Solve for x 3^(x+2)-3^(x+1)=18

$3^{x + 2} - 3^{x + 1} = 18$

Factor out $3^{x + 1}$ from the expression.

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

Subtract $1$ from $3$ .

$3^{x + 1} \cdot 2 = 18$

Move $2$ to the left of $3^{x + 1}$ .

$2 \cdot 3^{x + 1} = 18$

Divide each term in $(2 \cdot 3^{x + 1}) = 18$ by $2$ and simplify.

Divide each term in $(2 \cdot 3^{x + 1}) = 18$ by $2$ .

$\frac{(2 \cdot 3^{x + 1})}{2} = \frac{18}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \cdot 3^{x + 1}}{2} = \frac{18}{2}$

Divide $3^{x + 1}$ by $1$ .

$3^{x + 1} = \frac{18}{2}$

Simplify the right side.

Divide $18$ by $2$ .

$3^{x + 1} = 9$

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

$3^{x + 1} = 3^{2}$

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

$x + 1 = 2$

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

Subtract $1$ from both sides of the equation.

$x = 2 - 1$

Subtract $1$ from $2$ .

$x = 1$