Precalculus Examples

${(\frac{1}{9})}^{2 x} = 81^{x + 2}$

Step 1

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

$\frac{1^{2 x}}{9^{2 x}} = 81^{x + 2}$

Step 2

One to any power is one.

$\frac{1}{9^{2 x}} = 81^{x + 2}$

Step 3

Move $9^{2 x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$9^{- 2 x} = 81^{x + 2}$

Step 4

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

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

Step 5

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

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

Step 6

Solve for $x$ .

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Step 6.1

Simplify $2 (x + 2)$ .

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Step 6.1.1

Apply the distributive property.

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

Step 6.1.2

Multiply $2$ by $2$ .

$- 2 x = 2 x + 4$

Step 6.2

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

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Step 6.2.1

Subtract $2 x$ from both sides of the equation.

$- 2 x - 2 x = 4$

Step 6.2.2

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

$- 4 x = 4$

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Cancel the common factor of $- 4$ .

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Step 6.3.2.1.1

Cancel the common factor.

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

Step 6.3.2.1.2

Divide $x$ by $1$ .

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

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Divide $4$ by $- 4$ .

$x = - 1$