Understanding Logarithmic and Exponential Functions Graphs

In this lesson, exponential functions with base

e

and the key features of their graphs will be discussed. Furthermore, logarithmic functions will be introduced and graphed. Finally, the relationship between exponential and logarithmic functions will be explained both algebraically and graphically.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Function
Function notation
Inverse Function and Finding the Inverse of a Function
Exponential Function
Logarithms and Power Property of Logarithms
Common Logarithm
Natural Logarithm

Explore

Identifying Change

As soon as Zosia entered the room for her Algebra lesson, her teacher asked a question to the class.

Help Zosia find the answer by considering the following questions.

How do the $y -$ values change for one unit increase in $x ?$ Is it a constant change?
Does the table represent a linear relationship?

Discussion

Constant Multiplier

The rate of change for linear functions is constant. For each step in the $x -$ direction, the change between $y -$ values is the same. Therefore, the difference between $y -$ values for consecutive $x -$ values is the same. Conversely, the rate of change for exponential functions is not constant. This means that the differences between $y -$ values for consecutive $x -$ values are not the same.

In the table on the left, the difference between $y -$ values for consecutive $x -$ values is $2 .$ This means that the rate of change is constant. Conversely, in the table on the right, the differences between $y -$ values for consecutive $x -$ values are not the same. This means that for this table the rate of change is not constant.

The table on the left represents a linear function because it shows a constant rate of change. Conversely, the table on the right represents a non-linear function. Notice how the $y -$ values, for the table on the right, are doubled for each step in the $x -$ direction. To obtain a $y -$ value, the previous value is multiplied by $2 .$

The table on the left corresponds to a function with a constant increase. In the table on the right, the increase is not constant. However, the ratio between

y -

values for consecutive

x -

values is always the same. This means that the table on the right corresponds to a function with a constant multiplier. Functions with constant multipliers are exponential functions.

Pop Quiz

Identifying the Constant Multiplier

Does the table below correspond to an exponential function? If yes, write only the constant multiplier. If not, write only no.

Discussion

Natural Base Exponential Functions

Natural base exponential functions are exponential functions whose base, or constant factor, is the number

e

f (x) = a e^{r x}

Here,

a

and

r

are real numbers that are not equal to

0 .

In considering their positive or negative sign, two observations can be made.

If both $a$ and $r$ are positive, the function is an exponential growth function.
If $a$ is positive and $r$ is negative, the resulting function is instead an exponential decay function.

The graphs of $f (x) = e^{x}$ and $g (x) = e^{- x}$ are shown.

In general, the graph of a natural base exponential function is always a smooth curve.

Example

Graphing and Using a Continuously Compounded Interest Graph

Zosia wants to visit family in Argentina, and while there she hopes to climb Mount Aconcagua! This is going to cost a pretty penny so she needs to increase her savings. Zosia knows that banks do not offer continuously compounded interest, but she is daydreaming about opening an account that does use it.

Recently, she made

$ 17

from selling some old stuff at a garage sale. She would deposit this amount into the account. The amount

A

in the account after

t

years is calculated by using the following formula.

A = 17 e^{t}

a Graph the given function.

b Use the formula to calculate the balance of this account after

1

year and

9

months. Suppose that no deposits nor withdrawals are made and round the answer to two decimal places. Check the answer with the graph.

c Use the graph to estimate in how many years the balance of this account would be

$ 120 .

Suppose that no deposits nor withdrawals are made. Round the answer to the nearest integer.

Answer

b About

$ 97.83

c About

2

years

Hint

a Start by identifying the initial value and plotting the

y -

intercept.

b Keep in mind that

1

year and

9

months are

1.75

years.

c Identify the point on the curve whose second coordinate is

120 .

Solution

a The given function is a natural base exponential function whose initial value is

17 .

A = 17 e^{t}

Therefore, the

y -

intercept of the graph occurs at

(0, 17) .

Since time cannot be negative and the balance of the account will always be non-negative, only the first quadrant of the coordinate plane will be considered.

