Understanding Logarithmic and Exponential Functions Graphs

$x$	$1013 e^{- 0.000128 x}$	$y = 1013 e^{- 0.000128 x}$
$0$	$1013 e^{- 0.000128 (0)}$	$1013$
$1000$	$1013 e^{- 0.000128 (1000)}$	$\approx 891$
$2000$	$1013 e^{- 0.000128 (2000)}$	$\approx 784$
$3000$	$1013 e^{- 0.000128 (3000)}$	$\approx 690$
$4000$	$1013 e^{- 0.000128 (4000)}$	$\approx 607$
$5000$	$1013 e^{- 0.000128 (5000)}$	$\approx 534$
$6000$	$1013 e^{- 0.000128 (6000)}$	$\approx 470$
$7000$	$1013 e^{- 0.000128 (7000)}$	$\approx 414$
$8000$	$1013 e^{- 0.000128 (8000)}$	$\approx 364$
$9000$	$1013 e^{- 0.000128 (9000)}$	$\approx 320$
$10000$	$1013 e^{- 0.000128 (10000)}$	$\approx 282$

x

1013 e^{- 0.000128 x}

y = 1013 e^{- 0.000128 x}

0

1013 e^{- 0.000128 (0)}

1013

1000

1013 e^{- 0.000128 (1000)}

\approx 891

2000

1013 e^{- 0.000128 (2000)}

\approx 784

3000

1013 e^{- 0.000128 (3000)}

\approx 690

4000

1013 e^{- 0.000128 (4000)}

\approx 607

5000

1013 e^{- 0.000128 (5000)}

\approx 534

6000

1013 e^{- 0.000128 (6000)}

\approx 470

7000

1013 e^{- 0.000128 (7000)}

\approx 414

8000

1013 e^{- 0.000128 (8000)}

\approx 364

9000

1013 e^{- 0.000128 (9000)}

\approx 320

10000

1013 e^{- 0.000128 (10000)}

\approx 282

$x$	$2 lo g (x - 0.5)$	$c (x) = 2 lo g (x - 0.5)$
$1$	$2 lo g (1 - 0.5)$	$\approx - 0.6$
$2$	$2 lo g (2 - 0.5)$	$\approx 0.35$
$3$	$2 lo g (3 - 0.5)$	$\approx 0.8$
$4$	$2 lo g (4 - 0.5)$	$\approx 1.1$
$5$	$2 lo g (5 - 0.5)$	$\approx 1.3$

x

2 lo g (x - 0.5)

c (x) = 2 lo g (x - 0.5)

1

2 lo g (1 - 0.5)

\approx - 0.6

2

2 lo g (2 - 0.5)

\approx 0.35

3

2 lo g (3 - 0.5)

\approx 0.8

4

2 lo g (4 - 0.5)

\approx 1.1

5

2 lo g (5 - 0.5)

\approx 1.3

$x$	$2 lo g (x - 1)$	$f (x)$
$1.5$	$2 lo g (1.5 - 1)$	$\approx - 0.6$
$5$	$2 lo g (5 - 1)$	$\approx 1.2$
$10$	$2 lo g (10 - 1)$	$\approx 1.9$
$12$	$2 lo g (12 - 1)$	$\approx 2.1$
$18$	$2 lo g (18 - 1)$	$\approx 2.5$

x

2 lo g (x - 1)

f (x)

1.5

2 lo g (1.5 - 1)

\approx - 0.6

5

2 lo g (5 - 1)

\approx 1.2

10

2 lo g (10 - 1)

\approx 1.9

12

2 lo g (12 - 1)

\approx 2.1

18

2 lo g (18 - 1)

\approx 2.5

$x$	$1 0^{\frac{1}{2} x} + 1$	$f^{- 1} (x)$
$- 2$	$1 0^{\frac{1}{2} (- 2)} + 1$	$1.1$
$- 1$	$1 0^{\frac{1}{2} (- 1)} + 1$	$\approx 1.3$
$0$	$1 0^{\frac{1}{2} (0)} + 1$	$2$
$1$	$1 0^{\frac{1}{2} (1)} + 1$	$\approx 4.2$
$2$	$1 0^{\frac{1}{2} (2)} + 1$	$11$

x

1 0^{\frac{1}{2} x} + 1

f^{- 1} (x)

