SL 2.9 Connect exponential and logarithmic functions as inverses | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

Connecting equations, graphs, values, and real-world situations

An exponential function answers a question such as, “What value results when a base is raised to this exponent?” A logarithmic function reverses that question: “What exponent gives this value?” This lesson connects the two functions as inverses. You will use the connection to interpret values, solve equations, and understand graphs. We assume familiarity with powers and with solving simple equations by applying the same operation to both sides.

What you will learn

Explain what it means for two functions to be inverses.
Convert between exponential and logarithmic forms and evaluate expressions using this connection.
Describe how the graphs of an exponential function and its logarithmic inverse are related.
Use logarithms to find an unknown exponent and check results analytically or with graphing technology.

1. Prior knowledge and the inverse relationship

A function takes an input and produces an output. Its inverse reverses that process: it takes the original output and returns the original input. For example, if an exponential function sends an input of

3

to an output of

8

, its inverse sends

8

back to

3

.

For a base

a

with

a>0

and

a\ne 1

, the exponential statement

a^x=y

is equivalent to the logarithmic statement

\log_a(y)=x

. The notation

\log_a(y)

means “the exponent on

a

that gives

y

.” The base must be positive and cannot be

1

, because powers of

1

do not distinguish different exponents.

The input to a real logarithm must be positive:

y>0

. Thus

a^x

has domain all real numbers and range positive real numbers;

\log_a(x)

has domain positive real numbers and range all real numbers. These restrictions are part of the inverse relationship.

a^x=y\iff\log_a(y)=x

Exponential and logarithmic forms of the same statement are equivalent.
A logarithm returns an exponent.
The argument of a real logarithm must be positive.

2. Values and graphs

To evaluate a logarithm, ask which exponent produces its argument. For instance,

\log_2(32)=5

, because

2^5=32

. To find an exact value, try expressing the argument as a power of the base.

Two identities follow directly from the inverse relationship: taking a logarithm with base

a

reverses raising

a

to a power, and raising

a

to a logarithm with base

a

returns the original positive input. These identities work because each operation undoes the other.

For

a>1

, the exponential function and its logarithmic inverse are increasing. For

0<a<1

, both are decreasing. In either case, their graphs reflect across the line

y=x

: a point

(p,q)

on one graph corresponds to

(q,p)

on the other. The exponential graph passes through

(0,1)

, so its inverse logarithmic graph passes through

(1,0)

.

\log_a(a^x)=x,\qquad a^{\log_a(x)}=x\quad(x>0)

Check a logarithm value by rewriting it as a power.
Inverse graphs exchange the coordinates of corresponding points.
The exponential and logarithmic graphs reflect across $y=x$ .

3. Solving equations, context, and technology

When an unknown appears as an exponent, rewriting both sides with the same base may give an exact answer. If that is not practical, use a logarithm: from

a^x=c

, take logarithms to obtain

x=\log_a(c)

, provided

a>0

,

a\ne1

, and

c>0

. Calculators commonly provide the natural logarithm, written

\ln

, and the common logarithm, written

\log

. A change-of-base calculation can evaluate logarithms with other bases.

A graphing calculator can check an intersection or estimate a logarithm that is not a familiar exact value. Enter the two sides of an equation as separate functions and locate their intersection. Explain why logarithms apply, then substitute the estimate into the original equation. Calculator output is a numerical check, not a replacement for the mathematical reasoning.

In a context, identify what the exponent represents before calculating. If a quantity is multiplied by a constant factor over equal time intervals, an exponential model may describe it. Finding when it reaches a specified positive amount involves finding an exponent, so a logarithm is appropriate. Keep units attached to the time and check that the starting value and factor match the situation.

a^x=c\iff x=\log_a(c)

Try expressing both sides with a common base before using a calculator.
Use logarithms to isolate an unknown exponent.
Check numerical answers in the original equation and report appropriate units.

Worked example

Evaluate a logarithm exactly

Find

\log_3(81)

and verify the answer.

Interpret the logarithm
The logarithm asks which exponent on $3$ produces $81$ .
Rewrite as a power
Since $81=3^4$ , the required exponent is $4$ .
$\log_3(81)=4$
Verify
Convert back to exponential form. This confirms the value.
$3^4=81$

Answer:

\log_3(81)=4

.

