tower.

17. The Sasquatch ran $200\cot(6^{\circ}) - 200\cot(6.5^{\circ}) \approx 147$ feet in those 10 seconds. This translates to $\approx 10$ miles per hour. At the scene of the second sighting, the Sasquatch was $\approx 1755$ feet from the tower, which means, if it keeps up this pace, it will reach the tower in about $2$ minutes.

18. The tree is about 41 feet tall.

19. The boat has traveled about 244 feet.

20. The tower is about 682 feet tall. The guy wire hits the ground about 731 feet away from the base of the tower.

21. $68.9^{\circ}$

22. $7.7^{\circ}$

23. $51^{\circ}$

24. $19.5^{\circ}$

25. $41.81^{\circ}$

26. $75.5^{\circ}$.

Chapter 2: Trigonometric Ratios

Exercises: 2.2 Right Triangle Trigonometry

Exercises homework 2.2.

• Use measurements to calculate the trigonometric ratios for acute angles #1-10, 57-60
• Use trigonometric ratios to find unknown sides of right triangles #11-26
• Solve problems using trigonometric ratios #27-34, 41-46
• Use trig ratios to write equations relating the sides of a right triangle #35-40
• Use relationships among the trigonometric ratios #47-56, 61-68

Suggested Homework Problems

Here are two right triangles with a [latex]65 °[/latex] angle.

• Measure the sides [latex]AB[/latex] and [latex]BC[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
• Measure the sides [latex]AD[/latex] and [latex]DE[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
• Use your calculator to look up [latex]\sin 65°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 1 to calculate two estimates each for the cosine and tangent of [latex]65 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\cos 65°[/latex] and [latex]\tan 65°{.}[/latex]

Here are two right triangles with a [latex]40 °[/latex] angle.

• Measure the sides [latex]AB[/latex] and [latex]AC[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
• Measure the sides [latex]AD[/latex] and [latex]AE[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
• Use your calculator to look up [latex]\cos 40°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 2 to calculate two estimates each for the cosine and tangent of [latex]40 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\sin 40°[/latex] and [latex]\tan 40°{.}[/latex]

Exercise Group

For the right triangles in Problems 5–10,

• Find the length of the unknown side.
• Find the sine, cosine, and tangent of [latex]\theta \text{.}[/latex] Round your answers to four decimal places.

For Problems 11–16,

• Sketch and label the sides of a right triangle with angle [latex]\theta\text{.}[/latex]
• Sketch and label another right triangle with angle [latex]\theta[/latex] and longer sides.

[latex]\cos \theta = \dfrac{3}{5}[/latex]

[latex]\tan \theta = \dfrac{7}{2}[/latex]

[latex]\tan \theta = \dfrac{11}{4}[/latex]

[latex]\sin \theta = \dfrac{4}{9}[/latex]

[latex]\sin \theta = \dfrac{1}{9}[/latex]

[latex]\cos \theta = \dfrac{7}{8}[/latex]

For Problems 17–22, use one of the three trigonometric ratios to find the unknown side of the triangle. Round your answer to hundredths.

For Problems 23–26, sketch and label a right triangle with the given properties.

One angle is [latex]40°{,}[/latex] the side opposite that angle is 8 inches

One angle is [latex]65°{,}[/latex] the side adjacent to that angle is 30 yards

One angle is [latex]28°{,}[/latex] the hypotenuse is 56 feet

One leg is [latex]15[/latex] meters, the hypotenuse is [latex]18[/latex] meters

For Problems 27–34,

• Sketch a right triangle that illustrates the situation. Label your sketch with the given information.
• Choose the appropriate trig ratio and write an equation, then solve the problem.

To measure the height of cloud cover, airport controllers fix a searchlight to shine a vertical beam on the clouds. The searchlight is [latex]120[/latex] yards from the office. A technician in the office measures the angle of elevation to the light on the cloud cover at [latex]54.8°{.}[/latex] What is the height of the cloud cover?

To measure the distance across a canyon, Evel first sights an interesting rock directly opposite on the other side. He then walks [latex]200[/latex] yards down the rim of the canyon and sights the rock again, this time at an angle of [latex]18.5°[/latex] from the canyon rim. What is the width of the canyon?

A salvage ship is searching for the wreck of a pirate vessel on the ocean floor. Using sonar, they locate the wreck at an angle of depression of [latex]36.2°{.}[/latex] The depth of the ocean at their location is [latex]260[/latex] feet. How far should they move so that they are directly above the wrecked vessel?

Ramps for wheelchairs should be no steeper than an angle of [latex]6°{.}[/latex] How much horizontal distance should be allowed for a ramp that rises [latex]5[/latex] feet in height?

The radio signal from a weather balloon indicates that it is [latex]1500[/latex] meters from a meteorologist on the ground. The angle of elevation to the balloon is [latex]48°{.}[/latex] What is the balloon’s altitude?

