Solving Equations with Variable Exponents | Grade 8 Math | LevelKite

You already know how to find the value of a power, an expression made of a base and an exponent. In $2^3$ , the base is $2$ and the exponent is $3$ , so $2^3 = 2 \cdot 2 \cdot 2 = 8$ .

Now flip the question. In an equation with a variable exponent, the exponent is a variable, a letter that stands for an unknown number, as in $2^x = 8$ . To solve the equation, you find the value of $x$ that makes it true.

Read $2^x = 8$ as a question: which exponent on $2$ gives $8$ ? You need three factors of $2$ (three $2$ s multiplied together) to make $8$ , so $x = 3$ .

In this lesson, the unknown exponent is always an integer: zero, a counting number such as $1$ , $2$ or $3$ , or the opposite of a counting number, such as $-1$ , $-2$ or $-3$ . The table below lists powers of $2$ whose values you already know how to find.

Powers of 2 for the integer exponents from 3 down to negative 3

Exponent	Power of 2	Value
$3$	$2^3 = 2 \cdot 2 \cdot 2$	$8$
$2$	$2^2 = 2 \cdot 2$	$4$
$1$	$2^1$	$2$
$0$	$2^0$	$1$
$-1$	$2^{-1} = \frac{1}{2^1}$	$\frac{1}{2}$
$-2$	$2^{-2} = \frac{1}{2^2}$	$\frac{1}{4}$
$-3$	$2^{-3} = \frac{1}{2^3}$	$\frac{1}{8}$

Each row of the table shows one power of $2$ . The rows for the exponent $0$ and for the negative exponents use rules you already know. Those rules need a nonzero base, and $2$ is not zero.

Read down the Value column. Each time the exponent goes down by $1$ , you divide the value by $2$ , so it gets smaller. Each time the exponent goes up by $1$ , you multiply the value by $2$ , so it gets larger. The pattern continues above and below the table. Because the values always move in one direction, no value ever appears twice.

So each value in this pattern comes from exactly one integer exponent, and an equation such as $2^x = \frac{1}{4}$ has exactly one solution. To find it, look for $\frac{1}{4}$ in the Value column. It is in the row for $-2$ and in no other row, so $x = -2$ .

Key concept

Same base, equal powers, equal exponents

Let $b$ be a base that is positive and not equal to $1$ , and let $x$ and $y$ be integers. Just as with base $2$ , going up one exponent multiplies the value by $b$ . Every power of $b$ is positive, so the value gets larger every time when $b > 1$ and smaller every time when $b$ is between $0$ and $1$ . Either way, different exponents give different values. So if two powers of the same base $b$ are equal, their exponents are equal.

To solve an equation with a variable exponent, write both sides as powers of the same base $b$ , then set the exponents equal.

\text{If } b^x = b^y \text{ with } b > 0 \text{ and } b \neq 1, \text{ then } x = y

Here is a plan for an equation such as $b^x = c$ , where $b$ is a known base, $c$ is a known number and $x$ is an integer.

Check the base. Make sure $b > 0$ and $b \neq 1$ .
Rewrite the other side. Write $c$ as a power of $b$ . When $c$ is $1$ or a fraction, use two facts you know: for a nonzero base $b$ and a positive integer $n$ , $b^0 = 1$ and $b^{-n} = \frac{1}{b^n}$ .
Set the exponents equal. The bases match and the powers are equal, so the exponents are equal.
Check. Put your value into the original equation.

The two facts in step 2 need a nonzero base, because a base of $0$ would give an undefined expression, one that has no value. For example, $0^{-1}$ would mean $\frac{1}{0}$ , and you cannot divide by $0$ .

Step 2 does not work for every number $c$ . No integer exponent on $2$ gives $5$ , because $2^2 = 4$ is too small, $2^3 = 8$ is too large, and no integer lies between $2$ and $3$ . Every equation in this lesson has an integer solution, so here you can always do step 2.

Worked example

A positive exponent

Solve $3^x = 81$ , where $x$ is an integer.

Check the base. The base is 3. It is positive and it is not 1, so equal powers of 3 must have equal exponents.
$3 > 0 \quad \text{and} \quad 3 \neq 1$
Write 81 as a power of 3. Multiply by 3 until you reach 81 and count the factors: 3 times 3 is 9, 9 times 3 is 27, and 27 times 3 is 81. That uses four factors of 3.
$81 = 3 \cdot 3 \cdot 3 \cdot 3 = 3^4$
Replace 81 with the power you found. The two are the same number, so the equation still says the same thing.
$3^x = 3^4$
Both sides are now powers of the same base, 3. Different exponents give different powers of 3, so the exponents must be equal.
$x = 4$
Check by putting 4 in place of the variable in the original equation.
$3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81$

Answer

The solution is $x = 4$ , because $3^4 = 81$ .

Worked example

When one side is 1

Solve $7^x = 1$ , where $x$ is an integer.

Write 1 as a power of 7. A nonzero base raised to the exponent 0 equals 1, and 7 is not zero.
$1 = 7^0$
Replace 1 with this power. It is the same number, so the equation still says the same thing.
$7^x = 7^0$
The bases match, and 7 is positive and not 1, so the exponents must be equal.
$x = 0$
Check by putting 0 in place of the variable in the original equation.
$7^0 = 1$

Answer

The solution is $x = 0$ , because $7^0 = 1$ .

Worked example

A fraction on one side

Solve $\frac{1}{64} = 4^n$ , where $n$ is an integer.

Start with the denominator, the bottom number of the fraction, which is 64. Write it as a power of 4: 4 times 4 is 16, and 16 times 4 is 64. That uses three factors of 4.
$64 = 4 \cdot 4 \cdot 4 = 4^3$
Rewrite the fraction. Because the base 4 is not zero, 1 divided by a power of 4 equals 4 raised to the opposite exponent.
$\frac{1}{64} = \frac{1}{4^3} = 4^{-3}$
Replace the fraction with this power. It is the same number, so the equation still says the same thing.
$4^{-3} = 4^n$
The bases match, and 4 is positive and not 1, so the exponents must be equal.
$n = -3$
Check by putting negative 3 in place of the variable in the original equation.
$4^{-3} = \frac{1}{4^3} = \frac{1}{64}$

Answer

The solution is $n = -3$ , because $4^{-3} = \frac{1}{64}$ .

An equation with a variable exponent asks one question: which exponent on this base gives this number? When the base is positive and not $1$ , different integer exponents always give different values, so once you find an exponent that works, it is the only solution. Write both sides as powers of that same base. When you see $1$ or a fraction, use $b^0 = 1$ and $b^{-n} = \frac{1}{b^n}$ , which hold for a nonzero base $b$ and a positive integer $n$ . Then set the exponents equal and check your answer in the original equation.