Solve for x 9^(2x+4)*9^(2x)=1/81

$9^{2 x + 4} \cdot 9^{2 x} = \frac{1}{81}$

Step 1

Raise $81$ to the power of $1$ .

$9^{2 x + 4} \cdot 9^{2 x} = \frac{1}{81^{1}}$

Step 2

Move $81^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$9^{2 x + 4} \cdot 9^{2 x} = 81^{- 1}$

Step 3

Multiply $9^{2 x + 4}$ by $9^{2 x}$ by adding the exponents.

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

$9^{2 x + 4 + 2 x} = 81^{- 1}$

Add $2 x$ and $2 x$ .

$9^{4 x + 4} = 81^{- 1}$

Step 4

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

$9^{4 x + 4} = 9^{- 2}$

Step 5

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

$4 x + 4 = - 2$

Step 6

Solve for $x$ .

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

Subtract $4$ from both sides of the equation.

$4 x = - 2 - 4$

Subtract $4$ from $- 2$ .

$4 x = - 6$

Divide each term in $4 x = - 6$ by $4$ and simplify.

Divide each term in $4 x = - 6$ by $4$ .

$\frac{4 x}{4} = \frac{- 6}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 6}{4}$

Divide $x$ by $1$ .

$x = \frac{- 6}{4}$

Simplify the right side.

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

$x = \frac{2 (- 3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$x = \frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{- 3}{2}$

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3}{2}$

Decimal Form:

$x = - 1.5$

Mixed Number Form:

$x = - 1 \frac{1}{2}$

Solve for x 9^(2x+4)*9^(2x)=1/81

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