According to Chinese legend, around 200 BC, the general Han Xin used a kite to determine the distance from his location to an enemy palace. He then dug a secret tunnel which emerged inside the palace. When the kite was directly above the palace, its angle of elevation was [latex]27°[/latex] and the string to the kite was [latex]1850[/latex] feet long. How far did Han Xin’s troops have to dig?

A cable car on a ski lift traverses a horizontal distance of [latex]1800[/latex] meters at an angle of [latex]38°{.}[/latex] How long is the cable?

Zelda is building the loft on her summer cottage. At its central point, the height of the loft is [latex]8[/latex] feet, and the pitch of the roof should be [latex]24°{.}[/latex] How long should the rafters be?

For Problems 35–40, use a trig ratio to write an equation for [latex]x[/latex] in terms of [latex]\theta{.}[/latex]

For Problems 41–44, find the altitude of the triangle. Round your answer to two decimal places.

For Problems 45 and 46, find the length of the chord [latex]AB{.}[/latex] Round your answer to two decimal places.

For Problems 47–50, fill in the table.

• In each of the figures for Problems 47-50, what is the relationship between the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex]
• Study the tables for Problems 47-50. What do you notice about the values of sine and cosine for the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex] Explain why this is true.

There is a relationship between the tangent, the sine, and the cosine of any angle. Study the tables for Problems 47-50 to discover this relationship. Write your answer as an equation.

• Use the figure to explain what happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases, and why.
• Use the figure to explain what happens to [latex]\cos \theta[/latex] as [latex]\theta[/latex] increases, and why.

• What happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases?
• What value does your calculator give for [latex]\tan 90°{?}[/latex] Why?

Explain why it makes sense that [latex]\sin 0° = 0[/latex] and [latex]\sin 90° = 1{.}[/latex] Use a figure to illustrate your explanation.

Explain why it makes sense that [latex]\cos 0° = 1[/latex] and [latex]\cos 90° = 0{.}[/latex] Use a figure to illustrate your explanation

For Problems 57–60, explain why the trigonometric ratio is not correct.

[latex]\sin \theta = \dfrac{5}{9}[/latex]

[latex]\tan \theta = \dfrac{4}{7}[/latex]

[latex]\cos \theta = \dfrac{21}{20}[/latex]

[latex]\sin \theta = \dfrac{8}{10}[/latex]

For Problems 61–64, sketch and label a right triangle, then fill in the blank.

If [latex]\sin \theta = 0.2358\text{,}[/latex]  then [latex]\cos (90^{o} - \theta)=\underline{\qquad} \text{,}[/latex]

• If [latex]\cos \alpha = \dfrac{3}{11} \text{,}[/latex] then [latex]\underline{\qquad} (90° - \alpha) = \dfrac{3}{11}\text{.}[/latex]
• If [latex]\sin 42° = n\text{,}[/latex] then [latex]\cos \underline{\qquad} = n\text{.}[/latex]
• If [latex]\cos 13° = z\text{,}[/latex] then [latex]\sin \underline{\qquad} = z\text{.}[/latex]
• If [latex]\cos \beta = \dfrac{2}{\sqrt{7}}{,}[/latex] then [latex]\sin (90° - \beta) =\underline{\qquad} {.}[/latex]
• If [latex]\sin \phi = 0.693{,}[/latex] then [latex](90° - \phi) = 0.693{.}[/latex]
• If [latex]\cos 87° = p{,}[/latex] then [latex]\sin \underline{\qquad} = p{.}[/latex]
• If [latex]\sin 59° =w{,}[/latex] then [latex]\cos \underline{\qquad} = w{.}[/latex]
• If [latex]\sin \phi = \dfrac{5}{13}[/latex] and [latex]\cos \phi = \dfrac{12}{13}{,}[/latex] then [latex]\tan \phi =\underline{\qquad} {.}[/latex]
• If [latex]\cos \beta = \dfrac{1}{\sqrt{10}}{,}[/latex] and [latex]\sin \beta = \dfrac{3}{\sqrt{10}}{,}[/latex] then [latex]\tan \beta = \underline{\qquad}{.}[/latex]
• If [latex]\tan B = \dfrac{2}{\sqrt{5}}[/latex] and [latex]\cos B = \dfrac{\sqrt{5}}{3}{,}[/latex] then [latex]\sin B =\underline{\qquad}{.}[/latex]
• If [latex]\sin W = \sqrt{\dfrac{3}{7}}[/latex] and [latex]\tan W = \dfrac{\sqrt{3}}{2}{,}[/latex] then [latex]\cos W =\underline{\qquad}{.}[/latex]
• If [latex]\cos \theta = \dfrac{2}{\sqrt{10}}[/latex] and [latex]\sin \theta = \sqrt{\dfrac{3}{5}}{,}[/latex] then [latex]\tan \theta = \underline{\qquad}{.}[/latex]
• If [latex]\sin \alpha = \dfrac{\sqrt{2}}{4}{,}[/latex] and [latex]\cos \alpha =\dfrac{\sqrt{14}}{4}{,}[/latex] then [latex]\tan \alpha = \underline{\qquad}{.}[/latex]
• If [latex]\tan A = \dfrac{\sqrt{7}}{3}[/latex] and [latex]\cos A = \dfrac{3}{4}{,}[/latex] then [latex]\sin A =\underline{\qquad}{.}[/latex]
• If [latex]\sin V = \sqrt{\dfrac{10}{5}}[/latex] and [latex]\tan V = \dfrac{2}{5}{,}[/latex] then [latex]\cos V =\underline{\qquad}{.}[/latex]

Explain why the cosine of a [latex]73°[/latex] angle is always the same, no matter what size triangle the angle is in. Illustrate your explanation with a sketch.

