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

$3^{x - 2} {(\frac{1}{9})}^{1 + x} = 9$

Apply the product rule to $\frac{1}{9}$ .

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

One to any power is one.

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

Move $9^{1 + x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

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

Rewrite $9$ as $3^{2}$ .

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

Multiply the exponents in ${(3^{2})}^{- (1 + x)}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Apply the distributive property.

$3^{x - 2} \cdot 3^{2 (- 1 \cdot 1 - x)} = 9$

Multiply $- 1$ by $1$ .

$3^{x - 2} \cdot 3^{2 (- 1 - x)} = 9$

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

$3^{x - 2} \cdot 3^{- 2 + 2 (- x)} = 9$

Multiply $- 1$ by $2$ .

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

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

$3^{x - 2 - 2 - 2 x} = 9$

Subtract $2 x$ from $x$ .

$3^{- x - 2 - 2} = 9$

Subtract $2$ from $- 2$ .

$3^{- x - 4} = 9$

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

$3^{- x - 4} = 3^{2}$

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

$- x - 4 = 2$

Solve for $x$ .

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

Add $4$ to both sides of the equation.

$- x = 2 + 4$

Add $2$ and $4$ .

$- x = 6$

Divide each term in $- x = 6$ by $- 1$ and simplify.

Divide each term in $- x = 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{6}{- 1}$

Simplify the right side.

Divide $6$ by $- 1$ .

$x = - 6$

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

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