If 8^{2x - 1} = \left(\frac{1}{4}\right)^{x - 3}, what is the value of x? — ACT

If $8^{2x - 1} = \left(\frac{1}{4}\right)^{x - 3}$ , what is the value of $x$ ?

The correct answer is

\frac{9}{8}

By writing both sides of the equation with a common base of

2

, we get

(2^3)^{2x-1} = (2^{-2})^{x-3}

. Applying the power-to-a-power exponent rule, this simplifies to

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

. Since the bases are now identical, their exponents must be equal:

6x-3 = -2x+6

. Adding

2x

and

3

to both sides results in

8x = 9

, which gives

x = \frac{9}{8}

Express both bases as powers of

2

8 = 2^3

and

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

, so the equation becomes

(2^3)^{2x-1} = (2^{-2})^{x-3}

Finding a common base allows us to equate the exponents directly.

Apply the power-to-a-power rule

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

to simplify the exponents.

2^{3(2x-1)} = 2^{-2(x-3)} \Rightarrow 2^{6x-3} = 2^{-2x+6}

This simplifies the exponential expressions on both sides of the equation.

Equate the exponents and solve for

x

6x - 3 = -2x + 6 \Rightarrow 8x = 9 \Rightarrow x = \frac{9}{8}

Since the bases are equal, their exponents must be equal.

Solving exponential equations by finding a common base and applying exponent properties.

Estimated Time:1m 30s

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