- 2

1 0^{\frac{1}{2} (- 2)} + 1

1.1

- 1

1 0^{\frac{1}{2} (- 1)} + 1

\approx 1.3

0

1 0^{\frac{1}{2} (0)} + 1

2

1

1 0^{\frac{1}{2} (1)} + 1

\approx 4.2

2

1 0^{\frac{1}{2} (2)} + 1

11

$x$	$\frac{1}{5} e^{3 x}$	$g (x)$
$- 1$	$\frac{1}{5} e^{3 (- 1)}$	$\approx 0.01$
$- 0.5$	$\frac{1}{5} e^{3 (- 0.5)}$	$\approx 0.04$
$0$	$\frac{1}{5} e^{3 (0)}$	$0.2$
$0.5$	$\frac{1}{5} e^{3 (0.5)}$	$\approx 0.9$
$1$	$\frac{1}{5} e^{3 (1)}$	$\approx 4$

x

\frac{1}{5} e^{3 x}

g (x)

- 1

\frac{1}{5} e^{3 (- 1)}

\approx 0.01

- 0.5

\frac{1}{5} e^{3 (- 0.5)}

\approx 0.04

0

\frac{1}{5} e^{3 (0)}

0.2

0.5

\frac{1}{5} e^{3 (0.5)}

\approx 0.9

1

\frac{1}{5} e^{3 (1)}

\approx 4

$x$	$\frac{1}{3} ln 5 x$	$g^{- 1} (x)$
$0.1$	$\frac{1}{3} ln 5 (0.5)$	$\approx - 0.2$
$1$	$\frac{1}{3} ln 5 (1)$	$\approx 0.5$
$2$	$\frac{1}{3} ln 5 (2)$	$\approx 0.8$
$3$	$\frac{1}{3} ln 5 (3)$	$\approx 0.9$
$4$	$\frac{1}{3} ln 5 (4)$	$\approx 1$

x

\frac{1}{3} ln 5 x

g^{- 1} (x)

0.1

\frac{1}{3} ln 5 (0.5)

\approx - 0.2

1

\frac{1}{3} ln 5 (1)

\approx 0.5

2

\frac{1}{3} ln 5 (2)

\approx 0.8

3

\frac{1}{3} ln 5 (3)

\approx 0.9

4

\frac{1}{3} ln 5 (4)

\approx 1

	Definition of the First Function	Substitute the Second Function	Simplify
$g (g^{- 1} (x))$	$\frac{1}{5} e^{3 g^{- 1} (x)}$	$\frac{1}{5} e^{3 (\frac{1}{3} l n 5 x)}$	$x ✓$
$g^{- 1} (g (x))$	$\frac{1}{3} ln 5 g (x)$	$\frac{1}{3} ln 5 (\frac{1}{5} e^{3 x})$	$x ✓$

Definition of the First Function

Substitute the Second Function

Simplify

g (g^{- 1} (x))

\frac{1}{5} e^{3 g^{- 1} (x)}

\frac{1}{5} e^{3 (\frac{1}{3} l n 5 x)}

x ✓

g^{- 1} (g (x))

\frac{1}{3} ln 5 g (x)

\frac{1}{3} ln 5 (\frac{1}{5} e^{3 x})

x ✓

Suppose that the ratio between consecutive $y -$ values in a table is always the same. In such a case, the function represented by the table is called an exponential function. Additionally, its constant multiplier $c$ is the quotient between two consecutive $y -$ values.

Determine whether the following tables represent an exponential function. If yes, write the constant multiplier. Otherwise, just write no.

$x$	$y$
$1$	$2$
$2$	$3$
$3$	$4.5$
$4$	$6.75$

$x$	$y$
$1$	$5$
$2$	$10$
$3$	$15$
$4$	$20$

Let's find the ratio between consecutive y-values.

We can see that the ratio between consecutive y-values is constant and equal to 1.5. Therefore, the table does represent an exponential function with a constant multiplier of 1.5

Let's find the ratio between consecutive y-values.

This time, the ratios are not constant. Therefore, the table does not represent an exponential function.

An investment account for a house earns interest that is compounded continuously. The balance

B

of the account, in dollars, after

t

years is modeled by the following function.

B = 3224 e^{0.05 t}

What is the balance after

10

years? Write the answer to the nearest dollar.

We want to find the balance of the account after 10 years. To do so, we will substitute 10 for t in the given continuously compounded interest formula. Let's do it!

After 10 years, the balance of the account will be $5315 when rounded to the nearest dollar.

An investment account for a house earns annual interest compounded continuously. The balance

B

of the account in dollars after

t

years is modeled by the following function.

B = 3224 e^{0.05 t}

Which of the following graphs is the graph of this function?

The given formula represents a continuously compounded interest. In this formula, the interest rate and the principal are both positive.

B=3224e^(0.05t)
Initial Value	Interest Rate
3224	0.05

This means that the graph of this function is increasing in its entire domain. Therefore, we can disregard options A and C, since they represent decreasing functions.

Next, let's find the y-intercept of the graph. To do so, we will substitute 0 for t in our formula. Let's do it!

