Solve for x 9^(-3x)*9^x=27

$9^{- 3 x} \cdot 9^{x} = 27$

Multiply $9^{- 3 x}$ by $9^{x}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$9^{- 3 x + x} = 27$

Add $- 3 x$ and $x$ .

$9^{- 2 x} = 27$

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

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

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

$2 (- 2 x) = 3$

Solve for $x$ .

Multiply $- 2$ by $2$ .

$- 4 x = 3$

Divide each term in $- 4 x = 3$ by $- 4$ and simplify.

Divide each term in $- 4 x = 3$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{3}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{3}{- 4}$

Divide $x$ by $1$ .

$x = \frac{3}{- 4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3}{4}$

Decimal Form:

$x = - 0.75$