Solve for x (2^(x+1))^2=64

${(2^{x + 1})}^{2} = 64$

Step 1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2^{(x + 1) \cdot 2} = 64$

Step 2

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

$2^{(x + 1) \cdot 2} = 2^{6}$

Step 3

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

$(x + 1) \cdot 2 = 6$

Step 4

Solve for $x$ .

Divide each term in $((x + 1) \cdot 2) = 6$ by $2$ and simplify.

Divide each term in $((x + 1) \cdot 2) = 6$ by $2$ .

$\frac{((x + 1) \cdot 2)}{2} = \frac{6}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{(x + 1) \cdot 2}{2} = \frac{6}{2}$

Divide $x + 1$ by $1$ .

$x + 1 = \frac{6}{2}$

Simplify the right side.

Divide $6$ by $2$ .

$x + 1 = 3$

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

Subtract $1$ from both sides of the equation.

$x = 3 - 1$

Subtract $1$ from $3$ .

$x = 2$