The constant multiplier is $e .$ This means that when $x -$ values increase by $1,$ the $y -$ values are multiplied by $e .$

Finally, the curve can be graphed by connecting these points.

b First,

1

year and

9

months must be expressed in years. Recall that

1

year has

12

months. This means that

9

months are

\frac{9}{1 2},

0.75,

years.

1 year and 9 months ↓ 1 \frac{9}{1 2} = 1.75 years

Therefore,

t = 1.75

will be substituted into the equation.

A = 17 e^{t}

Substitute

$t = 1.75$

A = 17 e^{1.75}

UseCalc

Use a calculator

A = 97.828245 \dots

RoundDec

Round to $2$ decimal place(s)

A \approx 97.83

After

1

year and

9

months, the balance of the account would be about

$ 97.83 .

This can be verified by using the graph. To do so, the point whose first coordinate is

1.75

should be determined on the graph of the function.

The second coordinate of this point is a bit less than $100 .$ Therefore, it is reasonable to have a balance of $$ 97.83$ after $1$ year and $9$ months.

c This time the graph will be used to estimate after how many years the balance would be

$ 120 .

To do so, the point with the second coordinate

120

must be spotted on the graph of the function, and its first coordinate needs to be identified.

The first coordinate of the point is almost equal to $2 .$ Therefore, the balance would be $$ 120$ after about $2$ years. Zosia is realizing that it might take her quite some time to meet her financial goal, but she is making a valiant effort and certainly on her way!

Example

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

Zosia finally saved enough money to visit family in Argentina and set out to climb Mount Aconcagua while there! She is as excited as ever.

Here, her knowledge about natural base exponential functions will be extremely useful to her pursuit of reaching the peak. Zosia knows that the atmospheric pressure decreases as the 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}

To avoid unforeseen problems, Zosia realizes that she can share some responsibilities if she recruits a team to solve such issues. Give her a hand and become a valuable member of her team by answering the following questions. If unable to, the team could get altitude sickness!

a The altitude of the mountain's peak, called the summit, is

8849

meters. Calculate the atmospheric pressure at the summit. Round the answer to the nearest integer.

b Draw a graph of the given exponential function.

c Estimate the altitude where the atmospheric pressure is

700

hecto Pascals. At this altitude, Zosia will rest for a night or two. Round the answer to the nearest thousand meters.

Answer

a About

326

hPa

c About

3000

Hint

a Substitute

8849

for

x

in the given formula.

b Make a table of values.

c Use the graph from Part B.

Solution

a To calculate the atmospheric pressure at an altitude of

8849

meters, this number will be substituted for

x

into the given function.

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

Substitute

$x = 8849$

P (8849) = 1013 e^{- 0.000128 (8849)}

Evaluate right-hand side

MultNegPos

$(- a) b = - a b$

P (8849) = 1013 e^{- 1.132672}

UseCalc

Use a calculator

P (8849) = 326.359490 \dots

RoundInt

Round to nearest integer

P (8849) \approx 326

The pressure at the summit of Mount Aconcagua, which has an altitude of

8849

meters, is about

326

hecto Pascals.

b To graph the given function, a table of values will be made first. Only non-negative values will be considered since it is a mountain climb under consideration.

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

The obtained points will now be plotted on a coordinate plane and connected with a smooth curve. Since only non-negative values are considered, the graph just needs to be drawn in the first quadrant.

c To estimate the altitude where the atmospheric pressure is

700

hecto Pascals, the graph from Part B will be used. The point with the second coordinate

700

will be spotted and its

x -

coordinate identified.

Tracing a pen or pencil vertically down from the point of the graph that is even, horizontally, with $y - value$ of $700$ leads to the $x - value$ of $3 .$ This means that the altitude with an atmospheric pressure of $700$ hecto Pascals is about $3000$ meters. Zosia feels comfortable knowing she will be able to rest for a few nights at this level.

