1.9.5 Solving exponential equations

What you'll learn

How to recognise an exponential equation.
How to solve by rewriting both sides with the same base.
How logarithms solve equations of the form $a^x=b$ .
How to decide whether a solution exists and present suitable exact or decimal answers.

Prerequisites: powers and indices

An index (plural indices), also called an exponent or power, tells you how many times a base is used as a factor. In $2^5$ , the base is 2 and the index is 5.

You should remember these index laws:

am×an=am+naman=am−n(am)n=amna−n=1ana0=1\begin{aligned} a^m \times a^n &= a^{m+n} \\ \frac{a^m}{a^n} &= a^{m-n} \\ \left(a^m\right)^n &= a^{mn} \\ a^{-n} &= \frac{1}{a^n} \\ a^0 &= 1 \end{aligned}

These laws let you rewrite expressions so that both sides of an equation have the same base.

Example

Rewriting powers with the same base

Solve $8^x=32$ .

Write both 8 and 32 as powers of 2:
$8=23and32=25.8=2^3 \qquad\text{and}\qquad 32=2^5.$
Substitute these forms into the equation and use the power-of-a-power rule:
$(23)x=25⇒23x=25.\left(2^3\right)^x=2^5 \quad\Rightarrow\quad 2^{3x}=2^5.$
Equal powers with the same valid base have equal indices, so:
$\quad\Rightarrow\quad x=\frac{5}{3}.$

Key Idea

Same base, same index

If $a^p=a^q$ , where $a > 0$ and $a≠1a\neq1$ , then $p = q$ . Always check whether both sides can be written as powers of the same base before using logarithms.

What is an exponential equation?

An exponential equation is an equation in which the unknown appears in an exponent. For example, $3^x=17$ is exponential because $x$ is the power.

The AQA specification requires you to solve equations of the form:

a^x=b.

For a standard exponential function, the base must satisfy $a > 0$ and $a≠1a\neq1$ .

Understanding the graph

For $a > 1$ , the graph of $y=a^x$ is increasing and always lies above the $x$ -axis. The solution of $a^x=b$ is the $x$ -coordinate where the exponential curve intersects the horizontal line $y = b$ .

An increasing exponential curve meeting the horizontal line y equals b at x equals log base a of b

When $b > 0$ , there is exactly one intersection and therefore exactly one real solution.

If $0 < a < 1$ , the exponential graph is decreasing, but it still takes every positive value exactly once. There is again one real solution when $b > 0$ .

Common Mistake

When no real solution exists

For a valid positive base, $a^x$ is always positive. Therefore, $a^x=b$ has no real solution if $b≤0b\leq0$ .

Solving by inspection

Sometimes you can recognise $b$ as a familiar power of $a$ . This is called solving by inspection because no logarithms or calculator work are needed.

Example

Recognising a familiar power

Solve $5x=11255^x=\frac{1}{125}$ .

Express 125 as a power of 5:
$125=5^3.$
Use the negative-index law to rewrite the reciprocal:
$1125=153=5−3.\frac{1}{125}=\frac{1}{5^3}=5^{-3}.$
The equation is now $5^x=5^{-3}$ , so the indices must be equal:
$x = - 3.$

Tip

Check simple powers first

Before reaching for your calculator, look for positive, zero, fractional and negative powers. In particular, remember that $a^0=1$ and $a^{-n}=1/a^n$ .

Logarithms as inverse operations

Most numbers cannot be written as convenient powers of the base. For example, there is no obvious whole-number power of 3 equal to 17. This is where logarithms are needed.

A logarithm gives the exponent to which a base must be raised to produce a particular number.

Definition

Logarithm

For $a > 0$ , $a≠1a\neq1$ and $b > 0$ ,

ax=b⟺x=log⁡ab.a^x=b \quad\Longleftrightarrow\quad x=\log_a b.

The expression $log_a b$ means “the power of $a$ that gives $b$ ”.

For example, $log_2 8=3$ because $2^3=8$ . Exponentials and logarithms are inverse functions, meaning that each reverses the effect of the other.

Example

Using a logarithm with the required base

Solve $3^x=17$ .

Rewrite the exponential equation directly in logarithmic form:
$x=\log_3 17.$
This is the exact answer.
A calculator may not have a direct $log_3$ button, so use the change-of-base formula:
$log⁡317=log⁡17log⁡3.\log_3 17=\frac{\log 17}{\log 3}.$
Evaluate and round appropriately:
$x≈2.579.x\approx2.579.$

The change-of-base formula

Calculators usually provide $log⁡\log$ , the logarithm to base 10, and $ln⁡\ln$ , the natural logarithm to base $e$ .

