Solve for n 81^(2n-3)=9^(-n)

$81^{2 n - 3} = 9^{- n}$

Step 1

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

$9^{2 (2 n - 3)} = 9^{- n}$

Step 2

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

$2 (2 n - 3) = - n$

Step 3

Solve for $n$ .

Simplify $2 (2 n - 3)$ .

Start simplifying.

$0 + 0 + 2 (2 n - 3) = - n$

Simplify by adding zeros.

$2 (2 n - 3) = - n$

Apply the distributive property.

$2 (2 n) + 2 \cdot - 3 = - n$

Multiply.

Multiply $2$ by $2$ .

$4 n + 2 \cdot - 3 = - n$

Multiply $2$ by $- 3$ .

$4 n - 6 = - n$

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

Add $n$ to both sides of the equation.

$4 n - 6 + n = 0$

Add $4 n$ and $n$ .

$5 n - 6 = 0$

Add $6$ to both sides of the equation.

$5 n = 6$

Divide each term in $5 n = 6$ by $5$ and simplify.

Divide each term in $5 n = 6$ by $5$ .

$\frac{5 n}{5} = \frac{6}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 n}{5} = \frac{6}{5}$

Divide $n$ by $1$ .

$n = \frac{6}{5}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$n = \frac{6}{5}$

Decimal Form:

$n = 1.2$

Mixed Number Form:

$n = 1 \frac{1}{5}$