Solve for x 4^(2x)=(1/2)^(x+5)

$4^{2 x} = {(\frac{1}{2})}^{x + 5}$

Apply the product rule to $\frac{1}{2}$ .

$4^{2 x} = \frac{1^{x + 5}}{2^{x + 5}}$

One to any power is one.

$4^{2 x} = \frac{1}{2^{x + 5}}$

Move $2^{x + 5}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$4^{2 x} = 2^{- (x + 5)}$

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

$2^{2 (2 x)} = 2^{- (x + 5)}$

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

$2 (2 x) = - (x + 5)$

Solve for $x$ .

Multiply $2$ by $2$ .

$4 x = - (x + 5)$

Simplify $- (x + 5)$ .

Apply the distributive property.

$4 x = - x - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$4 x = - x - 5$

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

Add $x$ to both sides of the equation.

$4 x + x = - 5$

Add $4 x$ and $x$ .

$5 x = - 5$

Divide each term in $5 x = - 5$ by $5$ and simplify.

Divide each term in $5 x = - 5$ by $5$ .

$\frac{5 x}{5} = \frac{- 5}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 5}{5}$

Divide $x$ by $1$ .

$x = \frac{- 5}{5}$

Simplify the right side.

Divide $- 5$ by $5$ .

$x = - 1$