You may use either one:

log⁡ab=log⁡blog⁡aorlog⁡ab=ln⁡bln⁡a.\log_a b=\frac{\log b}{\log a} \qquad\text{or}\qquad \log_a b=\frac{\ln b}{\ln a}.

Therefore, the general solution of $a^x=b$ is:

x=log⁡blog⁡a.x=\frac{\log b}{\log a}.

The same answer results from natural logarithms:

x=ln⁡bln⁡a.x=\frac{\ln b}{\ln a}.

Why the formula works

You can derive the formula rather than simply memorising it. Starting from $a^x=b$ , take logarithms of both sides:

log⁡(ax)=log⁡b.\log\left(a^x\right)=\log b.

Using the logarithm law $log(a^x)=x\log a$ gives:

xlog⁡a=log⁡b.x\log a=\log b.

Dividing by $log⁡a\log a$ gives:

x=log⁡blog⁡a.x=\frac{\log b}{\log a}.

Example

Solving with natural logarithms

Solve $0.7^x=12$ , giving your answer to three significant figures.

Take natural logarithms of both sides:
$ln⁡(0.7x)=ln⁡12.\ln\left(0.7^x\right)=\ln 12.$
Bring the exponent down using $ln(a^x)=x\ln a$ :
$xln⁡0.7=ln⁡12.x\ln 0.7=\ln 12.$
Divide by $ln⁡0.7\ln 0.7$ and evaluate:
$x=ln⁡12ln⁡0.7≈−6.97.x=\frac{\ln 12}{\ln 0.7}\approx-6.97.$
The negative answer is sensible: because the base is between 0 and 1, a negative power produces a value greater than 1.

Common Mistake

Dividing the logarithms incorrectly

From $xlog⁡a=log⁡bx\log a=\log b$ , the correct result is $x=log⁡blog⁡ax=\frac{\log b}{\log a}$ . Keep the logarithm of the right-hand side in the numerator.

Exact and decimal answers

The form $log_a b$ or $ln⁡bln⁡a\frac{\ln b}{\ln a}$ is an exact value. A calculator value such as 2.579 is an approximation.

If a question asks for an exact answer, leave your result in logarithmic form. If it requests a given degree of accuracy, keep the exact value in your calculator and round only at the end.

Common Mistake

Rounding too early

Do not round separate logarithms before dividing them. Enter the full calculation, such as ln(17) ÷ ln(3), and round only the final answer.

Checking your solution

Substitute your answer into the original equation. For $3^x=17$ with $x≈2.579x\approx2.579$ :

32.579≈17.3^{2.579}\approx17.

A small difference may appear because the value of $x$ was rounded.

You can also check whether the sign is reasonable:

If $a > 1$ and $b > 1$ , then $x > 0$ .
If $a > 1$ and $0 < b < 1$ , then $x < 0$ .
If $0 < a < 1$ and $b > 1$ , then $x < 0$ .
For any valid base, if $b = 1$ , then $x = 0$ .

Exam technique

In the exam

Check whether both sides can be rewritten using the same base; this may give an exact solution quickly.
Otherwise, take logarithms and use $x=log⁡blog⁡ax=\frac{\log b}{\log a}$ or $x=ln⁡bln⁡ax=\frac{\ln b}{\ln a}$ .
Keep full calculator accuracy until the final line, round as requested, and substitute back to check that the result is sensible.

Self review

Check yourself

Can you solve $4x=184^x=\frac{1}{8}$ by first rewriting both sides with base 2?
How would you solve $7^x=20$ and give the answer to three significant figures?
Why does $5^x=-3$ have no real solution?

What you'll learn

How to recognise an exponential equation.
How to solve by rewriting both sides with the same base.
How logarithms solve equations of the form $a^x=b$ .
How to decide whether a solution exists and present suitable exact or decimal answers.

Prerequisites: powers and indices

An index (plural indices), also called an exponent or power, tells you how many times a base is used as a factor. In $2^5$ , the base is 2 and the index is 5.

