Solve for x 8^(2-x)=(1/4)^(2x)

$8^{2 - x} = {(\frac{1}{4})}^{2 x}$

Apply the product rule to $\frac{1}{4}$ .

$8^{2 - x} = \frac{1^{2 x}}{4^{2 x}}$

One to any power is one.

$8^{2 - x} = \frac{1}{4^{2 x}}$

Move $4^{2 x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$8^{2 - x} = 4^{- 2 x}$

Create equivalent expressions in the equation that all have equal bases.

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

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$3 (2 - x) = 2 (- 2 x)$

Solve for $x$ .

Simplify $3 (2 - x)$ .

Start simplifying.

$0 + 0 + 3 (2 - x) = 2 (- 2 x)$

Simplify by adding zeros.

$3 (2 - x) = 2 (- 2 x)$

Apply the distributive property.

$3 \cdot 2 + 3 (- x) = 2 (- 2 x)$

Multiply.

Multiply $3$ by $2$ .

$6 + 3 (- x) = 2 (- 2 x)$

Multiply $- 1$ by $3$ .

$6 - 3 x = 2 (- 2 x)$

Multiply $- 2$ by $2$ .

$6 - 3 x = - 4 x$

Move all terms containing $x$ to the left side of the equation.

Add $4 x$ to both sides of the equation.

$6 - 3 x + 4 x = 0$

Add $- 3 x$ and $4 x$ .

$6 + x = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$