# Question b143b

Oct 3, 2017

$\frac{1}{5} ^ \left(n + 7\right)$

#### Explanation:

In order to simplify exponents, make sure that the bases are all the same. Change numbers to the product of their prime factors.

$\frac{{\textcolor{red}{25}}^{2 n - 4}}{{5}^{3 n + 1} \times {5}^{2 n - 3} \times 5} = \frac{{\textcolor{red}{\left({5}^{2}\right)}}^{2 n - 4}}{{5}^{3 n + 1} \times {5}^{2 n - 3} \times 5}$

$= \frac{{\textcolor{red}{5}}^{4 n - 8}}{{5}^{3 n + 1} \times {5}^{2 n - 3} \times {5}^{1}}$

Add the indices of like bases when multiplying

(5^(4n-8))/(5^(5n-1)

Method 1

Subtract the indices when dividing

$= \frac{1}{5} ^ \left(5 n - 4 n - 1 - \left(- 8\right)\right)$

$= \frac{1}{5} ^ \left(n + 7\right)$

Method 2

${x}^{m} = \frac{1}{{x}^{-} m}$

(5^(4n-8))/(5^(5n-1)#

$= \frac{1}{{5}^{5 n - 1} \times {5}^{- 4 n + 8}}$

$= \frac{1}{5} ^ \left(n + 7\right)$

Give the answer without negative or zero indices.