Solve for x (1/27)^4=9^(2x)

${(\frac{1}{27})}^{4} = 9^{2 x}$

Step 1

Rewrite the equation as $9^{2 x} = {(\frac{1}{27})}^{4}$ .

$9^{2 x} = {(\frac{1}{27})}^{4}$

Step 2

Apply the product rule to $\frac{1}{27}$ .

$9^{2 x} = \frac{1^{4}}{27^{4}}$

Step 3

One to any power is one.

$9^{2 x} = \frac{1}{27^{4}}$

Step 4

Move $27^{4}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$9^{2 x} = 27^{- 4}$

Step 5

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

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

Step 6

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

$2 (2 x) = 3 \cdot - 4$

Step 7

Solve for $x$ .

Simplify.

Multiply $2$ by $2$ .

$4 \cdot x = 3 \cdot - 4$

Multiply $3$ by $- 4$ .

$4 \cdot x = - 12$

Divide each term in $(4 \cdot x) = - 12$ by $4$ and simplify.

Divide each term in $(4 \cdot x) = - 12$ by $4$ .

$\frac{(4 \cdot x)}{4} = \frac{- 12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 \cdot x}{4} = \frac{- 12}{4}$

Divide $x$ by $1$ .

$x = \frac{- 12}{4}$

Simplify the right side.

Divide $- 12$ by $4$ .

$x = - 3$