Solve exponential equations

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 25 minutes
Domains: Advanced Math
Techniques: Common-baseIsolate-firstEquate-exponentsMethod-choice

What you’ll learn

Common-base algebra. Get the power alone, write both sides as powers of the same number, and set the exponents equal.
Graphing. Find every real solution at once, then pick the one the question asks for.

Prerequisites

Why this matters on the SAT

Reveal one base, then solve a simpler equation

On the SAT, the bases in these equations are often the same number in disguise. $9$ and $27$ look different, but both are powers of $3$ . Once you see that, the exponential equation turns into a simpler one between the exponents.

Solution to the example

Write $9=3^2$ and $27=3^3$ . To raise a power to a power, multiply the exponents. Multiply by the whole outside exponent, not only its first term: $2(x+1)=2x+2$ and $3(2x-1)=6x-3$ .

\begin{aligned} (3^2)^{x+1}&=(3^3)^{2x-1},\\[1.4em] 3^{2x+2}&=3^{6x-3}. \end{aligned}

Now both sides are powers of $3$ , and they can only be equal if their exponents are equal. So $2x+2=6x-3$ . That gives $5=4x$ , so $x=\frac54$ . The answer is C.

SAT example

Which choice is the solution to

9^{x+1}=27^{2x-1}?

A
$\frac14$
B
$\frac34$
C
$\frac54$
D
$\frac74$

Example: Use a graph when there could be several solutions

A clean common base usually makes the algebra shorter, but it won’t always be enough. In the equation below, $2^{2x}$ is really $(2^x)^2$ , so the same power shows up twice. An equation like that can have more than one solution, and this question asks you to pick one. A graph shows you all of them before you choose.

Worked example

The equation

2^{2x}-9(2^x)+8=0

has two real solutions. What is the positive solution?

You can graph it as written, with no rearranging:

Enter y=2^(2x)-9(2^x)+8.
Enter y=0.
Click every point where the graph meets the $x$ -axis.
Only now look at the condition in the question. If you stop at the first point you find, you might grab one that doesn’t fit.

The graph meets the axis at $x=0$ and $x=3$ . The question wants the positive one, and $0$ isn’t positive, so the answer is

\boxed{3}.

So which method should you reach for? Look at the equation before you do any algebra:

Use common-base algebra when every base is a familiar power of one number, like $9$ and $27$ in the SAT example, and the question wants an exact value.
Graph first when a repeated or buried power could give more than one solution, like $2^{2x}$ next to $2^x$ . Graph first, too, when the power is messy to get alone, or when the numbers don’t share an obvious base, like $5$ and $7$ .

One catch with graphing: if the graph gives a rounded decimal, like $0.667$ , but the question wants an exact value, find or check the exact form, like $\frac23$ , before you submit.

Related: For more calculator practice, see Solve one-variable equations.

Calculator loads as you approach

Click both x-intercepts, then keep the positive one.

Use common-base algebra when the numbers are familiar powers

Which numbers should make you think "common base"? Watch for these:

If you see	Rewrite with base
$8,\ 16,\ 32,\ 64$	$2$
$9,\ 27,\ 81$	$3$
$25,\ 125$	$5$

Then it’s the same three moves every time:

\boxed{\text{isolate} \ \longrightarrow\ \text{rewrite with one base} \ \longrightarrow\ \text{equate exponents}.}

Isolate means get the power alone first. Often it isn’t, as in $5\left(16^{x-1}\right)+3=323$ . Treat the whole power as one block and undo what’s around it in reverse order: the $+3$ was the last thing done to the power, so it comes off first, and then the $5$ .

Equate exponents means set them equal. It works because equal powers of the same positive base have equal exponents. The base can’t be $1$ , though: $1^3$ and $1^5$ are both $1$ , but $3$ isn’t $5$ .

Common mistake:

It’s tempting to set the exponents of $9^{x+1}=27^{2x-1}$ equal right away. That gives $x+1=2x-1$ , so $x=2$ . But $9^3=729$ and $27^3=19683$ , so $x=2$ doesn’t work. Rewrite every base first, and set the exponents equal only once both sides have exactly the same positive base.