Check: Converting back gives

3^4=81

, as required.

Worked example

Solve an exponential equation

Solve

5^{x+1}=17

. Give the answer to three significant figures.

Apply the inverse
Take logarithm base $5$ on both sides. This reverses raising $5$ to a power.
$x+1=\log_5(17)$
Find the unknown
Subtract $1$ . A calculator can evaluate the logarithm; the logarithmic expression gives the exact value.
$x=\log_5(17)-1\approx0.760$
Check in the original equation
Substitute the rounded value into the original exponent, including the $+1$ . The result is close to $17$ because the solution has been rounded.
$5^{0.760+1}\approx17.0$

Answer:

x\approx0.760

to three significant figures.

Check: The original equation is

5^{x+1}=17

. Substitution gives

5^{0.760+1}\approx17.0

, consistent with

17

to three significant figures.

Worked example

Find a time from an exponential model

A quantity follows the model

Q=12(1.5)^t

, where

Q

is measured in grams and

t

is measured in hours. Find when the quantity reaches

30

grams. Give the time to three significant figures.

Substitute the target amount
Set the model output equal to $30$ grams. Divide by the starting amount, $12$ grams, to leave the exponential factor.
$1.5^t=2.5$
Use a logarithm
The unknown is an exponent, so take logarithm base $1.5$ to identify it.
$t=\log_{1.5}(2.5)$
Evaluate and interpret
A calculator gives the time in hours. Substitute the estimate into the original model to check the output.
$t\approx2.26\text{ h}$

Answer: The quantity reaches

30

grams after approximately

2.26

hours.

Check: Using

t\approx2.26

gives

12(1.5)^{2.26}\approx30.0

grams. The time unit is hours, as specified by the model.

Common mistakes and how to avoid them

Reading

\log_a(y)

as

a

multiplied by

y

.

Correction: Read it as the exponent on

a

that produces

y

. Rewrite it as

a^x=y

when interpreting the logarithm.

Taking a real logarithm of zero or a negative number.

Correction: Check that the logarithm’s argument is positive before using it.

Swapping the base and argument when converting forms.

Correction: In

a^x=y

, the base remains

a

, the result

y

becomes the logarithm’s argument, and the exponent becomes its value:

\log_a(y)=x

.

Rounding a calculator value too early.

Correction: Keep the exact logarithmic expression or retain extra calculator digits during working, then round the final result to the requested accuracy.

Lesson summary

The exponential equation $a^x=y$ and logarithmic equation $\log_a(y)=x$ describe the same relationship.
For real logarithms, the base satisfies $a>0$ and $a\ne1$ , and the argument is positive.
Inverse functions undo one another, and their graphs reflect across $y=x$ .
Logarithms turn an unknown exponent into a value that can be evaluated or estimated.

Check your understanding

Question 1

What is

\log_4(64)

?

$2$
$3$
$4$
$16$

Show answer and explanation

3

Since

4^3=64

, the exponent is

3

.

Question 2

Which equation is equivalent to

7^x=20

?

$\log_7(x)=20$
$\log_{20}(7)=x$
$\log_7(20)=x$
$20^x=7$

Show answer and explanation

\log_7(20)=x

Keep the base

7

as the logarithm’s base, put

20

in its argument, and make the exponent

x

the logarithm’s value.

Question 3

What is the domain of

\log_2(x)

as a real-valued function?

All real numbers
$x>0$
$x\geq0$
$x\ne2$

Show answer and explanation

x>0

A real logarithm requires a positive argument, so

x

must be greater than

0

.

Key terms

Inverse functions: Functions that undo one another by exchanging an input and its output.
Exponential function: A function in which the variable appears as an exponent, such as $a^x$ .
Logarithm: The exponent required on a specified base to produce a given positive value.
Argument: The value inside a logarithm, such as $y$ in $\log_a(y)$ .

Continue through IB AA SL

View the complete IB AA SL International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL, study topic SL 2.9. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

SL 2.9 · Connect exponential and logarithmic functions as inverses

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Connecting equations, graphs, values, and real-world situations

What you will learn

1. Prior knowledge and the inverse relationship

2. Values and graphs

3. Solving equations, context, and technology

Evaluate a logarithm exactly

Solve an exponential equation

Find a time from an exponential model

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review