Discussion

Logarithmic Functions

Logarithmic functions are functions that involve logarithms.

$f (x) = lo g_{b} x, b > 0 and b \neq = 1$

The function $f (x) = lo g_{b} x$ is the parent function of logarithmic functions. Since logarithms are defined for positive numbers, the domain of the function is $x > 0$ and its range is all real numbers. If $b$ is less than $1,$ the graph of the function is decreasing over its entire domain. Conversely, if $b$ is greater than $1,$ the graph is increasing over its entire domain.

Rule

Inverse Properties of Logarithms

A logarithm and a power with the same base undo each other.

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

In particular, the above equations also hold true for common and natural logarithms.

lo g 1 0^{x} = x ln e^{x} = x and and 1 0^{l o g x} = x e^{l n x} = x

Proof

The general equations will be proved one at a time.

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

This identity can be proved by using the Power Property of Logarithms and the definition of a logarithm.

lo g_{b} b^{x}

$lo g_{b} (a^{m}) = m \cdot lo g_{b} (a)$

x lo g_{b} b

$lo g_{b} (b) = 1$

x (1)

IdPropMult

\IdPropMult

x ✓

The logarithm of

b^{x}

with base

b

is equal to

x .

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

Let

lo g_{b} x = a .

Therefore, by the definition of a logarithm,

b^{a} = x .

lo g_{b} x = a \Leftrightarrow b^{a} = x

This will be used to prove the identity.

b^{l o g_{b} x}

Substitute

$lo g_{b} x = a$

b^{a}

Substitute

$b^{a} = x$

x ✓

Therefore,

b

to the power of

lo g_{b} x

is equal to

x .

Consider the following exponential and logarithmic functions.

f (x) = b^{x} and g (x) = l o g_{b} x

By using the Inverse Property of Logarithms, the composition of these functions results in the identity function.

f (g (x)) = b^{l o g_{b} x} ⇕ f (g (x)) = x g (f (x)) = l o g_{b} b^{x} ⇕ g (f (x)) = x

This means that exponential and logarithmic functions are inverse functions. Therefore, the graphs of these functions are each other's reflection across the line

y = x .

Example

Modeling Battery Charges Using Logarithmic Functions

Zosia is getting everything ready for the multi-day hike in Mount Aconcagua. Because her extra battery pack is quite limited, she wants her smartphone and camera to have their batteries fully charged.

The night before the hike, Zosia charges her smartphone and camera. She does not want to leave her devices plugged all night long because it is a waste of energy and could damage the batteries. She read in the user manuals of both devices that the charge in the batteries can be modeled by two logarithmic functions.

Phone ’ s Battery p (x) = ln x Camera ’ s Battery c (x) = 2 lo g (x - 0.5)

Here,

p (x)

and

c (x)

are the phone's and camera's charge, respectively,

x

hours after the devices were plugged in. Zosia wants to graph these two functions on separate coordinate planes. This will help her understand at what time to unplug her devices. Help Zosia graph these functions!

Answer

Phone's Battery:

Camera's Battery:

Hint

To graph $p (x) = ln x,$ first graph its inverse function $g (x) = e^{x}$ and then reflect the curve across the line $y = x .$ To graph $c (x) = 2 lo g (x - 0.5),$ make a table of values. Consider its domain first.

Solution

The logarithmic functions will be graphed one at a time.

Graphing $p (x) = ln x$

It is known that logarithms and exponents are inverse operations. Now, consider the natural base exponential function

g (x) = e^{x} .

By the Inverse Properties of Logarithms, it can be shown that the compositions of

p (x)

and

g (x)

result in the identity function.

p (g (x)) = ln e^{x} ⇕ p (g (x)) = x g (p (x)) = e^{l n x} ⇕ g (p (x)) = x

Therefore,

p (x)

and

g (x)

are inverse functions. This means that the graph of

p (x)

is a reflection across the line

y = x

of the graph of

g (x) .

Since

g (x)

is a natural base exponential function, the graph can be drawn by using its initial value and constant multiplier.

