Simplify ((7^5)(7^3))^-4

${((7^{5}) (7^{3}))}^{- 4}$

Step 1

Multiply $7^{5}$ by $7^{3}$ by adding the exponents.

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

${(7^{5 + 3})}^{- 4}$

Add $5$ and $3$ .

${(7^{8})}^{- 4}$

Step 2

Multiply the exponents in ${(7^{8})}^{- 4}$ .

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

$7^{8 \cdot - 4}$

Multiply $8$ by $- 4$ .

$7^{- 32}$

Step 3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{7^{32}}$

Step 4

Raise $7$ to the power of $32$ .

$\frac{1}{1104427674243920000000000000}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{1104427674243920000000000000}$

Decimal Form:

$9.05446344 \cdot 10^{- 28}$