Given below are questions related to exponents. All the questions are to standard of grade 9 math.

**Question 01**

Solve the expression and find the value of x.

\mathtt{( i) \ 7^{2x+3} =\ 1}\\\ \\ \mathtt{( ii) \ 2^{5x+3} =8^{x+3}}\\\ \\ \mathtt{( iii) \ 4^{2x} =\frac{1}{32}}

**Solution**

\mathtt{( i) \ 7^{2x+3} =\ 1}\\\ \\ \mathtt{7^{2x+3} \ =\ 7^{0}}\\\ \\ \mathtt{\ 2x\ +\ 3\ =\ 0}\\\ \\ \mathtt{x\ =\ \frac{-3}{2}}

\mathtt{( ii) \ 2^{5x+3} =8^{x+3}}\\\ \\ \mathtt{2^{5x+3} =2^{3\ ( x+3)}}\\\ \\ \mathtt{5x+3\ =\ 3( x+3)}\\\ \\ \mathtt{5x+3\ =\ 3x\ +\ 9}\\\ \\ \mathtt{2x\ =\ 6}\\\ \\ \mathtt{x\ =\ 3}

\mathtt{( iii) \ 4^{2x} =\frac{1}{32}}\\\ \\ \mathtt{( 2)^{2.2x} =\ \frac{1}{2^{5}}}\\\ \\ \mathtt{( 2)^{4x} =( 2)^{-5}}

Base are same so exponent will also be same.

\mathtt{4x\ =\ -5}\\\ \\ \mathtt{x\ =\ \frac{-5}{4}}

**Question 02**

Solve the expression and find the value of a, b and c.

\mathtt{1620\ =\ 2^{a} \times 3^{b} \times 5^{c}}

**Solution**

Calculating prime factorization of number 1620 and finding all the factors.

The above factorization can be expressed as follows;

\mathtt{1620\ =\ 2^{2} \times 3^{4} \times 5^{1}}

Hence, the value of a, b and c is given as;

a = 2

b = 4

c = 1

**Question 03**

Solve the expression and find the value of a, b and c.

\mathtt{4725\ =3^{a} \times 5^{b} \times 7^{c}}

**Solution**

Do the prime factorization of number 4725 and write all the factors.

The above factorization can be expressed as follows;

\mathtt{4725\ =3^{3} \times 5^{2} \times 7^{1}}

Hence, the value of a, b & c is given as;

a = 3

b = 2

c=1

**Question 04**

If \mathtt{a^{x} =b^{y} =c^{z}} and \mathtt{b^{2} =\ ac} .

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

**Solution**

Let \mathtt{a^{x} =b^{y} =c^{z} =\ k} .

Then we can write;

\mathtt{a\ =\ k^{\frac{1}{x}}}\\\ \\ \mathtt{b\ =\ k^{\frac{1}{y}}}\\\ \\ \mathtt{c\ =\ k^{\frac{1}{z}}}

Putting these value in the expression;

\mathtt{b^{2} =\ ac}\\\ \\ \mathtt{\left( k^{\frac{1}{y}}\right)^{2} =k^{\frac{1}{x}} .k^{\frac{1}{z}}}\\\ \\ \mathtt{k^{\frac{2}{y}} =k^{\frac{1}{x} +\frac{1}{z}}}

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

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

Hence Proved.

**Question 05**

Find the value of x for below expression;

\mathtt{27^{x} =\ \frac{9}{3^{x}}}

**Solution**

\mathtt{\left( 3^{3}\right)^{x} =\frac{3^{2}}{3^{x}}}\\\ \\ \mathtt{3^{3x} .3^{x} =\ 3^{2}}\\\ \\ \mathtt{3^{3x+x} =3^{2}}\\\ \\ \mathtt{3^{4x} =3^{2}}

Base are same so exponent will also be the same.

\mathtt{4x\ =\ 2}\\\ \\ \mathtt{x\ =\ \frac{2}{4}}\\\ \\ \mathtt{x\ =\frac{1}{2}}

Hence, value of x is 1/2.

**Next chapter :** **exponent grade 09 problems – set 03**