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

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

Step 1

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

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

Step 2

One to any power is one.

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

Step 3

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

$4^{- x} = 8^{1 - x}$

Step 4

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

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

Step 5

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

$2 (- x) = 3 (1 - x)$

Step 6

Solve for $x$ .

Multiply $- 1$ by $2$ .

$- 2 x = 3 (1 - x)$

Simplify $3 (1 - x)$ .

Apply the distributive property.

$- 2 x = 3 \cdot 1 + 3 (- x)$

Multiply.

Multiply $3$ by $1$ .

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

Multiply $- 1$ by $3$ .

$- 2 x = 3 - 3 x$

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

Add $3 x$ to both sides of the equation.

$- 2 x + 3 x = 3$

Add $- 2 x$ and $3 x$ .

$x = 3$