Solve for b 8^(b-1)=2^(4b+4)

$8^{b - 1} = 2^{4 b + 4}$

Step 1

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

$2^{3 (b - 1)} = 2^{4 b + 4}$

Step 2

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

$3 (b - 1) = 4 b + 4$

Step 3

Solve for $b$ .

Simplify $3 (b - 1)$ .

Rewrite.

$0 + 0 + 3 (b - 1) = 4 b + 4$

Simplify by adding zeros.

$3 (b - 1) = 4 b + 4$

Apply the distributive property.

$3 b + 3 \cdot - 1 = 4 b + 4$

Multiply $3$ by $- 1$ .

$3 b - 3 = 4 b + 4$

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

Subtract $4 b$ from both sides of the equation.

$3 b - 3 - 4 b = 4$

Subtract $4 b$ from $3 b$ .

$- b - 3 = 4$

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

Add $3$ to both sides of the equation.

$- b = 4 + 3$

Add $4$ and $3$ .

$- b = 7$

Divide each term in $- b = 7$ by $- 1$ and simplify.

Divide each term in $- b = 7$ by $- 1$ .

$\frac{- b}{- 1} = \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{b}{1} = \frac{7}{- 1}$

Divide $b$ by $1$ .

$b = \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$b = - 7$