Solve for b 256^(b+2)=4^(2-2b)

$256^{b + 2} = 4^{2 - 2 b}$

Step 1

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

$4^{4 (b + 2)} = 4^{2 - 2 b}$

Step 2

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

$4 (b + 2) = 2 - 2 b$

Step 3

Solve for $b$ .

Simplify $4 (b + 2)$ .

Rewrite.

$0 + 0 + 4 (b + 2) = 2 - 2 b$

Simplify by adding zeros.

$4 (b + 2) = 2 - 2 b$

Apply the distributive property.

$4 b + 4 \cdot 2 = 2 - 2 b$

Multiply $4$ by $2$ .

$4 b + 8 = 2 - 2 b$

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

Add $2 b$ to both sides of the equation.

$4 b + 8 + 2 b = 2$

Add $4 b$ and $2 b$ .

$6 b + 8 = 2$

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

Subtract $8$ from both sides of the equation.

$6 b = 2 - 8$

Subtract $8$ from $2$ .

$6 b = - 6$

Divide each term in $6 b = - 6$ by $6$ and simplify.

Divide each term in $6 b = - 6$ by $6$ .

$\frac{6 b}{6} = \frac{- 6}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 b}{6} = \frac{- 6}{6}$

Divide $b$ by $1$ .

$b = \frac{- 6}{6}$

Simplify the right side.

Divide $- 6$ by $6$ .

$b = - 1$