Solve for $x$ in the equation $(\frac{1}{4})^x = 8^{3x-2}$.

1. Problem Description

Solve for $x$ in the equation $(\frac{1}{4})^x = 8^{3x-2}$.

2. Solution Steps

The given equation is $(\frac{1}{4})^x = 8^{3x-2}$.

We can rewrite both sides of the equation using the base

2. $(\frac{1}{4})^x = (2^{-2})^x = 2^{-2x}$

$8^{3x-2} = (2^3)^{3x-2} = 2^{3(3x-2)} = 2^{9x-6}$

So we have $2^{-2x} = 2^{9x-6}$.

Since the bases are equal, we can equate the exponents:

$-2x = 9x - 6$

Add $2x$ to both sides:

$0 = 11x - 6$

Add 6 to both sides:

$6 = 11x$

Divide by 11:

$x = \frac{6}{11}$

3. Final Answer

$x = \frac{6}{11}$