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

To graph this function, a table of values will be used. Before that, however, the domain of

c (x)

should be determined. To do that, recall that logarithms are defined only for positive numbers. Therefore, whatever the expression is in the logarithm, it must be positive.

c (x) = 2 lo g (x - 0.5) x - 0.5 > 0 \Leftrightarrow x > 0.5

The domain of the function is the set of all real numbers greater than

0.5 .

To make the table, only values in the domain will be considered.

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

These points can now be plotted on a coordinate plane and connected with a smooth curve.

Example

Finding Inverse Functions of Logarithmic and Exponential Functions

The expedition was a complete success! What is more, some members of the hiking team felt so inspired by Zosia’s math skills in helping them reach the summit, that they decided to practice some logarithmic functions. They believe this knowledge will also help them in their next expedition to the Amazon Rainforest.

Touched by their passion to learn, Zosia found a few interesting exercises about exponential and logarithmic functions in her online textbook to share with the team.

a Find the inverse function of

f (x) = 2 lo g (x - 1) .

b Find the inverse function of

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

Hint

a The inverse of

f (x)

is a common base exponential function.

b The inverse of

g (x)

is a natural base logarithmic function.

Solution

a To find the inverse of the function, the first step is replacing

f (x)

with

y .

f (x) = 2 lo g (x - 1) ↓ ⏐ ⏐ y = 2 lo g (x - 1)

Now, the

x -

and

y -

variables need to be switched.

y = 2 lo g (x - 1) ↓ ⏐ ⏐ x = 2 lo g (y - 1)

Next, in the obtained equation,

y

will be isolated.

x = 2 lo g (y - 1)

DivEqn

$LHS / 2 = RHS / 2$

\frac{x}{2} = lo g (y - 1)

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

\frac{1}{2} x = lo g (y - 1)

The definition of a logarithm will be used to express the logarithmic equation as an exponential equation. Since it is a common logarithm, its base is

10 .

\frac{1}{2} x = lo g (y - 1) ⇕ 1 0^{\frac{1}{2} x} = y - 1

Finally,

1

can be added to both sides of the equation.

1 0^{\frac{1}{2} x} = y - 1

AddEqn

$LHS + 1 = RHS + 1$

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

RearrangeEqn

Rearrange equation

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

Substitute

$y = f^{- 1} (x)$

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

To verify that the obtained function is actually the inverse of the given function, both will be graphed on the same coordinate plane. To do so, two tables of values will be made. The first table corresponds to

f (x) .

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

Next, the table that corresponds to $f^{- 1} (x)$ will be constructed.

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

Now the points can be plotted on a coordinate plane and connected with smooth curves. Also, the line $y = x$ will be graphed.

The graphs are each other's reflection across the line $y = x .$ Therefore, $f^{- 1} (x) = 1 0^{\frac{1}{2} x} + 1$ is the inverse function of $f (x) = 2 lo g (x - 1) .$

b As in Part A, to find the inverse of the function the first step is replacing

g (x)

with

y .

g (x) = \frac{1}{5} e^{3 x} ⟶ y = \frac{1}{5} e^{3 x}

Now, the

x -

and

y -

variables need to be switched.

y = \frac{1}{5} e^{3 x} ⟶ x = \frac{1}{5} e^{3 y}

Next, in the obtained equation,

y

will be isolated.

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

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

5 x = e^{3 y}

The definition of a logarithm will be used to express the exponential equation as a logarithmic equation.

5 x = e^{3 y} \Leftrightarrow ln 5 x = 3 y

Finally, both sides of the equation can be divided by

3 .

ln 5 x = 3 y

DivEqn

$LHS / 3 = RHS / 3$

\frac{ln 5 x}{3} = y

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

\frac{1}{3} ln 5 x = y

RearrangeEqn

Rearrange equation

y = \frac{1}{3} ln 5 x

Substitute

$y = g^{- 1} (x)$

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

g (x) .

