Solve for x 9^(2x)*9^(1-3x)=27^(x+2)

$9^{2 x} \cdot 9^{1 - 3 x} = 27^{x + 2}$

Multiply $9^{2 x}$ by $9^{1 - 3 x}$ by adding the exponents.

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

$9^{2 x + 1 - 3 x} = 27^{x + 2}$

Subtract $3 x$ from $2 x$ .

$9^{- x + 1} = 27^{x + 2}$

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

$3^{2 (- x + 1)} = 3^{3 (x + 2)}$

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

$2 (- x + 1) = 3 (x + 2)$

Solve for $x$ .

Simplify $2 (- x + 1)$ .

Start simplifying.

$0 + 0 + 2 (- x + 1) = 3 (x + 2)$

Simplify by adding zeros.

$2 (- x + 1) = 3 (x + 2)$

Apply the distributive property.

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

Multiply.

Multiply $- 1$ by $2$ .

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

Multiply $2$ by $1$ .

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

Simplify $3 (x + 2)$ .

Apply the distributive property.

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

Multiply $3$ by $2$ .

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

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

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

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

Subtract $3 x$ from $- 2 x$ .

$- 5 x + 2 = 6$

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

Subtract $2$ from both sides of the equation.

$- 5 x = 6 - 2$

Subtract $2$ from $6$ .

$- 5 x = 4$

Divide each term in $- 5 x = 4$ by $- 5$ and simplify.

Divide each term in $- 5 x = 4$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{4}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{4}{- 5}$

Divide $x$ by $1$ .

$x = \frac{4}{- 5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{4}{5}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{4}{5}$

Decimal Form:

$x = - 0.8$

Solve for x 9^(2x)*9^(1-3x)=27^(x+2)

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