Question

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Q32. If $\sf{\dfrac{2^{m+n}}{2^{m-n}}=16}$ and $\sf{a=2^{\tiny\dfrac{1}{10}}}$ then $\sf{\dfrac{a^{2m+n-p}}{(a^{m-2n+2p})^{-1}}=}$ _____.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a = {\bigg(2 \bigg) }^{\dfrac{1}{10} }$

$\rm :\longmapsto\:\sf{\dfrac{2^{m+n}}{2^{m-n}}=16}$

We know that

$\underbrace{ \boxed{ \bf \: {x}^{m} \div {x}^{n} = {x}^{m - n}}}$

So, using this identity, we get

$\rm :\longmapsto\: {2}^{m + n - (m - n)} = 16$

$\rm :\longmapsto\: {2}^{m + n -m + n} = {2}^{4}$

$\rm :\longmapsto\: {2}^{2n} = {2}^{4}$

We know,

$\bf :\longmapsto\: {x}^{m} = {x}^{n} \implies \: m = n$

So, using this we get

$\rm :\longmapsto\:2n = 4$

$\bf\implies \:n = 2$

Now, Consider

$\rm :\longmapsto\:{\dfrac{a^{2m+n-p}}{(a^{m-2n+2p})^{-1}}}$

$\rm \: = \:{\dfrac{a^{2m+n-p}}{(a^{ - m + 2n - 2p})}}$

$\rm \: = \: {a}^{2m + n - p - ( - m + 2n - 2p)}$

$\rm \: = \: {a}^{2m + n - p + m - 2n + 2p}$

$\rm \: = \: {a}^{3m - n + p }$

$\rm \: = \: {a}^{3m - 2 + p }$

$\rm \: = \: { { \bigg( \bigg(2} \bigg)^{ \dfrac{1}{10} } \bigg)}^{3m - 2 + p }$

$\rm \: = \: {\bigg(2\bigg) }^{ \dfrac{3m - 2 + p}{10} }$

Hence,

$\boxed{\bf \: {\dfrac{a^{2m+n-p}}{(a^{m-2n+2p})^{-1}}=} \: {\bigg(2\bigg) }^{ \dfrac{3m + p - 2}{10} } }$

Answer 2

CORRECT QUESTION:

Q32. If $\sf{\dfrac{2^{m+n}}{2^{n-m}}=16}$ and $\sf{a=2^{\tiny\dfrac{1}{10}}}$ then $\sf{\dfrac{(a^{2m+n-p})^2}{(a^{m-2n+2p})^{-1}}=}$ _____.

ANSWER:

Given:

$\:\:\bullet\:\:\:\sf\dfrac{2^{m+n}}{2^{n-m}}=16$

$\:\:\bullet\:\:\:\sf a=2^{\frac{1}{10}}$

To Find:

• Value of:

$\sf\dfrac{(a^{2m+n-p})^2}{(a^{m-2n+2p})^{-1}}$

Solution:

We are given that,

$\implies\sf\dfrac{2^{m+n}}{2^{n-m}}=16$

We know that,

$\hookrightarrow \dfrac{x^q}{x^r}=x^{q-r}$

So,

$\implies\sf\dfrac{2^{m+n}}{2^{n-m}}=16$

$\implies\sf2^{m+n\!\!\!/-n\!\!\!/+m}=16$

$\implies\sf2^{2m}=2^4$

Comparing the powers,

$\implies\sf2\!\!\!/m=4\!\!\!/$

So,

$\implies\sf m=2$

Now, we need to find the value of,

$\sf\implies \dfrac{(a^{2m+n-p})^2}{(a^{m-2n+2p})^{-1}}$

We know that,

$\hookrightarrow \dfrac{1}{x^{-1}}=x^1$

So,

$\sf\implies\dfrac{(a^{2m+n-p})^2}{(a^{m-2n+2p})^{-1}}$

$\sf\implies(a^{2m+n-p})^2\times(a^{m-2n+2p})$

We know that,

$\hookrightarrow (x^q)^r=x^{qr}$

So,

$\implies\sf(a^{2m+n-p})^2\times(a^{m-2n+2p})$

$\implies\sf(a^{2(2m+n-p)})\times(a^{m-2n+2p})$

$\implies\sf(a^{4m+2n-2p})\times(a^{m-2n+2p})$

$\hookrightarrow (x^q)(x^r)=x^{q+r}$

$\implies\sf(a^{4m+2n-2p})\times(a^{m-2n+2p})$

$\implies\sf a^{4m+2n-2p+m-2n+2p}$

$\implies\sf a^{4m+m+2n-2n-2p+2p}$

So,

$\implies\sf a^{5m}$

But, we are also given that,

$\implies\sf a=2^{\frac{1}{10}}$

So,

$\implies\sf a^{5m}$

$\implies\sf (2^{\frac{1}{10}})^{5m}$

$\implies\sf 2^{\frac{5m}{10}}$

$\implies\sf 2^{\frac{m}{2}}$

Substituting the value of m, as 2,

We get,

$\implies\sf 2^{\frac{2}{2}}$

$\implies\sf 2^1$

$\implies\bf 2$

HENCE,

$\bf\implies\dfrac{(a^{2m+n-p})^2}{(a^{m-2n+2p})^{-1}}=2$