Solve for b (7^4)/(7^b)=7^-2

$\frac{7^{4}}{7^{b}} = 7^{- 2}$

Move $7^{b}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$7^{4} \cdot 7^{- b} = 7^{- 2}$

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

$7^{4 - b} = 7^{- 2}$

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

$4 - b = - 2$

Solve for $b$ .

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

Subtract $4$ from both sides of the equation.

$- b = - 2 - 4$

Subtract $4$ from $- 2$ .

$- b = - 6$

Divide each term in $- b = - 6$ by $- 1$ and simplify.

Divide each term in $- b = - 6$ by $- 1$ .

$\frac{- b}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{b}{1} = \frac{- 6}{- 1}$

Divide $b$ by $1$ .

$b = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$b = 6$