You should remember these index laws:

am×an=am+naman=am−n(am)n=amna−n=1ana0=1\begin{aligned} a^m \times a^n &= a^{m+n} \\ \frac{a^m}{a^n} &= a^{m-n} \\ \left(a^m\right)^n &= a^{mn} \\ a^{-n} &= \frac{1}{a^n} \\ a^0 &= 1 \end{aligned}

These laws let you rewrite expressions so that both sides of an equation have the same base.

Example

Rewriting powers with the same base

Solve $8^x=32$ .

Write both 8 and 32 as powers of 2:
$8=23and32=25.8=2^3 \qquad\text{and}\qquad 32=2^5.$
Substitute these forms into the equation and use the power-of-a-power rule:
$(23)x=25⇒23x=25.\left(2^3\right)^x=2^5 \quad\Rightarrow\quad 2^{3x}=2^5.$
Equal powers with the same valid base have equal indices, so:
$\quad\Rightarrow\quad x=\frac{5}{3}.$

Key Idea

Same base, same index

If $a^p=a^q$ , where $a > 0$ and $a≠1a\neq1$ , then $p = q$ . Always check whether both sides can be written as powers of the same base before using logarithms.

What is an exponential equation?

An exponential equation is an equation in which the unknown appears in an exponent. For example, $3^x=17$ is exponential because $x$ is the power.

The AQA specification requires you to solve equations of the form:

a^x=b.

For a standard exponential function, the base must satisfy $a > 0$ and $a≠1a\neq1$ .

Understanding the graph

An increasing exponential curve meeting the horizontal line y equals b at x equals log base a of b

When $b > 0$ , there is exactly one intersection and therefore exactly one real solution.

If $0 < a < 1$ , the exponential graph is decreasing, but it still takes every positive value exactly once. There is again one real solution when $b > 0$ .

Common Mistake

When no real solution exists

For a valid positive base, $a^x$ is always positive. Therefore, $a^x=b$ has no real solution if $b≤0b\leq0$ .

Solving by inspection

Sometimes you can recognise $b$ as a familiar power of $a$ . This is called solving by inspection because no logarithms or calculator work are needed.

Example

Recognising a familiar power

Solve $5x=11255^x=\frac{1}{125}$ .

Express 125 as a power of 5:
$125=5^3.$
Use the negative-index law to rewrite the reciprocal:
$1125=153=5−3.\frac{1}{125}=\frac{1}{5^3}=5^{-3}.$
The equation is now $5^x=5^{-3}$ , so the indices must be equal:
$x = - 3.$

Tip

Check simple powers first

Before reaching for your calculator, look for positive, zero, fractional and negative powers. In particular, remember that $a^0=1$ and $a^{-n}=1/a^n$ .

Logarithms as inverse operations

Most numbers cannot be written as convenient powers of the base. For example, there is no obvious whole-number power of 3 equal to 17. This is where logarithms are needed.

A logarithm gives the exponent to which a base must be raised to produce a particular number.

Definition

Logarithm

For $a > 0$ , $a≠1a\neq1$ and $b > 0$ ,

ax=b⟺x=log⁡ab.a^x=b \quad\Longleftrightarrow\quad x=\log_a b.

The expression $log_a b$ means “the power of $a$ that gives $b$ ”.

For example, $log_2 8=3$ because $2^3=8$ . Exponentials and logarithms are inverse functions, meaning that each reverses the effect of the other.

Example

Using a logarithm with the required base

Solve $3^x=17$ .

Rewrite the exponential equation directly in logarithmic form:
$x=\log_3 17.$
This is the exact answer.
A calculator may not have a direct $log_3$ button, so use the change-of-base formula:
$log⁡317=log⁡17log⁡3.\log_3 17=\frac{\log 17}{\log 3}.$
Evaluate and round appropriately:
$x≈2.579.x\approx2.579.$

The change-of-base formula

Calculators usually provide $log⁡\log$ , the logarithm to base 10, and $ln⁡\ln$ , the natural logarithm to base $e$ .

You may use either one:

log⁡ab=log⁡blog⁡aorlog⁡ab=ln⁡bln⁡a.\log_a b=\frac{\log b}{\log a} \qquad\text{or}\qquad \log_a b=\frac{\ln b}{\ln a}.

Therefore, the general solution of $a^x=b$ is:

x=log⁡blog⁡a.x=\frac{\log b}{\log a}.

The same answer results from natural logarithms:

x=ln⁡bln⁡a.x=\frac{\ln b}{\ln a}.