• If you plotted the points in your table, would they lie on a straight line? Why or why not?
• What is the slope of the line through the origin and point [latex]P{?}[/latex]
• What is the tangent of the angle [latex]\theta{?}[/latex]
• On the same grid, sketch an angle whose tangent is [latex]\dfrac{8}{5}.[/latex]
• Use your calculator to complete the table. Round your answers to hundredths.

• Use the values of tan [latex]\theta[/latex] to sketch all the angles listed in the table. Locate the vertex of each angle at the origin and the initial side along the positive [latex]x[/latex]-axis.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Chapter 4.2: Right Triangle Trigonometry

Solving right triangle problems.

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem . We begin with the familiar Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], and model an equation to fit a situation.

Example 18: Using the Pythagorean Theorem to Model an Equation

Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)?

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

The angle of elevation is [latex]\theta [/latex], formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

The angle of elevation is approximately [latex]{71.7}^{\circ }[/latex], and the length of the cable is 73.2 meters.

Example 19: Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “ a” feet from the wall, the length of the ladder will be 4 a feet.

The side adjacent to [latex]\theta [/latex] is a and the hypotenuse is [latex]4a[/latex]. Thus,

The elevation of the ladder forms an angle of [latex]{75.5}^{\circ }[/latex] with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

Thus, the ladder touches the wall at [latex]\sqrt{15}a[/latex] feet from the ground.

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

4.1 Exponential Functions

g ( x ) = 0.875 x g ( x ) = 0.875 x and j ( x ) = 1095.6 − 2 x j ( x ) = 1095.6 − 2 x represent exponential functions.

5.5556 5.5556

About 1.548 1.548 billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.

( 0 , 129 ) ( 0 , 129 ) and ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t

f ( x ) = 2 ( 1.5 ) x f ( x ) = 2 ( 1.5 ) x

f ( x ) = 2 ( 2 ) x . f ( x ) = 2 ( 2 ) x . Answers may vary due to round-off error. The answer should be very close to 1.4142 ( 1.4142 ) x . 1.4142 ( 1.4142 ) x .

y ≈ 12 ⋅ 1.85 x y ≈ 12 ⋅ 1.85 x

about $3,644,675.88

e − 0.5 ≈ 0.60653 e − 0.5 ≈ 0.60653

$3,659,823.44

3.77E-26 (This is calculator notation for the number written as 3.77 × 10 − 26 3.77 × 10 − 26 in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

4.2 Graphs of Exponential Functions

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 0 , ∞ ) ; ( 0 , ∞ ) ; the horizontal asymptote is y = 0. y = 0.

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 3 , ∞ ) ; ( 3 , ∞ ) ; the horizontal asymptote is y = 3. y = 3.

x ≈ − 1.608 x ≈ − 1.608

f ( x ) = − 1 3 e x − 2 ; f ( x ) = − 1 3 e x − 2 ; the domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( − ∞ , −2 ) ; ( − ∞ , −2 ) ; the horizontal asymptote is y = −2. y = −2.

4.3 Logarithmic Functions

• ⓐ log 10 ( 1 , 000 , 000 ) = 6 log 10 ( 1 , 000 , 000 ) = 6 is equivalent to 10 6 = 1 , 000 , 000 10 6 = 1 , 000 , 000
• ⓑ log 5 ( 25 ) = 2 log 5 ( 25 ) = 2 is equivalent to 5 2 = 25 5 2 = 25
• ⓐ 3 2 = 9 3 2 = 9 is equivalent to log 3 ( 9 ) = 2 log 3 ( 9 ) = 2
• ⓑ 5 3 = 125 5 3 = 125 is equivalent to log 5 ( 125 ) = 3 log 5 ( 125 ) = 3
• ⓒ 2 − 1 = 1 2 2 − 1 = 1 2 is equivalent to log 2 ( 1 2 ) = − 1 log 2 ( 1 2 ) = − 1

log 121 ( 11 ) = 1 2 log 121 ( 11 ) = 1 2 (recalling that 121 = ( 121 ) 1 2 = 11 121 = ( 121 ) 1 2 = 11 )

log 2 ( 1 32 ) = − 5 log 2 ( 1 32 ) = − 5

log ( 1 , 000 , 000 ) = 6 log ( 1 , 000 , 000 ) = 6

log ( 123 ) ≈ 2.0899 log ( 123 ) ≈ 2.0899

The difference in magnitudes was about 3.929. 3.929.