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

Next, the table that corresponds to $g^{- 1} (x)$ will be constructed.

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

The obtained points can be plotted on a coordinate plane and connected with smooth curves. Also, the line $y = x$ will be graphed.

It can be seen that the graphs are each other's reflection across the line $y = x .$ Therefore, $g^{- 1} (x) = \frac{1}{3} ln 5 x$ is the inverse function of $g (x) = \frac{1}{5} e^{3 x} .$

Checking Our Answer

a To verify that the obtained function is actually the inverse of the given function, instead of graphing it can be checked that both composite functions

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

and

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

simplify to

x .

Start by calculating

f (f^{- 1} (x)) .

f (x) = 2 lo g (x - 1) ⇓ f (f^{- 1} (x)) = 2 lo g (f^{- 1} (x) - 1)

Now, substitute

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

in the above equation, and simplify as much as possible.

f (f^{- 1} (x)) = 2 lo g (f^{- 1} (x) - 1)

Substitute

$f^{- 1} (x) = 1 0^{\frac{1}{2} x} + 1$

f (f^{- 1} (x)) = 2 lo g (1 0^{\frac{1}{2} x} + 1 - 1)

Simplify right-hand side

SubTerm

Subtract term

f (f^{- 1} (x)) = 2 lo g (1 0^{\frac{1}{2} x})

$lo g (1 0^{m}) = m$

f (f^{- 1} (x)) = 2 (\frac{1}{2} x)

AssociativePropMult

\AssociativePropMult

f (f^{- 1} (x)) = 2 (\frac{1}{2}) x

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

f (f^{- 1} (x)) = 1 x

IdPropMult

\IdPropMult

f (f^{- 1} (x)) = x ✓

Next, by following the same procedure, check that

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

also simplifies to the identity function.

f^{- 1} (x) = 1 0^{\frac{1}{2} x} + 1 ⇓ f^{- 1} (f (x)) = 1 0^{\frac{1}{2} f (x)} + 1

Now, substitute

f (x) = 2 lo g (x - 1)

in the above equation, and simplify as much as possible.

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

Substitute

$f (x) = 2 l o g (x - 1)$

f^{- 1} (f (x)) = 1 0^{\frac{1}{2} (2 l o g (x - 1))} + 1

Simplify right-hand side

AssociativePropMult

\AssociativePropMult

f^{- 1} (f (x)) = 1 0^{\frac{1}{2} (2) l o g (x - 1)} + 1

FracMultDenomToNumber

$\frac{a}{2} \cdot 2 = a$

f^{- 1} (f (x)) = 1 0^{1 l o g (x - 1)} + 1

IdPropMult

\IdPropMult

f^{- 1} (f (x)) = 1 0^{l o g (x - 1)} + 1

$1 0^{l o g (m)} = m$

f^{- 1} (f (x)) = x - 1 + 1

AddTerms

Add terms

f^{- 1} (f (x)) = x ✓

Since both compositions simplify to the identity function,

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

is the inverse function of

f (x) = 2 lo g (x - 1) .

b As in the previous part, to check whether

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

is the inverse function of

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

the compositions

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

and

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

must both simplify to

x .

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

Since both compositions simplify to the identity function, $g^{- 1} (x) = \frac{1}{3} ln 5 x$ is the inverse function of $g (x) = \frac{1}{5} e^{3 x} .$

Closure

Horizontal and Vertical Asymptotes

In this lesson, exponential functions, logarithmic functions, and the relationship between their graphs have been discussed. One more characteristic of these graphs is that they have asymptotes.

Exponential Graphs

Every exponential graph has a horizontal asymptote. In the applet below, both graphs have the horizontal line

y = 0,

which is the

x -

axis, as an asymptote.

Logarithmic Graphs

Every logarithmic graph has a vertical asymptote. In the applet below, both graphs have the vertical line $x = 0,$ which is the $y -$ axis, as an asymptote.

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Catch-Up and Review

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

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

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

Graphing $p (x) = ln x$

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