If 8^{x - 2} = \left(\frac{1}{4}\right)^{x - 7}, what is the value of x? — ACT

If $8^{x - 2} = \left(\frac{1}{4}\right)^{x - 7}$ , what is the value of $x$ ?

Answer: 4

The correct answer is

4

To solve

8^{x - 2} = \left(\frac{1}{4}\right)^{x - 7}

, express both bases as powers of

2

. Since

8 = 2^3

and

\frac{1}{4} = 2^{-2}

, the equation can be written as

(2^3)^{x-2} = (2^{-2})^{x-7}

. Using the power of a power property

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

, we simplify this to

2^{3x - 6} = 2^{-2x + 14}

. Because the bases are equal, their exponents must be equal:

3x - 6 = -2x + 14

. Adding

2x

and

6

to both sides yields

5x = 20

, which simplifies to

x = 4

Express both bases as powers of 2

8 = 2^3

and

\frac{1}{4} = 2^{-2}

To solve exponential equations with different bases, rewrite them using a common base.

Substitute the powers back into the equation and simplify exponents

2^{3(x-2)} = 2^{-2(x-7)}

Apply the exponent power rule

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

Equate the exponents since the bases are identical

3(x-2) = -2(x-7)

b^y = b^z

and

b > 0, b \neq 1

, then

y = z

Solve the linear equation for

x

3x - 6 = -2x + 14 \implies 5x = 20 \implies x = 4

Distribute, group like terms, and isolate the variable.

Solving exponential equations using a common base

Question