It is not possible to take the logarithm of a negative number in the set of real numbers.

4.4 Graphs of Logarithmic Functions

( 2 , ∞ ) ( 2 , ∞ )

( 5 , ∞ ) ( 5 , ∞ )

The domain is ( 0 , ∞ ) , ( 0 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

The domain is ( − 4 , ∞ ) , ( − 4 , ∞ ) , the range ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the asymptote x = – 4. x = – 4.

The domain is ( 2 , ∞ ) , ( 2 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 2. x = 2.

The domain is ( − ∞ , 0 ) , ( − ∞ , 0 ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

x ≈ 3.049 x ≈ 3.049

x = 1 x = 1

f ( x ) = 2 ln ( x + 3 ) − 1 f ( x ) = 2 ln ( x + 3 ) − 1

4.5 Logarithmic Properties

log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k

log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 ) log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 )

2 ln x 2 ln x

− 2 ln ( x ) − 2 ln ( x )

log 3 16 log 3 16

2 log x + 3 log y − 4 log z 2 log x + 3 log y − 4 log z

2 3 ln x 2 3 ln x

1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 ) 1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 )

log ( 3 ⋅ 5 4 ⋅ 6 ) ; log ( 3 ⋅ 5 4 ⋅ 6 ) ; can also be written log ( 5 8 ) log ( 5 8 ) by reducing the fraction to lowest terms.

log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) ) log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) )

log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; this answer could also be written log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 . log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 .

The pH increases by about 0.301.

ln 8 ln 0.5 ln 8 ln 0.5

ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861 ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861

4.6 Exponential and Logarithmic Equations

x = − 2 x = − 2

x = − 1 x = − 1

x = 1 2 x = 1 2

The equation has no solution.

x = ln 3 ln ( 2 3 ) x = ln 3 ln ( 2 3 )

t = 2 ln ( 11 3 ) t = 2 ln ( 11 3 ) or ln ( 11 3 ) 2 ln ( 11 3 ) 2

t = ln ( 1 2 ) = − 1 2 ln ( 2 ) t = ln ( 1 2 ) = − 1 2 ln ( 2 )

x = ln 2 x = ln 2

x = e 4 x = e 4

x = e 5 − 1 x = e 5 − 1

x ≈ 9.97 x ≈ 9.97

x = 1 x = 1 or x = − 1 x = − 1

t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years . t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years .

4.7 Exponential and Logarithmic Models

f ( t ) = A 0 e − 0.0000000087 t f ( t ) = A 0 e − 0.0000000087 t

less than 230 years, 229.3157 to be exact

f ( t ) = A 0 e ln 2 3 t f ( t ) = A 0 e ln 2 3 t

6.026 hours

895 cases on day 15

Exponential. y = 2 e 0.5 x . y = 2 e 0.5 x .

y = 3 e ( ln 0.5 ) x y = 3 e ( ln 0.5 ) x

4.8 Fitting Exponential Models to Data

• ⓐ The exponential regression model that fits these data is y = 522.88585984 ( 1.19645256 ) x . y = 522.88585984 ( 1.19645256 ) x .
• ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.
• ⓐ The logarithmic regression model that fits these data is y = 141.91242949 + 10.45366573 ln ( x ) y = 141.91242949 + 10.45366573 ln ( x )
• ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.
• ⓐ The logistic regression model that fits these data is y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x . y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x .
• ⓑ If the population continues to grow at this rate, there will be about 25,634   25,634   seals in 2020.
• ⓒ To the nearest whole number, the carrying capacity is 25,657.

4.1 Section Exercises

Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal .

exponential; the population decreases by a proportional rate. .

not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

The forest represented by the function B ( t ) = 82 ( 1.029 ) t . B ( t ) = 82 ( 1.029 ) t .

After t = 20 t = 20 years, forest A will have 43 43 more trees than forest B.

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

exponential growth; The growth factor, 1.06 , 1.06 , is greater than 1. 1.

exponential decay; The decay factor, 0.97 , 0.97 , is between 0 0 and 1. 1.

f ( x ) = 2000 ( 0.1 ) x f ( x ) = 2000 ( 0.1 ) x

f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x

$ 10 , 250 $ 10 , 250

$ 13 , 268.58 $ 13 , 268.58

P = A ( t ) ⋅ ( 1 + r n ) − n t P = A ( t ) ⋅ ( 1 + r n ) − n t

$ 4,572.56 $ 4,572.56

continuous growth; the growth rate is greater than 0. 0.

continuous decay; the growth rate is less than 0. 0.

