Question

Chapter - Laws of Indices

Simplify :

$\bigg(\frac{9^{n+\frac{1}{4}}.\sqrt{3.3^{n}}}{3\sqrt{3^{-n}}}\bigg)^{\frac{1}{n}}$​

Answer 1

Answer:

Required value of $\bigg(\frac{9^{n+\frac{1}{4}}.\sqrt{3.3^{n}}}{3\sqrt{3^{-n}}}\bigg)^{\frac{1}{n}}$ is 27.

Step-by-step explanation:

$\implies \bigg(\dfrac{9^{n+\dfrac{1}{4}}.\sqrt{3.3^n}}{3\sqrt{3^{-n}}}\bigg)^{\dfrac{1}{n}}\\\\\\\implies \bigg( \dfrac{(3\times3)^{n+\dfrac{1}{4}}.\sqrt{3.3^n}}{3\sqrt{3^{-n}}} \bigg)^{\dfrac{1}{n}} \\\\\\\implies\bigg( \dfrac{(3^2)^{n+\dfrac{1}{4}}\times 3^{\dfrac{1}{2}}\times 3^{(n\times\dfrac{1}{2})}}{3\times3^{-n\times \dfrac{1}{2}}}\bigg)^{\dfrac{1}{n}}\\\\\\\implies \bigg( \dfrac{3^{2n +\dfrac{1}{2}}\times 3^{\dfrac{1}{2}}\times3^{\dfrac{n}{2}}}{3\times 3^{-\dfrac{n}{2}}}\bigg)^{\dfrac{1}{n}}$$\\\\\\\implies \bigg(\dfrac{3^{2n +\dfrac{1}{2} +\dfrac{1}{2}+\dfrac{n}{2}}}{3^{1-\dfrac{n}{2}}}\bigg)^{\dfrac{1}{n}}\\\\\\\implies \bigg(\dfrac{3^{\dfrac{4n+1+1+n}{2}}}{3^{\dfrac{2-n}{2}}}\bigg)^{\dfrac{1}{n}}\\\\\\\implies 3^{\bigg(\dfrac{5n+2}{2}-\dfrac{2-n}{2} \bigg)\dfrac{1}{n}}}\\\\\\\implies 3^{(3n)(1/n)}\\\\\\\implies 3^{3\times n\times\dfrac{1}{n}}\\\\\\\implies 3^3\\\\\implies 27$

Hence the required value of $\bigg(\frac{9^{n+\frac{1}{4}}.\sqrt{3.3^{n}}}{3\sqrt{3^{-n}}}\bigg)^{\frac{1}{n}}$ is 27.

Properties of indices( used in answer ) :  

• a^n / a^m = a^( n - m )
• a^n x a^m = a^( n + m )
• √a = a^( 1 / 2 )
• ( a^n )^m = a^mn

Answer 2

Answer:

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