# How do you simplify (b(b^-5)b^-2)/(b(b^8))?

Nov 9, 2017

See a solution process below:

#### Explanation:

First, cancel common terms in the numerator and denominator:

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{b}}} \left({b}^{-} 5\right) {b}^{-} 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{b}}} \left({b}^{8}\right)} \implies \frac{{b}^{-} 5 {b}^{-} 2}{b} ^ 8$

Next, use this rule for exponents to simplify the numerator:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\frac{{b}^{\textcolor{red}{- 5}} {b}^{\textcolor{b l u e}{- 2}}}{b} ^ 8 \implies {b}^{\textcolor{red}{- 5} + \textcolor{b l u e}{- 2}} / {b}^{8} \implies {b}^{-} \frac{7}{b} ^ 8$

Now, use this rule for exponents to complete the simplification:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

${b}^{\textcolor{red}{- 7}} / {b}^{\textcolor{b l u e}{8}} \implies \frac{1}{b} ^ \left(\textcolor{b l u e}{8} - \textcolor{red}{- 7}\right) \implies \frac{1}{b} ^ \left(\textcolor{b l u e}{8} + \textcolor{red}{7}\right) \implies \frac{1}{b} ^ 15$