# How do you simplify ((4a^2b)/(a^3b^2))((5a^2b)/(2b^4)) and write it using only positive exponents?

Jul 31, 2017

See a solution process below:

#### Explanation:

First, rewrite this expression as:

$\frac{4 \cdot 5}{2} \left(\frac{{a}^{2} \cdot {a}^{2}}{a} ^ 3\right) \left(\frac{b \cdot b}{{b}^{2} \cdot {b}^{4}}\right) \implies$

$\frac{20}{2} \left(\frac{{a}^{2} \cdot {a}^{2}}{a} ^ 3\right) \left(\frac{b \cdot b}{{b}^{2} \cdot {b}^{4}}\right) \implies$

$10 \left(\frac{{a}^{2} \cdot {a}^{2}}{a} ^ 3\right) \left(\frac{b \cdot b}{{b}^{2} \cdot {b}^{4}}\right)$

Next, use this rule of exponents to rewrite the numerator for the $b$ terms:

$a = {a}^{\textcolor{red}{1}}$

$10 \left(\frac{{a}^{2} \cdot {a}^{2}}{a} ^ 3\right) \left(\frac{{b}^{\textcolor{red}{1}} \cdot {b}^{\textcolor{red}{1}}}{{b}^{2} \cdot {b}^{4}}\right)$

Then, use this rule of exponents to multiply the numerators for both the $a$ and $b$ terms and the denominator for the $b$ term:

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

$10 \left(\frac{{a}^{\textcolor{red}{2}} \cdot {a}^{\textcolor{b l u e}{2}}}{a} ^ 3\right) \left(\frac{{b}^{\textcolor{red}{1}} \cdot {b}^{\textcolor{b l u e}{1}}}{{b}^{\textcolor{red}{2}} \cdot {b}^{\textcolor{b l u e}{4}}}\right) \implies$

$10 \left({a}^{\textcolor{red}{2} + \textcolor{b l u e}{2}} / {a}^{3}\right) \left({b}^{\textcolor{red}{1} + \textcolor{b l u e}{1}} / {b}^{\textcolor{red}{2} + \textcolor{b l u e}{4}}\right) \implies$

$10 \left({a}^{4} / {a}^{3}\right) \left({b}^{2} / {b}^{6}\right)$

Next, use these rules of exponents to complete the simplification of the $a$ term:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${a}^{\textcolor{red}{1}} = a$

$10 \left({a}^{\textcolor{red}{4}} / {a}^{\textcolor{b l u e}{3}}\right) \left({b}^{2} / {b}^{6}\right) \implies$

$10 {a}^{\textcolor{red}{4} - \textcolor{b l u e}{3}} \left({b}^{2} / {b}^{6}\right) \implies$

$10 {a}^{\textcolor{red}{1}} \left({b}^{2} / {b}^{6}\right) \implies$

$10 a \left({b}^{2} / {b}^{6}\right)$

Now, use this rule of exponents to complete the simplification for the $b$ term:

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

$10 a \left({b}^{\textcolor{red}{2}} / {b}^{\textcolor{b l u e}{6}}\right) \implies$

$10 a \left(\frac{1}{b} ^ \left(\textcolor{b l u e}{6} - \textcolor{red}{2}\right)\right) \implies$

$10 a \left(\frac{1}{b} ^ 4\right) \implies$

$\frac{10 a}{b} ^ 4$