Solve for x 9^(2x-8)=27^(x-4)

$9^{2 x - 8} = 27^{x - 4}$

Step 1

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

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

Step 2

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

$2 (2 x - 8) = 3 (x - 4)$

Step 3

Solve for $x$ .

Simplify $2 (2 x - 8)$ .

Rewrite.

$0 + 0 + 2 (2 x - 8) = 3 (x - 4)$

Simplify by adding zeros.

$2 (2 x - 8) = 3 (x - 4)$

Apply the distributive property.

$2 (2 x) + 2 \cdot - 8 = 3 (x - 4)$

Multiply.

Multiply $2$ by $2$ .

$4 x + 2 \cdot - 8 = 3 (x - 4)$

Multiply $2$ by $- 8$ .

$4 x - 16 = 3 (x - 4)$

Simplify $3 (x - 4)$ .

Apply the distributive property.

$4 x - 16 = 3 x + 3 \cdot - 4$

Multiply $3$ by $- 4$ .

$4 x - 16 = 3 x - 12$

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

Subtract $3 x$ from both sides of the equation.

$4 x - 16 - 3 x = - 12$

Subtract $3 x$ from $4 x$ .

$x - 16 = - 12$

Move all terms not containing $x$ to the right side of the equation.

Add $16$ to both sides of the equation.

$x = - 12 + 16$

Add $- 12$ and $16$ .

$x = 4$