Solve for x (1/4)^(3-2x)=8^(-x)

${(\frac{1}{4})}^{3 - 2 x} = 8^{- x}$

Step 1

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

$\frac{1^{3 - 2 x}}{4^{3 - 2 x}} = 8^{- x}$

Step 2

One to any power is one.

$\frac{1}{4^{3 - 2 x}} = 8^{- x}$

Step 3

Move $4^{3 - 2 x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$4^{- (3 - 2 x)} = 8^{- x}$

Step 4

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

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

Step 5

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

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

Step 6

Solve for $x$ .

Simplify $2 (- (3 - 2 x))$ .

Start simplifying.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $- 1$ by $3$ .

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

Multiply $- 2$ by $- 1$ .

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

Apply the distributive property.

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

Multiply.

Multiply $2$ by $- 3$ .

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

Multiply $2$ by $2$ .

$- 6 + 4 x = 3 (- x)$

Multiply $- 1$ by $3$ .

$- 6 + 4 x = - 3 x$

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

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

$- 6 + 4 x + 3 x = 0$

Add $4 x$ and $3 x$ .

$- 6 + 7 x = 0$

Add $6$ to both sides of the equation.

$7 x = 6$

Divide each term in $7 x = 6$ by $7$ and simplify.

Divide each term in $7 x = 6$ by $7$ .

$\frac{7 x}{7} = \frac{6}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{6}{7}$

Divide $x$ by $1$ .

$x = \frac{6}{7}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{6}{7}$

Decimal Form:

$x = 0.\overline{857142}$

Solve for x (1/4)^(3-2x)=8^(-x)

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