# How do you simplify \frac { b ^ { 95} c ^ { - 4} } { b ^ { 58} c \cdot b ^ { - 8} c ^ { - 1} }?

Apr 17, 2017

${b}^{45} / {c}^{4}$

#### Explanation:

First of all simplify the numerator and the denominator by multiplying numbers with same base (i.e. $c$ with $c$ and $b$ with $b$). (Remember, when you are multiplying you are adding the exponents, and when you are dividing you are subtracting them).

$\frac{{b}^{95} {c}^{-} 4}{{b}^{58} \textcolor{m a \ge n t a}{c} {b}^{-} 8 {c}^{-} 1}$

This $\textcolor{m a \ge n t a}{c}$ is the same as ${c}^{1}$

$\frac{{b}^{95} {c}^{-} 4}{{b}^{\textcolor{red}{58}} {b}^{\textcolor{g r e e n}{- 8}} {c}^{-} 1 {c}^{\textcolor{b l u e}{1}}}$

$\frac{{b}^{95} {c}^{-} 4}{\left({b}^{\textcolor{red}{58} \textcolor{g r e e n}{- 8}}\right) \left({c}^{- 1 + \textcolor{b l u e}{1}}\right)}$

$= \frac{{b}^{95} {c}^{-} 4}{{b}^{50} {c}^{0}}$

Anything to the power of $0$ is one, so it can be removed

$= \frac{{b}^{\textcolor{red}{95}} {c}^{-} 4}{b} ^ \textcolor{red}{50}$

Subtract the smaller exponent from the same base

$= \frac{{b}^{\textcolor{red}{95} - 50} {c}^{-} 4}{b} ^ \left(\textcolor{red}{50} - 50\right)$

$= \frac{{b}^{45} {c}^{-} 4}{b} ^ 0$

$= {b}^{45} {c}^{-} 4$

The definition of negative exponents is ${a}^{-} m = \frac{1}{a} ^ m$ so,

${b}^{45} {c}^{-} 4 = {b}^{45} \cdot \frac{1}{c} ^ 4 = \textcolor{p u r p \le}{{b}^{45} / {c}^{4}}$

This is the final answer, note that when you simplify expressions with negative exponents you need to remove them and convert them to a positive exponents.