# Question #75cc2

Apr 14, 2017

$\frac{8}{r}$

#### Explanation:

$4 {r}^{-} 3 \cdot 2 {r}^{2} = \frac{8}{r}$

${a}^{-} 3 \cdot {a}^{4} = {a}^{- 3 + 4} = {a}^{1} = a$

${a}^{-} 1 = \frac{1}{a}$

$\therefore 4 {r}^{- 3 + 2} \cdot 2 = \frac{8}{r}$

$\therefore 4 {r}^{-} 1 \cdot 2 = \frac{8}{r}$

$\therefore 4 \cdot \frac{1}{r} \cdot 2 = \frac{8}{r}$

$\therefore \frac{8}{r} = \frac{8}{r}$

Apr 14, 2017

YOur working is correct until the very last step:

Multiply the numbers and add the indices of like bases.
Remember that the numbers and variables are independent of each other. You can think of it as:

$4 {r}^{-} 3 \times 2 {r}^{2}$

$= \left(4 \times 2\right) \times \left({r}^{-} 3 \times {r}^{2}\right)$

$= 8 \times \textcolor{b l u e}{{r}^{-} 1} \text{ } \leftarrow$ only the $r$ is raised to a negative index

$= \frac{8}{\textcolor{b l u e}{r}}$

There is a difference between ${\left(8 r\right)}^{-} 1 \mathmr{and} 8 {r}^{-} 1$

${\left(8 r\right)}^{-} 1 = \frac{1}{8 r} \mathmr{and} 8 {r}^{-} 1 = \frac{8}{r}$

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An alternative method is:

$4 \textcolor{red}{{r}^{-} 3} \times 2 {r}^{2}$

$= \frac{4}{\textcolor{red}{{r}^{3}}} \times 2 {r}^{2} = \frac{8 {r}^{2}}{r} ^ 3 = \frac{8}{r}$

Hope this helps to explain your problem.