Solve for b 2^a*4^(3b)=128

$2^{a} \cdot 4^{3 b} = 128$

Step 1

Rewrite $4$ as $2^{2}$ .

$2^{a} \cdot {(2^{2})}^{3 b} = 128$

Step 2

Multiply the exponents in ${(2^{2})}^{3 b}$ .

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

$2^{a} \cdot 2^{2 (3 b)} = 128$

Multiply $3$ by $2$ .

$2^{a} \cdot 2^{6 b} = 128$

Step 3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{a + 6 b} = 128$

Step 4

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

$2^{a + 6 b} = 2^{7}$

Step 5

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

$a + 6 b = 7$

Step 6

Solve for $b$ .

Subtract $a$ from both sides of the equation.

$6 b = 7 - a$

Divide each term in $6 b = 7 - a$ by $6$ and simplify.

Divide each term in $6 b = 7 - a$ by $6$ .

$\frac{6 b}{6} = \frac{7}{6} + \frac{- a}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 b}{6} = \frac{7}{6} + \frac{- a}{6}$

Divide $b$ by $1$ .

$b = \frac{7}{6} + \frac{- a}{6}$

Simplify the right side.

Move the negative in front of the fraction.

$b = \frac{7}{6} - \frac{a}{6}$