Example: Isolate, rewrite, and equate

Worked example

The equation

5\left(16^{x-1}\right)+3=323

is true. What is the value of $x$ ?

Step 1

Isolate the whole power

Subtract $3$ first, then divide by $5$ :

\begin{aligned} 5\left(16^{x-1}\right)&=320,\\[1.4em] 16^{x-1}&=64. \end{aligned}

Step 2

Rewrite both sides with base 2

Both numbers are powers of $2$ : $16=2^4$ and $64=2^6$ . Multiply the exponents, and let the $4$ multiply all of $x-1$ :

\begin{aligned} (2^4)^{x-1}&=2^6,\\[1.4em] 2^{4(x-1)}&=2^6,\\[1.4em] 2^{4x-4}&=2^6. \end{aligned}

Step 3

Set the exponents equal and solve

The bases match, so the exponents must too:

4x-4=6.

Add $4$ to get $4x=10$ , so

\boxed{x=\frac52}.

Step 4

Check the original equation

Put $x=\frac52$ back in. The exponent becomes $x-1=\frac32$ , and a $\frac32$ power means take the square root, then cube it:

16^{x-1}=16^{3/2}=(\sqrt{16})^3=64.

Then $5(64)+3=323$ , so the answer checks.

Why a repeated power can create two solutions

Back to $2^{2x}-9(2^x)+8=0$ . The graph found its two solutions quickly, but why are there two? Substitution shows you. Since

2^{2x}=(2^x)^2,

let $u=2^x$ stand for the repeated power. Keep one fact in mind: $2^x$ is always positive, so only a positive $u$ can lead to a real solution. Now the equation is a plain quadratic in $u$ , and it factors:

\begin{aligned} u^2-9u+8&=0,\\[1.4em] (u-1)(u-8)&=0. \end{aligned}

Both $1$ and $8$ are positive, so both can work. Switch back from $u$ to $2^x$ :

\begin{aligned} 2^x=1=2^0 &\Longrightarrow x=0,\\[1.4em] 2^x=8=2^3 &\Longrightarrow x=3. \end{aligned}

That’s the same pair the graph found, in more steps. Substitute when the exact structure matters, or when substitution is clearly shorter than entering the graph.

Common mistake:

Getting $u=8$ doesn’t mean $x=8$ , because $u$ stands for $2^x$ , not $x$ . Switch every positive value of $u$ back to $2^x$ and solve for $x$ . Then apply the condition in the question.

Practice problems

Isolate before matching bases

Practice problem

Which choice is the solution to

7\left(4^{x-1}\right)+2=450?

Calculator loads as you approach

Familiar powers, so common-base algebra is the quicker method here.

Multiply the whole outside exponent

Practice problem

The equation

81^{x+1}=27^{2x+3}

is true. What is the value of $x$ ?

Your answer

Calculator loads as you approach

Rewrite 81 and 27 with base 3 to keep the fraction exact.

Pick the greater solution

Practice problem

The equation

3^{2x}-10(3^x)+9=0

has two real solutions. What is the greater solution?

Your answer

Calculator loads as you approach

A repeated power, so graph it and find both intersections before you pick.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Pick your method from the equation before you do any algebra.
Familiar powers of one base and an exact answer point to common-base algebra.
A repeated or buried power that could give several solutions, a power that’s messy to isolate, or no visible common base points to a graph.
Get the whole power alone before you rewrite its base.
For a power of a power, multiply by the whole outside exponent: $(a^m)^n=a^{mn}$ .
Set the exponents equal only once both sides have the same positive base, and that base isn’t $1$ .
Find every solution before you apply a condition like positive, greater, or integer.
After a substitution, switch back to $x$ before you answer.

Related lesson

When a question asks you to build or interpret a model of repeated multiplicative change, like a population that grows by the same percent each year, see Build, identify, and interpret exponential models.

Next lesson

Solve linear and nonlinear systems

Combine equations to find values that satisfy a line and a curve or two nonlinear conditions.

Start next lesson

Practice

Practice this lesson

128 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice