Solve for x 2(3^(x+1))+4=22

$2 (3^{x + 1}) + 4 = 22$

Step 1

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

Subtract $4$ from both sides of the equation.

$2 \cdot 3^{x + 1} = 22 - 4$

Subtract $4$ from $22$ .

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

Step 2

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 $2$ .

$3^{x + 1} = 9$

Step 3

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

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

Step 4

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

$x + 1 = 2$

Step 5

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 $2$ .

$x = 1$