$ 669.42 $ 669.42

f ( − 1 ) = − 4 f ( − 1 ) = − 4

f ( − 1 ) ≈ − 0.2707 f ( − 1 ) ≈ − 0.2707

f ( 3 ) ≈ 483.8146 f ( 3 ) ≈ 483.8146

y = 3 ⋅ 5 x y = 3 ⋅ 5 x

y ≈ 18 ⋅ 1.025 x y ≈ 18 ⋅ 1.025 x

y ≈ 0.2 ⋅ 1.95 x y ≈ 0.2 ⋅ 1.95 x

APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; I ( n ) = ( 1 + r n ) n − 1 I ( n ) = ( 1 + r n ) n − 1

Let f f be the exponential decay function f ( x ) = a ⋅ ( 1 b ) x f ( x ) = a ⋅ ( 1 b ) x such that b > 1. b > 1. Then for some number n > 0 , n > 0 , f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x . f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x .

47 , 622 47 , 622 fox

1.39 % ; 1.39 % ; $ 155 , 368.09 $ 155 , 368.09

$ 35 , 838.76 $ 35 , 838.76

$ 82 , 247.78 ; $ 82 , 247.78 ; $ 449.75 $ 449.75

4.2 Section Exercises

An asymptote is a line that the graph of a function approaches, as x x either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

g ( x ) = 4 ( 3 ) − x ; g ( x ) = 4 ( 3 ) − x ; y -intercept: ( 0 , 4 ) ; ( 0 , 4 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

g ( x ) = − 10 x + 7 ; g ( x ) = − 10 x + 7 ; y -intercept: ( 0 , 6 ) ; ( 0 , 6 ) ; Domain: all real numbers; Range: all real numbers less than 7. 7.

g ( x ) = 2 ( 1 4 ) x ; g ( x ) = 2 ( 1 4 ) x ; y -intercept: ( 0 , 2 ) ; ( 0 , 2 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

y -intercept: ( 0 , − 2 ) ( 0 , − 2 )

Horizontal asymptote: h ( x ) = 3 ; h ( x ) = 3 ; Domain: all real numbers; Range: all real numbers strictly greater than 3. 3.

As x → ∞ x → ∞ , f ( x ) → − ∞ f ( x ) → − ∞ ; As x → − ∞ x → − ∞ , f ( x ) → − 1 f ( x ) → − 1

As x → ∞ x → ∞ , f ( x ) → 2 f ( x ) → 2 ; As x → − ∞ x → − ∞ , f ( x ) → ∞ f ( x ) → ∞

f ( x ) = 4 x − 3 f ( x ) = 4 x − 3

f ( x ) = 4 x − 5 f ( x ) = 4 x − 5

f ( x ) = 4 − x f ( x ) = 4 − x

y = − 2 x + 3 y = − 2 x + 3

y = − 2 ( 3 ) x + 7 y = − 2 ( 3 ) x + 7

g ( 6 ) = 800 + 1 3 ≈ 800.3333 g ( 6 ) = 800 + 1 3 ≈ 800.3333

h ( − 7 ) = − 58 h ( − 7 ) = − 58

x ≈ − 2.953 x ≈ − 2.953

x ≈ − 0.222 x ≈ − 0.222

The graph of G ( x ) = ( 1 b ) x G ( x ) = ( 1 b ) x is the refelction about the y -axis of the graph of F ( x ) = b x ; F ( x ) = b x ; For any real number b > 0 b > 0 and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b ) x ( 1 b ) x is the the reflection about the y -axis, F ( − x ) . F ( − x ) .

The graphs of g ( x ) g ( x ) and h ( x ) h ( x ) are the same and are a horizontal shift to the right of the graph of f ( x ) ; f ( x ) ; For any real number n , real number b > 0 , b > 0 , and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b n ) b x ( 1 b n ) b x is the horizontal shift f ( x − n ) . f ( x − n ) .

4.3 Section Exercises

A logarithm is an exponent. Specifically, it is the exponent to which a base b b is raised to produce a given value. In the expressions given, the base b b has the same value. The exponent, y , y , in the expression b y b y can also be written as the logarithm, log b x , log b x , and the value of x x is the result of raising b b to the power of y . y .

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation b y = x , b y = x , and then properties of exponents can be applied to solve for x . x .

The natural logarithm is a special case of the logarithm with base b b in that the natural log always has base e . e . Rather than notating the natural logarithm as log e ( x ) , log e ( x ) , the notation used is ln ( x ) . ln ( x ) .

a c = b a c = b

x y = 64 x y = 64

15 b = a 15 b = a

13 a = 142 13 a = 142

e n = w e n = w

log c ( k ) = d log c ( k ) = d

log 19 y = x log 19 y = x

log n ( 103 ) = 4 log n ( 103 ) = 4

log y ( 39 100 ) = x log y ( 39 100 ) = x

ln ( h ) = k ln ( h ) = k

x = 2 − 3 = 1 8 x = 2 − 3 = 1 8

x = 3 3 = 27 x = 3 3 = 27

x = 9 1 2 = 3 x = 9 1 2 = 3

x = 6 − 3 = 1 216 x = 6 − 3 = 1 216

x = e 2 x = e 2

14.125 14.125

2 . 7 0 8 2 . 7 0 8

0.151 0.151

No, the function has no defined value for x = 0. x = 0. To verify, suppose x = 0 x = 0 is in the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) . Then there is some number n n such that n = log ( 0 ) . n = log ( 0 ) . Rewriting as an exponential equation gives: 10 n = 0 , 10 n = 0 , which is impossible since no such real number n n exists. Therefore, x = 0 x = 0 is not the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) .

