A3.2 Solve exponential equations using common bases or logarithms

Ontario Grade 12 Mathematics

Exponential and Logarithmic Functions

Choosing between common bases and logarithms

An exponential equation is an equation in which the variable appears in an exponent. For example,

2^x=16

is exponential because

x

is an exponent. First, look for a way to express both sides with the same base. If that is not practical, use logarithms to bring the exponent down so you can solve for the variable. This lesson reviews the needed exponent rules, explains both methods, and shows how to check a solution.

What you will learn

Identify when an exponential equation can be solved by rewriting both sides with the same base.
Use logarithms to solve an exponential equation when a common base is not practical.
Check a solution in the original equation and state it appropriately.

1. Prerequisite bridge: powers and exponents

A power has a base and an exponent. In

a^n

,

a

is the base and

n

is the exponent. When the exponent is a positive whole number, it tells how many times the base is used as a factor. For instance,

3^2=3\times3=9

.

You can rewrite a number as a power in more than one way. Since

16=2^4

, the equation

2^x=16

can be written as

2^x=2^4

. Both sides now have the same base. For a positive base other than

1

, equal powers with that base have equal exponents, so

x=4

.

An exponent rule that will help is

(a^m)^n=a^{mn}

. For example,

(5^2)^3=5^6

. This rule lets you rewrite powers using a chosen base. It does not allow you to add exponents when powers are multiplied unless the bases are the same.

(a^m)^n=a^{mn}

The variable in an exponential equation is in an exponent.
A common base is a base that appears in the power expressions on both sides.
For the common-base method, use equal exponents when the bases match and are positive and not equal to $1$ .

2. Method one: rewrite using common bases

Try common bases first when each side can be written as a power of the same number. Common bases are especially useful when the numbers are familiar powers, such as powers of

2

,

3

, or

10

.

For example, in

3^{x+1}=27

, recognize that

27=3^3

. The equation becomes

3^{x+1}=3^3

, so the exponents must match:

x+1=3

. Solving this linear equation gives

x=2

. The method works because both sides are powers of the same positive base, and that base is not

1

.

A coefficient may need to be rewritten too. In

4^x=8

, write

4=2^2

and

8=2^3

. Then

(2^2)^x=2^3

, so

2^{2x}=2^3

and

2x=3

. Thus

x=\frac{3}{2}

. The exponent rule explains why the exponent on the left becomes

2x

.

Sometimes the sides cannot be conveniently written as powers of one common base. In that case, do not force a common-base approach. Use logarithms instead.

(a^u=a^v)\;\Longrightarrow\;u=v\quad(a>0,\ a\ne1)

Rewrite each side completely before matching exponents.
Apply exponent rules carefully when a power is raised to another power.
After matching exponents, solve the resulting equation using familiar algebra.

3. Method two: use logarithms

A logarithm answers a question about an exponent. The expression

\log_b N

means “the exponent on

b

that gives

N

.” For example,

\log_2 8=3

because

2^3=8

. The base

b

must be positive and not equal to

1

, and the number

N

must be positive.

A key relationship is that taking a logarithm with base

b

reverses raising

b

to a power. Thus, if

b^x=N

, then

x=\log_b N

. This is useful when the right side is not an easy power of the base.

On many calculators, the common logarithm key is written

\log

and means logarithm base

10

. The natural logarithm key is written

\ln

and means logarithm base

e

, where

e

is a positive constant. Either can be used to solve an equation like

5^x=17

, as long as the same type of logarithm is used on both sides. The change-of-base relationship gives

\log_b N=\frac{\log N}{\log b}

, so a calculator can evaluate logarithms with other bases.

If the equation is

b^{g(x)}=N

, take a logarithm of both sides. The exponent can then be brought down as a factor:

\log(b^{g(x)})=g(x)\log b

. Solve the resulting equation for the variable. Since logarithms require positive inputs, check that the expressions used as logarithm inputs are positive.

\log(b^u)=u\log b

A logarithm tells you which exponent produces a given positive number.
When using logarithms to solve, take the same type of logarithm on both sides.
After using logarithms, isolate the variable and check the result in the original equation.

4. Choosing a method and checking a solution

Start by asking whether the numbers can be rewritten as powers of one common base. If they can, matching exponents often gives an exact answer with little calculation. If they cannot, logarithms give a general method.

