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

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

Step 1

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

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

Step 2

One to any power is one.

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

Step 3

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

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

Step 4

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

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

Step 5

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

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

Step 6

Solve for $x$ .

Simplify $3 (2 - x)$ .

Start simplifying.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $3$ by $2$ .

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

Multiply $- 1$ by $3$ .

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

Multiply $- 1$ by $2$ .

$6 - 3 x = - 2 x$

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

Add $2 x$ to both sides of the equation.

$6 - 3 x + 2 x = 0$

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

$6 - x = 0$

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

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

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