Yes. Suppose there exists a real number x x such that ln x = 2. ln x = 2. Rewriting as an exponential equation gives x = e 2 , x = e 2 , which is a real number. To verify, let x = e 2 . x = e 2 . Then, by definition, ln ( x ) = ln ( e 2 ) = 2. ln ( x ) = ln ( e 2 ) = 2.

No; ln ( 1 ) = 0 , ln ( 1 ) = 0 , so ln ( e 1.725 ) ln ( 1 ) ln ( e 1.725 ) ln ( 1 ) is undefined.

4.4 Section Exercises

Since the functions are inverses, their graphs are mirror images about the line y = x . y = x . So for every point ( a , b ) ( a , b ) on the graph of a logarithmic function, there is a corresponding point ( b , a ) ( b , a ) on the graph of its inverse exponential function.

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

Domain: ( − ∞ , 1 2 ) ; ( − ∞ , 1 2 ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( − 17 4 , ∞ ) ; ( − 17 4 , ∞ ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( 5 , ∞ ) ; ( 5 , ∞ ) ; Vertical asymptote: x = 5 x = 5

Domain: ( − 1 3 , ∞ ) ; ( − 1 3 , ∞ ) ; Vertical asymptote: x = − 1 3 x = − 1 3

Domain: ( − 3 , ∞ ) ; ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3

Domain: ( 3 7 , ∞ ) ( 3 7 , ∞ ) ; Vertical asymptote: x = 3 7 x = 3 7 ; End behavior: as x → ( 3 7 ) + , f ( x ) → − ∞ x → ( 3 7 ) + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ x → ∞ , f ( x ) → ∞

Domain: ( − 3 , ∞ ) ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3 ; End behavior: as x → − 3 + x → − 3 + , f ( x ) → − ∞ f ( x ) → − ∞ and as x → ∞ x → ∞ , f ( x ) → ∞ f ( x ) → ∞

Domain: ( 1 , ∞ ) ; ( 1 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 1 ; x = 1 ; x -intercept: ( 5 4 , 0 ) ; ( 5 4 , 0 ) ; y -intercept: DNE

Domain: ( − ∞ , 0 ) ; ( − ∞ , 0 ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( − e 2 , 0 ) ; ( − e 2 , 0 ) ; y -intercept: DNE

Domain: ( 0 , ∞ ) ; ( 0 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( e 3 , 0 ) ; ( e 3 , 0 ) ; y -intercept: DNE

f ( x ) = log 2 ( − ( x − 1 ) ) f ( x ) = log 2 ( − ( x − 1 ) )

f ( x ) = 3 log 4 ( x + 2 ) f ( x ) = 3 log 4 ( x + 2 )

x = 2 x = 2

x ≈ 2 .303 x ≈ 2 .303

x ≈ − 0.472 x ≈ − 0.472

The graphs of f ( x ) = log 1 2 ( x ) f ( x ) = log 1 2 ( x ) and g ( x ) = − log 2 ( x ) g ( x ) = − log 2 ( x ) appear to be the same; Conjecture: for any positive base b ≠ 1 , b ≠ 1 , log b ( x ) = − log 1 b ( x ) . log b ( x ) = − log 1 b ( x ) .

Recall that the argument of a logarithmic function must be positive, so we determine where x + 2 x − 4 > 0 x + 2 x − 4 > 0 . From the graph of the function f ( x ) = x + 2 x − 4 , f ( x ) = x + 2 x − 4 , note that the graph lies above the x -axis on the interval ( − ∞ , − 2 ) ( − ∞ , − 2 ) and again to the right of the vertical asymptote, that is ( 4 , ∞ ) . ( 4 , ∞ ) . Therefore, the domain is ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) . ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) .

4.5 Section Exercises

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, log b ( x 1 n ) = 1 n log b ( x ) . log b ( x 1 n ) = 1 n log b ( x ) .

log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y ) log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y )

log b ( 13 ) − log b ( 17 ) log b ( 13 ) − log b ( 17 )

− k ln ( 4 ) − k ln ( 4 )

ln ( 7 x y ) ln ( 7 x y )

log b ( 4 ) log b ( 4 )

log b ( 7 ) log b ( 7 )

15 log ( x ) + 13 log ( y ) − 19 log ( z ) 15 log ( x ) + 13 log ( y ) − 19 log ( z )

3 2 log ( x ) − 2 log ( y ) 3 2 log ( x ) − 2 log ( y )

8 3 log ( x ) + 14 3 log ( y ) 8 3 log ( x ) + 14 3 log ( y )

ln ( 2 x 7 ) ln ( 2 x 7 )

log ( x z 3 y ) log ( x z 3 y )

log 7 ( 15 ) = ln ( 15 ) ln ( 7 ) log 7 ( 15 ) = ln ( 15 ) ln ( 7 )

log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b

log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1 log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1

2.81359 2.81359

0.93913 0.93913

− 2.23266 − 2.23266

x = 4 ; x = 4 ; By the quotient rule: log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1. log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1.