A logarithm calculation may produce a decimal approximation. Use the full calculator value during your work, and round only the final answer to the requested precision. A rounded decimal may not make the original equation exactly true, so it is sensible to check with the unrounded value when possible.

To check, substitute the solution into the original equation and compare the two sides. For a decimal answer, compare their calculator values at a reasonable precision. If the sides do not agree closely, check for an arithmetic error, a calculator-entry error, or premature rounding.

A common mistake is to treat the exponent as if it can be separated from the base without a logarithm or a valid common-base rewrite. Another is to apply the exponent rule incorrectly:

(a^m)^n

is

a^{mn}

, not

a^{m+n}

. Keep the equation balanced by applying the same operation to both sides.

\log_b N=\frac{\log N}{\log b}

Prefer common bases when they are clear and convenient.
Use logarithms when matching bases is not practical.
Check the answer in the original equation, not only in the rearranged equation.

Worked example

Solving with logarithms

Solve

5^{2x-1}=17

. Give the answer to three decimal places.

Choose a method
The number $17$ is not a convenient whole-number power of $5$ . Use logarithms on both sides. The input $17$ is positive, so its logarithm is defined.
Take logarithms
Use the common logarithm on each side. The power rule for logarithms brings the exponent down as a factor.
$\log(5^{2x-1})=\log 17$
Bring down the exponent
Apply $\log(b^u)=u\log b$ . This gives an equation with $x$ outside the exponent.
$(2x-1)\log 5=\log 17$
Isolate the variable
Divide both sides by $\log 5$ , then add $1$ and divide by $2$ . Keep the calculator value unrounded until the final step.
$x=\frac{1+\frac{\log 17}{\log 5}}{2}\approx1.380$
Check
Using the unrounded solution, the exponent $2x-1$ is approximately $1.76036$ . Substitution gives a value approximately equal to $17$ ; the displayed rounded solution is intended to three decimal places.
$5^{2(1.380)-1}\approx17.0$

Answer:

x\approx1.380

Check: Using the unrounded value from the logarithm expression gives

5^{2x-1}=17

.

Common mistakes and how to avoid them

Matching exponents before the bases are the same.

Correction: Rewrite both sides with a common base first. If that is not practical, take logarithms.

Changing

(a^m)^n

to

a^{m+n}

.

Correction: When a power is raised to a power, multiply the exponents:

(a^m)^n=a^{mn}

.

Taking a logarithm of only one side of an equation.

Correction: Apply the same logarithm to both sides to keep the equation balanced.

Rounding a logarithm value too early.

Correction: Keep the calculator value through the calculation and round only the final answer.

Lesson summary

Rewrite both sides with a common base when possible, then equate their exponents.
When a common base is not practical, take logarithms of both sides and use the power rule.
Solve the resulting equation, round as directed, and check the answer in the original equation.

Check your understanding

Question 1

Solve

2^{x+1}=16

.

$x=2$
$x=3$
$x=4$
correctIndex

Show answer and explanation

x=3

Since

16=2^4

, match exponents:

x+1=4

, so

x=3

.

Question 2

Which equation correctly represents the result of taking a common logarithm of both sides of

7^x=20

?

$x\log 7=\log 20$
$\log x=20\log 7$
$7\log x=\log 20$
correctIndex

Show answer and explanation

x\log 7=\log 20

The logarithm power rule gives

\log(7^x)=x\log 7

, so the equation is

x\log 7=\log 20

.

Question 3

Which value is the solution of

3^x=10

to three decimal places?

$x\approx2.096$
$x\approx2.000$
$x\approx3.333$
correctIndex

Show answer and explanation

x\approx2.096

Taking logarithms gives

x=\frac{\log 10}{\log 3}\approx2.096

. Substitution gives a value close to

10

.

Key terms

Exponential equation: An equation in which the variable appears in an exponent.
Base: The number raised to a power, as $a$ is in $a^n$ .
Exponent: The number that indicates the power to which the base is raised.
Logarithm: The exponent that a stated base must be raised to in order to produce a given positive number.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation A3.2. It is a study resource, not an official curriculum publication.

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A3.2 · Solve exponential equations using common bases or logarithms

Ontario Grade 12 Mathematics

Choosing between common bases and logarithms

What you will learn

1. Prerequisite bridge: powers and exponents

2. Method one: rewrite using common bases

3. Method two: use logarithms

4. Choosing a method and checking a solution

Solving with logarithms

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MHF4U

About this lesson