Question

Simplify each of the following , expressing you answer in positive index from ( with step-by-step explanation )

(a) a^5 ÷ a^-2

(b) b^4 ÷ √b × b^-7

(c)

Answer 1

To Evalute:

$\displaystyle \rm 1) \frac{{a}^{5} }{ {a}^{ - 2} }$

$\displaystyle \rm2) \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7}$

$\displaystyle \rm 3) (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5}$

Solution :

$\displaystyle \rm 1) \frac{{a}^{5} }{ {a}^{ - 2} }$

Now, we will use the Identity:

$\displaystyle \rm \bull \frac{ {x}^{y} }{ {x}^{z} } = {x}^{y - z}$

So, We get that :

$\displaystyle \rm \implies \frac{ {a}^{5} }{ {a}^{ - 2} } = {a}^{5- ( - 2)}$

$\implies \displaystyle \rm \frac{ {a}^{ 5} }{ {a}^{ - 2} } = {a}^{5 + 3}$

$\red{ \underline{ \boxed{ \implies \displaystyle \rm \frac{ {a}^{ 5} }{ {a}^{ - 2} } = {a}^{8} }}}$

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$\displaystyle \rm2) \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7}$

Using The Identities:

$\bull \displaystyle \rm \sqrt{x} = {x}^{ \frac{1}{2} }$

$\bull \displaystyle \rm \frac{ {x}^{y} }{ {x}^{z} } = {x}^{y - z}$

$\bull \displaystyle \rm {x}^{y } \times {x}^{z} = {x}^{y + z}$

So, we get that :

$\displaystyle \rm \implies \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7} = \frac{ {b}^{4} }{ {b}^{ \frac{1}{2} } } \times {b}^{ - 7}$

$\displaystyle \rm \implies \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7} = {b}^{(4 - \frac{1}{2}) } \times {b}^{ - 7}$

$\displaystyle \rm \implies \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7} = {b}^{ \frac{7}{2} } \times {b}^{ - 7}$

$\displaystyle \rm \implies \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7} = {b}^{ \frac{7}{2} + ( - 7) }$

$\displaystyle \rm \implies \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7} = {b}^{ \frac{ - 7}{2}}$

$\blue{ \underline{ \boxed{\displaystyle \rm \implies \frac{ {b}^{4} }{ \sqrt{b} } \times {b}^{ - 7} = {( \frac{1}{b}) }^{( \frac{7}{2} )} }}}$

You can further write it as :

$\displaystyle \rm \implies \sqrt{ {( \frac{1}{b} )}^{7} }$

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$\displaystyle \rm 3) (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5}$

All the Identities here are already used :

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = ( \frac{ {c}^{2} }{ {c}^{2} } \times { \frac{d}{ {d}^{ - 2} } )}^{ - 5}$

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = ( \cancel{ \frac{ {c}^{2} }{ {c}^{2} } }\times { {d}^{ 1 - ( - 2)} )}^{ - 5}$

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = (1\times { {d}^{ 1 + 2} )}^{ - 5}$

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = (1\times { {d}^{ 3} )}^{ - 5}$

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = ( { {d}^{ 3} )}^{ - 5}$

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = {d}^{ 3 \times ( - 5)}$

$\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = {d}^{ - 15}$

$\green{ \underline{ \boxed{\displaystyle \rm \implies (\frac{ {c}^{2}d }{ {c}^{2} {d}^{ - 2} } {)}^{ - 5} = { (\frac{1}{d}) }^{15} }}}$

Answer 2

1. a⁸

2. b⁷/²

3. 1/d¹⁵