# Exponent problems – Grade 09 – Set 05

In this chapter we will solved exponent questions related to Grade 9 math.

All the questions are provided with detailed solutions.

Question 01
Solve the below expression and find value of x.

\mathtt{2^{2x} -2^{x+3} +2^{4} =0}

Solution
\mathtt{\left( 2^{x}\right)^{2} -2^{x} .2^{3} +16=0}\\\ \\ \mathtt{\left( 2^{x}\right)^{2} -2^{x} .8+16=0} \\\ \\

Let \mathtt{2^{x} =y}

Now the expression can be written as;

\mathtt{y^{2} -8y+\ 16=0}\\\ \\ \mathtt{y^{2} -2.4.y+( 4)^{2} =0}

Referring to following formula;

\mathtt{( a+b)^{2} =a^{2} +2ab+b^{2}}

\mathtt{( y-4)^{2} =0}\\\ \\ \mathtt{y-4\ =\ 0}\\\ \\ \mathtt{y=\ 4}

Putting the original value of y.

\mathtt{2^{x} =4}\\\ \\ \mathtt{2^{x} =2^{2}}

Hence, value of x = 2.

Question 02
Solve the below expression and find value of x.

\mathtt{3^{2x+4} +1=2.3^{x+2}}

Solution
\mathtt{3^{2x+4} +1=2.3^{x+2}}\\\ \\ \mathtt{3^{2x} .3^{4} +1=2.3^{x} .3^{2}}\\\ \\ \mathtt{81.3^{2x} +1-2.3^{2} .3^{x} =0}\\\ \\ \mathtt{9^{2} .3^{2x} -2.3^{2} .3^{x} +1\ =\ 0}

\mathtt{let\ 3^{x} =\ y}

Now the expression can be written as;

\mathtt{9^{2} y^{2} -2.9.y+1=\ 0}\\\ \\ \mathtt{( 9y)^{2} -2.9.y\ +1\ =0}

Referring to following formula;

\mathtt{( a-b)^{2} =a^{2} -2ab+b^{2}}

\mathtt{( 9y-1)^{2} =\ 0}\\\ \\ \mathtt{9y-1=0}\\\ \\ \mathtt{y\ =\ \frac{1}{9}}

Putting the original value of y.

\mathtt{3^{x} =\frac{1}{3^{2}}}\\\ \\ \mathtt{3^{x} =\ 3^{-2}}\\\ \\ \mathtt{x\ =\ -2}

Hence, value of x is -2.

Example 03
Simplify the below expression;
\mathtt{\sqrt{32^{-3}}}

Solution
\mathtt{\Longrightarrow \sqrt{\frac{1}{32^{3}}} \ }\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{32^{\frac{3}{5}}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{2^{5\times \frac{3}{5}}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{2^{3}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{8}}

Example 04
If \mathtt{3^{x} =5^{y} =75^{z}}

Show that \mathtt{z\ =\frac{xy}{2x+y}}

Solution
\mathtt{Let\ 3^{x} =5^{y} =75^{z} =k}

The above expression can be written as;

\mathtt{3=k^{\frac{1}{x}}}\\\ \\ \mathtt{5=k^{\frac{1}{y}}}\\\ \\ \mathtt{75=k^{\frac{1}{z}}}

Solving one of the above expression;

\mathtt{75=k^{\frac{1}{z}}}\\\ \\ \mathtt{5^{2} \times 3=k^{\frac{1}{z}}}\\\ \\ \mathtt{\left( k^{\frac{1}{y}}\right)^{2} \times k^{\frac{1}{x}} =k^{\frac{1}{z}}}\\\ \\ \mathtt{k^{\frac{2}{y}} \times k^{\frac{1}{x}} =k^{\frac{1}{z}}}\\\ \\ \mathtt{k^{\frac{2}{y} +\frac{1}{x}} =k^{\frac{1}{z}}}

Since base are same exponent will also be the same.

\mathtt{\frac{2}{y} +\frac{1}{x} =\frac{1}{z}}\\\ \\ \mathtt{\frac{2x+y}{xy} =\frac{1}{z}}\\\ \\ \mathtt{x\ =\ \frac{xy}{2x+y}}

Hence Proved.

Question 05
Solve the below expression and find value of x.

\mathtt{\left(\sqrt{\frac{3}{5}}\right)^{x+1} =\ \frac{125}{27}}

Solution
\mathtt{\left(\frac{3}{5}\right)^{\frac{x+1}{2}} =\frac{5^{3}}{3^{3}}}\\\ \\ \mathtt{\left(\frac{3}{5}\right)^{\frac{x+1}{2}} =\frac{3^{-3}}{5^{-3}}}

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

\mathtt{\frac{x+1}{2} =-3}\\\ \\ \mathtt{x+1=-6}\\\ \\ \mathtt{x\ =-7}

Hence, value of x is -7.