Rewriting as an exponential equation and solving for x : x :

6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4 6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4

Checking, we find that log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) is defined, so x = 4. x = 4.

Let b b and n n be positive integers greater than 1. 1. Then, by the change-of-base formula, log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) . log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) .

4.6 Section Exercises

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

x = − 1 3 x = − 1 3

n = − 1 n = − 1

b = 6 5 b = 6 5

x = 10 x = 10

No solution

p = log ( 17 8 ) − 7 p = log ( 17 8 ) − 7

k = − ln ( 38 ) 3 k = − ln ( 38 ) 3

x = ln ( 38 3 ) − 8 9 x = ln ( 38 3 ) − 8 9

x = ln 12 x = ln 12

x = ln ( 3 5 ) − 3 8 x = ln ( 3 5 ) − 3 8

no solution

x = ln ( 3 ) x = ln ( 3 )

10 − 2 = 1 100 10 − 2 = 1 100

n = 49 n = 49

k = 1 36 k = 1 36

x = 9 − e 8 x = 9 − e 8

n = 1 n = 1

x = ± 10 3 x = ± 10 3

x = 0 x = 0

x = 3 4 x = 3 4

x = 9 x = 9

x = e 2 3 ≈ 2.5 x = e 2 3 ≈ 2.5

x = − 5 x = − 5

x = e + 10 4 ≈ 3.2 x = e + 10 4 ≈ 3.2

x = 11 5 ≈ 2.2 x = 11 5 ≈ 2.2

x = 101 11 ≈ 9.2 x = 101 11 ≈ 9.2

about $ 27 , 710.24 $ 27 , 710.24

about 5 years

ln ( 17 ) 5 ≈ 0.567 ln ( 17 ) 5 ≈ 0.567

x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078 x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078

x ≈ 2.2401 x ≈ 2.2401

x ≈ − 44655 . 7143 x ≈ − 44655 . 7143

about 5.83 5.83

t = ln ( ( y A ) 1 k ) t = ln ( ( y A ) 1 k )

t = ln ( ( T − T s T 0 − T s ) − 1 k ) t = ln ( ( T − T s T 0 − T s ) − 1 k )

4.7 Section Exercises

Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 10 2 10 2 times, or 2 orders of magnitude greater, than the mass of Earth.

f ( 0 ) ≈ 16.7 ; f ( 0 ) ≈ 16.7 ; The amount initially present is about 16.7 units.

exponential; f ( x ) = 1.2 x f ( x ) = 1.2 x

logarithmic

about 1.4 1.4 years

about 7.3 7.3 years

4 4 half-lives; 8.18 8.18 minutes

M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2 M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2

Let y = b x y = b x for some non-negative real number b b such that b ≠ 1. b ≠ 1. Then,

ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b ) ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b )

A = 125 e ( − 0.3567 t ) ; A ≈ 43 A = 125 e ( − 0.3567 t ) ; A ≈ 43 mg

about 60 60 days

A ( t ) = 250 e ( − 0.00822 t ) ; A ( t ) = 250 e ( − 0.00822 t ) ; half-life: about 84 84 minutes

r ≈ − 0.0667 , r ≈ − 0.0667 , So the hourly decay rate is about 6.67 % 6.67 %

f ( t ) = 1350 e ( 0.03466 t ) ; f ( t ) = 1350 e ( 0.03466 t ) ; after 3 hours: P ( 180 ) ≈ 691 , 200 P ( 180 ) ≈ 691 , 200

f ( t ) = 256 e ( 0.068110 t ) ; f ( t ) = 256 e ( 0.068110 t ) ; doubling time: about 10 10 minutes

about 88 88 minutes

T ( t ) = 90 e ( − 0.008377 t ) + 75 , T ( t ) = 90 e ( − 0.008377 t ) + 75 , where t t is in minutes.

about 113 113 minutes

log ( x ) = 1.5 ; x ≈ 31.623 log ( x ) = 1.5 ; x ≈ 31.623

MMS magnitude: 5.82 5.82

N ( 3 ) ≈ 71 N ( 3 ) ≈ 71

4.8 Section Exercises

Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

The y -intercept on the graph of a logistic equation corresponds to the initial population for the population model.

P ( 0 ) = 22 P ( 0 ) = 22 ; 175

p ≈ 2.67 p ≈ 2.67

y -intercept: ( 0 , 15 ) ( 0 , 15 )

about 6.8 6.8 months.