The y-intercept of the graph occurs at (0,3224), so the graph intersects the y-axis above the x-axis. This characteristic matches the graph in choice B.

A savings account earns compound interest. The balance of the account $B$ after $t$ years is represented by the following graph.

What will be the balance of the account after

3

years? Round the answer to the nearest thousand.

After how many years the balance will be

$ 8000 ?

Round the answer to the nearest integer.

To find the balance of the account after 3 years, we will identify the point on the curve whose x-coordinate is 3. The balance after 3 years is the y-coordinate of this point.

The exact value of the y-coordinate cannot be determined by only looking at the graph. However, we can see that, to the nearest thousand, the balance of the account after 3 years will be about $6000.

To find after how many years the balance of the account will be $ 8000, we will identify the point on the curve whose y-coordinate is 8000. Then, we will determine its x-coordinate.

The exact value of the x-coordinate cannot be determined by only using the graph. However, we can see that, to the nearest integer, the x-coordinate is 6. Therefore, the balance of the account will be $8000 after about 6 years.

Atmospheric pressure decreases as altitude increases. The pressure at sea level is about

1013

hecto Pascals (hPa), and the pressure

P (x)

at an altitude of

x

meters is given by the following formula.

P (x) = 1013 e^{- 0.000128 x}

Find the atmospheric pressure in the city of La Paz, Bolivia, which is located at an altitude of

3625

meters above sea level. Round the answer to the nearest integer.

We want to find the atmospheric pressure in La Paz, which has an altitude of 3625 meters above sea level. To do so, we will substitute 3625 for x in the given natural base exponential function. Let's do it!

Therefore, the atmospheric pressure in La Paz is about 637 hecto Pascals.

Atmospheric pressure decreases as altitude increases. The pressure at sea level is about

1013

hecto Pascals (hPa), and the pressure

P (x)

at an altitude of

x

meters is given by the following formula.

P (x) = 1013 e^{- 0.000128 x}

Which of the following graphs can be the graph of this function?

The given function is a natural base exponential function. In this formula, the coefficient of the variable in the exponent is negative.

P(x)=1013e^(- 0.000128x)
Initial Value	Coefficient of x
1013	- 0.000128

This means that the graph of this function is decreasing in its entire domain. Therefore, we can disregard options B and D, since they represent increasing functions.

Next, let's find the y-intercept of the graph. To do so, we will substitute 0 for x in our formula. Let's do it!

The y-intercept of the graph occurs at (0,1013), so the graph intersects the y-axis above the x-axis. This characteristic matches the graph in choice C.

The atmospheric pressure $P (x),$ in hecto Pascals, at $x$ meters above sea level is represented by the following graph.

What is the atmospheric pressure at an altitude of

12000

meters above sea level? Round the answer to the nearest hundred.

What is the altitude when the atmospheric pressure is

600

hecto Pascals? Round the answer to the nearest thousand.

Let's use the graph to find the atmospheric pressure at an altitude of 12 000 meters above sea level. To do so, we will identify the point on the curve whose first coordinate is 12 000.

We cannot state the exact value for the second coordinate of this point. However, we can see that the second coordinate of this point, rounded to the nearest hundred, is 200. This means that the atmospheric pressure at an altitude of 12 000 meters is about 200 hecto Pascals.

This time we need to find the altitude when the atmospheric pressure is 600 hecto Pascals. To do so, we will identify the point on the curve whose second coordinate is 600.

We cannot state the exact value for the first coordinate of this point. However, we can see that the first coordinate of this point, rounded to the nearest thousand, is 4000. This means that the altitude when the atmospheric pressure is 600 hecto Pascals is about 4000 meters.

Consider the given logarithmic function.

f (x) = lo g (x - 1)

Which of the following is the graph of the function?

To determine the graph of the given logarithmic function, we will start by finding its domain. To do so, recall that the argument of a logarithm must be positive. log( x-1_(> 0) ) ⇒ x>1 At this point we can say that the domain of the function is the set of real numbers greater than 1. Therefore, the graph of the function must be located to the right of the vertical line x=1, which is a vertical asymptote. With this information, we can disregard graphs C and D.

Now, the possible choices are only A and B. Notice that the graph shown in A is an increasing curve, while the graph shown in B is a decreasing curve. Therefore, to determine which option is the correct choice, we will evaluate the function at two values of x. For convenience, we can choose x=2 and x=11. Let's start by evaluating the function at x=2.

We found that f(2)=0, so the x-intercept of the graph occurs at x=2. Now let's calculate the value of the function for x=11. If the result is positive, the graph shown in A is the correct choice because the values of this curve to the right of the x-intercept are positive. Otherwise, the graph shown in B is the correct choice, as the values of this curve to the right of the x-intercept are negative.

