Question

$\star \: {\red{\bold{Hello...! :}}}$

• Find the value of
$\large \bigg \{ \frac{ \big( {9}^{n + \frac{1}{4} } \big) \sqrt{3. {3}^{n} } }{3 \sqrt{ {3}^{ - n} } } \bigg \}^{ \frac{1}{n} }$​

Answer 1

Answer:

In LaTeX backslash is used to generate a special symbol or a command. ... 3√8=813=2, \sqrt[3]{8}=8^{\frac{1}{3}}=2. 2/34/

maths mode :

by H Voß · 2009 · Cited by 1 · Related articles

22-Jan-2009 — 3 \frac{1}{\sqrt{n}}= & \frac{\ sqrt{n}}{n}= & \frac{n. }{n\sqrt{n}}. 4 \end{ eqnarray*}.

Step-by-step explanation:

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Answer 2

To Find :-

• Value of $\sf \large \bigg \{ \frac{ \big( {9}^{n + \frac{1}{4} } \big) \sqrt{3. {3}^{n} } }{3 \sqrt{ {3}^{ - n} } } \bigg \}^{ \frac{1}{n} }$

Solution :-

$\sf \large \bigg \{ \frac{ \big( {9}^{n + \frac{1}{4} } \big) \sqrt{3. {3}^{n} } }{3 \sqrt{ {3}^{ - n} } } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ \frac{ \big( {3}^{2(n + \frac{1}{4}) } \big) \sqrt{ {3}^{n + 1} } }{3.3 ^{ (\frac{ - n}{2}) } } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ \frac{ \big( {3}^{(2n + \frac{1}{2}) } \big) {3}^{ \frac{n + 1}{2} } }{3 ^{ (\frac{ - n}{2} + 1 )} } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ {3}^{(2n + \frac{1}{2} + \frac{n + 1}{2} - (\frac{ - n}{2} + 1) } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ {3}^{(2n + \frac{n}{2} + \frac{n}{2} + \frac{1}{2} + \frac{1}{2} - 1 } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ {3}^{(2n +n + \cancel1- \cancel1 } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ {3}^{(3n) } \bigg \}^{ \frac{1}{n} }$

$= \sf \large \bigg \{ {3}^{( \frac{3n}{n} ) } \bigg \}$

$\sf \: = {3}^{3}$

$\sf \: = 27$

$\boxed {\sf {\purple {Required\ value\ is\ 27.}}}$

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