About 38 wolves

About 8.7 years

f ( x ) = 776.682 ( 1.426 ) x f ( x ) = 776.682 ( 1.426 ) x

f ( x ) = 731.92 e -0.3038 x f ( x ) = 731.92 e -0.3038 x

When f ( x ) = 250 , x ≈ 3.6 f ( x ) = 250 , x ≈ 3.6

y = 5.063 + 1.934 log ( x ) y = 5.063 + 1.934 log ( x )

When f ( 10 ) ≈ 2.3 f ( 10 ) ≈ 2.3

When f ( x ) = 8 , x ≈ 0.82 f ( x ) = 8 , x ≈ 0.82

f ( x ) = 25.081 1 + 3.182 e − 0.545 x f ( x ) = 25.081 1 + 3.182 e − 0.545 x

When f ( x ) = 68 , x ≈ 4.9 f ( x ) = 68 , x ≈ 4.9

f ( x ) = 1.034341 ( 1.281204 ) x f ( x ) = 1.034341 ( 1.281204 ) x ; g ( x ) = 4.035510 g ( x ) = 4.035510 ; the regression curves are symmetrical about y = x y = x , so it appears that they are inverse functions.

f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b

Review Exercises

exponential decay; The growth factor, 0.825 , 0.825 , is between 0 0 and 1. 1.

y = 0.25 ( 3 ) x y = 0.25 ( 3 ) x

$ 42 , 888.18 $ 42 , 888.18

continuous decay; the growth rate is negative.

domain: all real numbers; range: all real numbers strictly greater than zero; y -intercept: (0, 3.5);

g ( x ) = 7 ( 6.5 ) − x ; g ( x ) = 7 ( 6.5 ) − x ; y -intercept: ( 0 , 7 ) ; ( 0 , 7 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

17 x = 4913 17 x = 4913

log a b = − 2 5 log a b = − 2 5

x = 64 1 3 = 4 x = 64 1 3 = 4

log ( 0 .000001 ) = − 6 log ( 0 .000001 ) = − 6

ln ( e − 0.8648 ) = − 0.8648 ln ( e − 0.8648 ) = − 0.8648

Domain: x > − 5 ; x > − 5 ; Vertical asymptote: x = − 5 ; x = − 5 ; End behavior: as x → − 5 + , f ( x ) → − ∞ x → − 5 + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ . x → ∞ , f ( x ) → ∞ .

log 8 ( 65 x y ) log 8 ( 65 x y )

ln ( z x y ) ln ( z x y )

log y ( 12 ) log y ( 12 )

ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2 ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2

log 7 ( v 3 w 6 u 3 ) log 7 ( v 3 w 6 u 3 )

x = log ( 125 ) log ( 5 ) + 17 12 = 5 3 x = log ( 125 ) log ( 5 ) + 17 12 = 5 3

x = − 3 x = − 3

x = ln ( 11 ) x = ln ( 11 )

a = e 4 − 3 a = e 4 − 3

x = ± 9 5 x = ± 9 5

about 5.45 5.45 years

f − 1 ( x ) = 2 4 x − 1 3 f − 1 ( x ) = 2 4 x − 1 3

f ( t ) = 300 ( 0.83 ) t ; f ( t ) = 300 ( 0.83 ) t ; f ( 24 ) ≈ 3.43     g f ( 24 ) ≈ 3.43     g

about 45 45 minutes

about 8.5 8.5 days

exponential

y = 4 ( 0.2 ) x ; y = 4 ( 0.2 ) x ; y = 4 e -1.609438 x y = 4 e -1.609438 x

about 7.2 7.2 days

logarithmic; y = 16.68718 − 9.71860 ln ( x ) y = 16.68718 − 9.71860 ln ( x )

Practice Test

About 13 dolphins.

$ 1,947 $ 1,947

y -intercept: ( 0 , 5 ) ( 0 , 5 )

8.5 a = 614.125 8.5 a = 614.125

x = ( 1 7 ) 2 = 1 49 x = ( 1 7 ) 2 = 1 49

ln ( 0.716 ) ≈ − 0.334 ln ( 0.716 ) ≈ − 0.334

Domain: x < 3 ; x < 3 ; Vertical asymptote: x = 3 ; x = 3 ; End behavior: x → 3 − , f ( x ) → − ∞ x → 3 − , f ( x ) → − ∞ and x → − ∞ , f ( x ) → ∞ x → − ∞ , f ( x ) → ∞

log t ( 12 ) log t ( 12 )

3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3 3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3

x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497 x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497

a = ln ( 4 ) + 8 10 a = ln ( 4 ) + 8 10

x = ln ( 9 ) x = ln ( 9 )

x = ± 3 3 2 x = ± 3 3 2

f ( t ) = 112 e − .019792 t ; f ( t ) = 112 e − .019792 t ; half-life: about 35 35 days

T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F

exponential; y = 15.10062 ( 1.24621 ) x y = 15.10062 ( 1.24621 ) x

logistic; y = 18.41659 1 + 7.54644 e − 0.68375 x y = 18.41659 1 + 7.54644 e − 0.68375 x

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/chapter-4

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