How do you express $\frac{- 16 a^{- 3} b^{- 5}}{2 a^{- 4} b^{2}}$ $(-16a^ -3 b^ -5)/(2a^ -4 b^2)$ with positive exponents?

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How do you express $\frac{- 16 a^{- 3} b^{- 5}}{2 a^{- 4} b^{2}}$ $(-16a^ -3 b^ -5)/(2a^ -4 b^2)$ with positive exponents?

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Answer 1 · 2018-06-23T13:28:01

See a solution process below:

Explanation:

First, rewrite the expression as:

$\frac{- 16}{2} (\frac{a^{- 3}}{a^{- 4}}) (\frac{b^{- 5}}{b^{2}}) \Rightarrow - 8 (\frac{a^{- 3}}{a^{- 4}}) (\frac{b^{- 5}}{b^{2}})$ $(-16)/2(a^-3/a^-4)(b^-5/b^2) => -8(a^-3/a^-4)(b^-5/b^2)$

Next, use these rules for exponents to simplify the $a$ $a$ term:

$\frac{x^{a}}{x^{b}} = x^{a - b}$ $x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))$ and $a^{1} = a$ $a^color(red)(1) = a$

$- 8 (\frac{a^{- 3}}{a^{- 4}}) (\frac{b^{- 5}}{b^{2}}) \Rightarrow$ $-8(a^color(red)(-3)/a^color(blue)(-4))(b^-5/b^2) =>$

$- 8 a^{- 3 - - 4} (\frac{b^{- 5}}{b^{2}}) \Rightarrow$ $-8a^(color(red)(-3)-color(blue)(-4))(b^-5/b^2) =>$

$- 8 a^{- 3 + 4} (\frac{b^{- 5}}{b^{2}}) \Rightarrow$ $-8a^(color(red)(-3)+color(blue)(4))(b^-5/b^2) =>$

$- 8 a^{1} (\frac{b^{- 5}}{b^{2}}) \Rightarrow$ $-8a^color(red)(1)(b^-5/b^2) =>$

$- 8 a (\frac{b^{- 5}}{b^{2}})$ $-8a(b^-5/b^2)$

Now, use this rule for exponents to simplify the $b$ $b$ term:

$- 8 a (\frac{b^{- 5}}{b^{2}}) \Rightarrow$ $-8a(b^color(red)(-5)/b^color(blue)(2)) =>$

$- 8 a (\frac{1}{b^{2 - - 5}}) \Rightarrow$ $-8a(1/b^(color(blue)(2)-color(red)(-5))) =>$

$- 8 a (\frac{1}{b^{2 + 5}}) \Rightarrow$ $-8a(1/b^(color(blue)(2)+color(red)(5))) =>$

$- 8 a (\frac{1}{b^{7}}) \Rightarrow$ $-8a(1/b^7) =>$

$\frac{- 8 a}{b^{7}}$ $(-8a)/b^7$

Answer 2 · 2018-06-23T13:30:37

$- \frac{8 a}{b^{7}}$ $-(8a)/(b^7)$

Explanation:

$\frac{- 16 a^{- 3} b^{- 5}}{2 a^{- 4} b^{2}}$ $(-16a^ -3 b^ -5)/(2a^ -4 b^2)$

Use the rule for exponents: $a^{- n} = \frac{1}{a^{n}}$ $a^-n=1/a^n$

and the rule $\frac{a^{n}}{a^{m}} = a^{n - m}$ $a^n/a^m=a^(n-m)$

$\frac{- 16 a^{- 3} b^{- 5}}{2 a^{- 4} b^{2}}$ $(-16a^ -3 b^ -5)/(2a^ -4 b^2)$

first let's simplify:

$\frac{- 16 \cdot a^{- 3} \cdot b^{- 5}}{2 \cdot a^{- 4} \cdot b^{2}}$ $(-16*a^ -3*b^ -5)/(2*a^ -4*b^2)$

$- 8 \cdot a^{- 3 - (- 4)} \cdot b^{- 5 - 2}$ $-8*a^(-3-(-4))*b^(-5-2)$

$- 8 \cdot a^{- 3 + 4} \cdot b^{- 7}$ $-8*a^(-3+4)*b^(-7)$

$- 8 \cdot a \cdot b^{- 7}$ $-8*a*b^(-7)$

Now move the negative exponents:

$- \frac{8 a}{b^{7}}$ $-(8a)/(b^7)$

Answer 3 · 2018-06-23T13:33:10

$\frac{- 16 a^{- 3} b^{- 5}}{2 a^{- 4} b^{2}}$ $(-16a^-3b^-5)/(2a^-4b^2)$

Group the like terms.

$= \frac{- 16}{2} \cdot \frac{a^{- 3}}{a^{- 4}} \cdot \frac{b^{- 5}}{b^{2}}$ $=frac(-16)(2)*frac(a^-3)(a^-4)*frac(b^-5)(b^2)$

Use the rule $\frac{x^{p}}{x^{q}} = x^{p - q}$ $frac(x^p)(x^q)=x^(p-q)$

$= - 8 \cdot a^{- 3 + 4} \cdot b^{- 5 - 2}$ $=-8*a^(-3+4)*b^(-5-2)$

Simplify the exponents.

$= - 8 \cdot a \cdot b^{- 7}$ $=-8*a*b^-7$

Use the rule $x^{- n} = \frac{1}{x^{n}}$ $x^-n=frac(1)(x^n)$ to write it with positive exponents.

$= - 8 a \cdot \frac{1}{b^{7}}$ $=-8a*frac(1)(b^7)$

Simplify.

$= - \frac{8 a}{b^{7}}$ $=-frac(8a)(b^7)$

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