Solve for x 4^(x+1)=8

$4^{x + 1} = 8$

Step 1

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

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

Step 2

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

$2 (x + 1) = 3$

Step 3

Solve for $x$ .

Divide each term in $2 (x + 1) = 3$ by $2$ and simplify.

Divide each term in $2 (x + 1) = 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x + 1$ by $1$ .

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

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

Subtract $1$ from both sides of the equation.

$x = \frac{3}{2} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{3}{2} - 1 \cdot \frac{2}{2}$

Combine $- 1$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$x = \frac{3 - 1 \cdot 2}{2}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $3$ .

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.5$