Why the formula works

You can derive the formula rather than simply memorising it. Starting from $a^x=b$ , take logarithms of both sides:

log⁡(ax)=log⁡b.\log\left(a^x\right)=\log b.

Using the logarithm law $log(a^x)=x\log a$ gives:

xlog⁡a=log⁡b.x\log a=\log b.

Dividing by $log⁡a\log a$ gives:

x=log⁡blog⁡a.x=\frac{\log b}{\log a}.

Example

Solving with natural logarithms

Solve $0.7^x=12$ , giving your answer to three significant figures.

Take natural logarithms of both sides:
$ln⁡(0.7x)=ln⁡12.\ln\left(0.7^x\right)=\ln 12.$
Bring the exponent down using $ln(a^x)=x\ln a$ :
$xln⁡0.7=ln⁡12.x\ln 0.7=\ln 12.$
Divide by $ln⁡0.7\ln 0.7$ and evaluate:
$x=ln⁡12ln⁡0.7≈−6.97.x=\frac{\ln 12}{\ln 0.7}\approx-6.97.$
The negative answer is sensible: because the base is between 0 and 1, a negative power produces a value greater than 1.

Common Mistake

Dividing the logarithms incorrectly

From $xlog⁡a=log⁡bx\log a=\log b$ , the correct result is $x=log⁡blog⁡ax=\frac{\log b}{\log a}$ . Keep the logarithm of the right-hand side in the numerator.

Exact and decimal answers

The form $log_a b$ or $ln⁡bln⁡a\frac{\ln b}{\ln a}$ is an exact value. A calculator value such as 2.579 is an approximation.

If a question asks for an exact answer, leave your result in logarithmic form. If it requests a given degree of accuracy, keep the exact value in your calculator and round only at the end.

Common Mistake

Rounding too early

Do not round separate logarithms before dividing them. Enter the full calculation, such as ln(17) ÷ ln(3), and round only the final answer.

Checking your solution

Substitute your answer into the original equation. For $3^x=17$ with $x≈2.579x\approx2.579$ :

32.579≈17.3^{2.579}\approx17.

A small difference may appear because the value of $x$ was rounded.

You can also check whether the sign is reasonable:

If $a > 1$ and $b > 1$ , then $x > 0$ .
If $a > 1$ and $0 < b < 1$ , then $x < 0$ .
If $0 < a < 1$ and $b > 1$ , then $x < 0$ .
For any valid base, if $b = 1$ , then $x = 0$ .

Exam technique

In the exam

Check whether both sides can be rewritten using the same base; this may give an exact solution quickly.
Otherwise, take logarithms and use $x=log⁡blog⁡ax=\frac{\log b}{\log a}$ or $x=ln⁡bln⁡ax=\frac{\ln b}{\ln a}$ .
Keep full calculator accuracy until the final line, round as requested, and substitute back to check that the result is sensible.

Self review

Check yourself

Can you solve $4x=184^x=\frac{1}{8}$ by first rewriting both sides with base 2?
How would you solve $7^x=20$ and give the answer to three significant figures?
Why does $5^x=-3$ have no real solution?

What you'll learn

Prerequisites: powers and indices

What is an exponential equation?

Understanding the graph

Solving by inspection

Logarithms as inverse operations

The change-of-base formula

Why the formula works

Exact and decimal answers

Checking your solution

Review 1.9.5 Solving exponential equations by teaching Genie

1.9.5 Solving exponential equations

What you'll learn

Prerequisites: powers and indices

What is an exponential equation?

Understanding the graph

Solving by inspection

Logarithms as inverse operations

The change-of-base formula

Why the formula works

Exact and decimal answers

Checking your solution

Review 1.9.5 Solving exponential equations by teaching Genie

1.9.5 Solving exponential equations

What you'll learn

Prerequisites: powers and indices

What is an exponential equation?

Understanding the graph

Solving by inspection

Logarithms as inverse operations

The change-of-base formula

Why the formula works

Exact and decimal answers

Checking your solution

Review 1.9.5 Solving exponential equations by teaching Genie

1.9.5 Solving exponential equations Revision Guide

1.9.5 Solving exponential equations

What you'll learn

Prerequisites: powers and indices

What is an exponential equation?

Understanding the graph

Solving by inspection

Logarithms as inverse operations

The change-of-base formula

Why the formula works

Exact and decimal answers

Checking your solution

Review 1.9.5 Solving exponential equations by teaching Genie

1.9.5 Solving exponential equations Revision Guide