Find the Exact Value ((3^5)^2)/(3^-2)

$\frac{{(3^{5})}^{2}}{3^{- 2}}$

Step 1

Move $3^{- 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

${(3^{5})}^{2} \cdot 3^{2}$

Step 2

Multiply the exponents in ${(3^{5})}^{2}$ .

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

$3^{5 \cdot 2} \cdot 3^{2}$

Multiply $5$ by $2$ .

$3^{10} \cdot 3^{2}$

Step 3

Multiply $3^{10}$ by $3^{2}$ by adding the exponents.

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

$3^{10 + 2}$

Add $10$ and $2$ .

$3^{12}$

Step 4

Raise $3$ to the power of $12$ .

$531441$