# Exponents problems – Grade 09- Set 04

Given below are collection of exponent questions with detailed solutions.

All questions are to the level of grade 09.

Question 01
Prove the below expression;

\mathtt{\left(\frac{x^{a}}{x^{b}}\right)^{c} \times \left(\frac{x^{b}}{x^{c}}\right)^{a} \times \left(\frac{x^{c}}{x^{a}}\right)^{b} =1}

Solution
Solving the left side of the equation;

\mathtt{\Longrightarrow \ \frac{x^{ac}}{x^{bc}} \times \frac{x^{ba}}{x^{ca}} \times \frac{x^{cb}}{x^{ab}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\cancel{x^{ac}}}{\cancel{x^{bc}}} \times \frac{\cancel{x^{ba}}}{\cancel{x^{ac}}} \times \frac{\cancel{x^{cb}}}{\cancel{x^{ab}}}}\\\ \\ \mathtt{\Longrightarrow \ 1}

Hence Proved.

Question 02
Prove the following expression;

\mathtt{\frac{a+b+c}{a^{-1} b^{-1} +b^{-1} c^{-1} +c^{-1} a^{-1}} =\ abc}

Solution
Solving the left side of the equation.

\mathtt{\Longrightarrow \ \frac{a+b+c}{\frac{1}{ab} +\frac{1}{bc} +\frac{1}{ca}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{a+b+c}{\frac{c+a+b}{abc}}}\\\ \\ \mathtt{\Longrightarrow ( a+b+c) \times \frac{abc}{a+b+c}}\\\ \\ \mathtt{\Longrightarrow \ abc}

Hence Proved.

Question 03
If \mathtt{\sqrt{2^{n}} =\ 1024} ;

then find the value of \mathtt{3^{2\left(\frac{n}{4} -4\right)}}

Solution
Let’s do the prime factorization of 1024

The number 1024 can be expressed as;

\mathtt{1024\ =\ 2^{10}}

Now solving the main equation;

\mathtt{\sqrt{2^{n}} =\ 1024}\\\ \\ \mathtt{\sqrt{2^{n}} \ =\ 2^{10}}\\\ \\ \mathtt{\left( 2^{n}\right)^{\frac{1}{2}} =2^{10}}\\\ \\ \mathtt{2^{\frac{n}{2}} =\ 2^{10}}

Since base are same, exponent will also be the same.

\mathtt{\frac{n}{2} =\ 10}\\\ \\ \mathtt{n\ =\ 20}

\mathtt{\Longrightarrow \ 3^{2\left(\frac{n}{4} -4\right)}}\\\ \\ \mathtt{\Longrightarrow \ 3^{2\left(\frac{20}{4} -4\right)}}\\\ \\ \mathtt{\Longrightarrow \ 3^{2( 5-4)}}\\\ \\ \mathtt{\Longrightarrow \ 3^{2}}\\\ \\ \mathtt{\Longrightarrow \ 9}

Hence, 9 is the value of given expression.

Question 04
If \mathtt{16^{2x+3} =64^{x+3}} ;

then find value of \mathtt{4^{2x-2}}

Solution
Solving the main equation for value of x

\mathtt{16^{2x+3} =64^{x+3}}\\\ \\ \mathtt{\left( 2^{4}\right)^{2x+3} =\left( 2^{6}\right)^{x+3}}\\\ \\ \mathtt{2^{8x+12} =2^{6x+\ 18}}

If base are same then exponent will also be the same.

\mathtt{8x+\ 12\ =\ 6x\ +\ 18}\\\ \\ \mathtt{8x-6x\ =\ 18-12}\\\ \\ \mathtt{2x\ =\ 6}\\\ \\ \mathtt{x\ =\ 3}

Now put the value of x in given equation.

\mathtt{\Longrightarrow \ 4^{2x-2}}\\\ \\ \mathtt{\Longrightarrow \ 4^{2( 3) -2}}\\\ \\ \mathtt{\Longrightarrow \ 4^{6-2}}\\\ \\ \mathtt{\Longrightarrow \ 4^{4}}\\\ \\ \mathtt{\Longrightarrow \ 256}

Hence, 256 is the value of given expression

Question 05
Find the value of below expression;

\mathtt{64^{\frac{-1}{3}}\left( 64^{\frac{1}{3}} -64^{\frac{2}{3}}\right)}

Solution
\mathtt{\Longrightarrow \ \left( 4^{3}\right)^{\frac{-1}{3}}\left(\left( 4^{3}\right)^{\frac{1}{3}} -\left( 4^{3}\right)^{\frac{2}{3}}\right)}\\\ \\ \mathtt{\Longrightarrow \ 4^{-1}\left( 4-4^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 4^{-1}( 4-16)}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{4} \times -12}\\\ \\ \mathtt{\Longrightarrow \ -3}

Hence, -3 is the solution of given expression.

Next chapter : Exponent grade 09 problems – set 05