We found that f(11) is positive, so the graph that represents the given function is the graph shown in A.

Consider the given logarithmic function.

g (x) = ln (3 - x)

Which of the following is the graph of the function?

To determine which of the graphs is the graph of the logarithmic function, we will draw its graph and compare it with the given choices. To do so, we will start by finding the domain. Recall that the argument of a logarithm must be positive. Function g(x) = ln( 3-x) [0.7em] Domain 3-x>0 ⇔ x<3 The domain of the function is the set of real numbers less than 3. Let's use this to make a table of values.

x	ln (3-x)	g(x)
2.5	ln (3- 2.5)	≈ - 0.69
2	ln (3- 2)	0
1	ln (3- 1)	≈ 0.69
0	ln (3- 0)	≈ 1.1
- 1	ln (3-( - 1))	≈ 1.39

We can now plot the points obtained in the table and connect them with a smooth curve. Remember that the domain is the set of all real numbers less than 3.

This graph matches option A.

Consider the logarithmic function.

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

Find

f^{- 1} (x),

the inverse function of

f (x) .

To find the inverse of the logarithmic function, we will start by replacing f(x) with y. y=ln (2x+1) Next, we will switch the variables x and y. This will give us the inverse of the function. Function y=ln (2 x+1) [0.3em] ⇓ [0.3em] Switch Variables x=ln (2 y+1) Now we will isolate the y-variable. To do so, we can use the definition of a logarithm. Recall that a natural logarithm is a logarithm with base e. x=ln (2y+1) ⇔ e^x=2y+1 We can use inverse operations to finally isolate y.

Therefore, the inverse of the given function is f^(- 1)(x)= e^x-12.

Consider the given exponential function.

g (x) = \frac{1}{2} (1 0^{\frac{1}{3} x})

Find

g^{- 1} (x),

the inverse function of

g (x) .

To find the inverse of the logarithmic function, we will start by replacing g(x) with y. y=1/2 (10^(13x)) Next, we will switch the variables x and y. Function y=1/2 (10^(13 x)) [0.6em] ⇓ [0.4em] Switch Variables x=1/2 (10^(13 y)) Now we will isolate the y-variable. To do so, we can first multiply both sides of the equation by 2.

We can use the definition of a logarithm. Recall that a logarithm with base 10 is a common logarithm. 2x=10^(13y) ⇔ log 2x=1/3y Finally, we can multiply both sides of the equation by 3 and simplify by using the Power Property of Logarithms.

Therefore, the inverse of the given function is g^(- 1)(x)=log 8x^3.

Consider the logarithmic function.

y = lo g_{2} (4 x - 32)

Find the domain of this function. Write the answer as an inequality.

To find the domain of the logarithmic function, we have to keep in mind that the argument of a logarithm must be positive.

Therefore, the domain of the function is the set of all real numbers greater than 8.

Graphing Exponential and Logarithmic Functions

Catch-Up and Review

Identifying Change

Constant Multiplier

Identifying the Constant Multiplier

Natural Base Exponential Functions

Graphing Natural Base Functions

Continuously Compounded Interest

Proof

Graphing and Using a Continuously Compounded Interest Graph

Answer

Hint

Solution

Using a Natural Base Exponential Function to Model the Atmospheric Pressure at Different Altitudes

Answer

Hint

Solution

Logarithmic Functions

Inverse Properties of Logarithms

Proof

$lo g_{b} b^{x} = x$

$b^{l o g_{b} x} = x$

Modeling Battery Charges Using Logarithmic Functions

Answer

Hint

Solution

Graphing $p (x) = ln x$

Graphing $c (x) = 2 lo g (x - 0.5)$

Finding Inverse Functions of Logarithmic and Exponential Functions

Hint

Solution

Checking Our Answer

Horizontal and Vertical Asymptotes

Exponential Graphs

Logarithmic Graphs

Graphing Exponential and Logarithmic Functions

Recommended exercises

	11 Theory slides
	16 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Graphing Exponential and Logarithmic Functions

Catch-Up and Review

Proof

Answer

Hint

Solution

Answer

Hint

Solution

Proof

logb​bx=x

blogb​x=x

Answer

Hint

Solution

Graphing p(x)=lnx

Graphing c(x)=2log(x−0.5)

Hint

Solution

Checking Our Answer

Exponential Graphs

Logarithmic Graphs

Graphing Exponential and Logarithmic Functions

Recommended exercises

$lo g_{b} b^{x} = x$

$b^{l o g_{b} x} = x$

Graphing $p (x) = ln x$

Graphing $c (x) = 2 lo g (x - 0.5)$