SELINA Solution Class 9 Chapter 7 Indices (Exponents) Exercise 7C

Question 1

Evaluate : ${\displaystyle 9^{\frac{5}{2}}-3\times 8^0-{\left(\frac{1}{81}\right)}^{-\frac{1}{2}}}$

Sol:

${\displaystyle 9^{\frac{5}{2}}-3\times 8^0-{\left(\frac{1}{81}\right)}^{-\frac{1}{2}}}$

= ${\displaystyle {\left(3^2\right)}^{\frac{5}{2}}-3\times 1-{\left(\frac{1}{3^4}\right)}^{-\frac{1}{2}}}$

= ${\displaystyle 3^{2\times \frac{5}{2}}-3-3^{-4\times (-\frac{1}{2})}}$

= 3⁵ - 3 - 3²
= 243 - 3 - 9
= 231

Question 1.2

Evaluate : ${\displaystyle {\left(64\right)}^{\frac{2}{3}}-\sqrt[3]{125}-\frac{1}{2^{-5}}+{\left(27\right)}^{-\frac{2}{3}}\times {\left(\frac{25}{9}\right)}^{-\frac{1}{2}}}$

Sol:

${\displaystyle {\left(64\right)}^{\frac{2}{3}}-\sqrt[3]{125}-\frac{1}{2^{-5}}+{\left(27\right)}^{-\frac{2}{3}}\times {\left(\frac{25}{9}\right)}^{-\frac{1}{2}}}$

= ${\displaystyle {\left(4^3\right)}^{\frac{2}{3}}-\sqrt[3]{5^3}-2^5+{\left(3^3\right)}^{-\frac{2}{3}}\times {\left(\frac{5^2}{3^2}\right)}^{-\frac{1}{2}}}$

= ${\displaystyle 4^2-5-2^5+3^{-2}\times {\left(\frac{5}{3}\right)}^{2\times (-\frac{1}{2})}}$

= ${\displaystyle 16-5-32+\frac{1}{3^2}\times {\left(\frac{5}{3}\right)}^{-1}}$

= ${\displaystyle -21+\frac{1}{9}\times \frac{3}{5}}$

= ${\displaystyle -21+\frac{1}{15}}$

= ${\displaystyle \frac{-315+1}{15}}$

= ${\displaystyle -\frac{314}{15}}$

= ${\displaystyle -20\frac{14}{15}}$

Question 1.3

Evaluate : ${\displaystyle {\left(64\right)}^{\frac{2}{3}}-\sqrt[3]{125}-\frac{1}{2^{-5}}+{\left(27\right)}^{-\frac{2}{3}}\times {\left(\frac{25}{9}\right)}^{-\frac{1}{2}}}$

Sol:

${\displaystyle {\left(64\right)}^{\frac{2}{3}}-\sqrt[3]{125}-\frac{1}{2^{-5}}+{\left(27\right)}^{-\frac{2}{3}}\times {\left(\frac{25}{9}\right)}^{-\frac{1}{2}}}$

= ${\displaystyle {\left(4^3\right)}^{\frac{2}{3}}-\sqrt[3]{5^3}-2^5+{\left(3^3\right)}^{-\frac{2}{3}}\times {\left(\frac{5^2}{3^2}\right)}^{-\frac{1}{2}}}$

= ${\displaystyle 4^2-5-2^5+3^{-2}\times {\left(\frac{5}{3}\right)}^{2\times (-\frac{1}{2})}}$

= ${\displaystyle 16-5-32+\frac{1}{3^2}\times {\left(\frac{5}{3}\right)}^{-1}}$

= ${\displaystyle -21+\frac{1}{9}\times \frac{3}{5}}$

= ${\displaystyle -21+\frac{1}{15}}$

= ${\displaystyle \frac{-315+1}{15}}$

= ${\displaystyle -\frac{314}{15}}$

= ${\displaystyle -20\frac{14}{15}}$

Question 1.3

Evaluate : ${\displaystyle {\left[{(-\frac{2}{3})}^{-2}\right]}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}}$

Sol:

${\displaystyle {\left[{(-\frac{2}{3})}^{-2}\right]}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}}$

= ${\displaystyle {\left[{(-\frac{3}{2})}^2\right]}^3\times {\left(3\right)}^4\times \frac{1}{3}\times \frac{1}{3\times 2}}$

= ${\displaystyle {(-\frac{3}{2})}^6\times {\left(3\right)}^2\times \frac{1}{2}}$

= ${\displaystyle \frac{3^{6+2}}{2^{6+1}}}$

= ${\displaystyle \frac{3^8}{2^7}}$

Question 2

Simplify : ${\displaystyle \frac{3\times 9^{n+1}-9\times 3^{2n}}{3\times 3^{2n+3}-9^{n+1}}}$

Sol:

${\displaystyle \frac{3\times 9^{n+1}-9\times 3^{2n}}{3\times 3^{2n+3}-9^{n+1}}}$

= ${\displaystyle \frac{3\times {\left(3^2\right)}^{n+1}-3^2\times 3^{2n}}{3\times 3^{2n+3}-{\left(3^2\right)}^{n+1}}}$

= ${\displaystyle \frac{3^{1+2n+2}-3^{2+2n}}{3^{1+2n+3}-3^{2n+2}}}$

= ${\displaystyle \frac{3^{3+2n}-3^{2+2n}}{3^{4+2n}-3^{2n+2}}}$

= ${\displaystyle \frac{3^{2n}(3^3-3^2)}{3^{2n}(3^4-3^2)}}$

= ${\displaystyle \frac{27-9}{81-9}}$

= ${\displaystyle \frac{18}{72}}$

= ${\displaystyle \frac{1}{4}}$

Question 3

Solve : 3^x-1× 5^2y-3 = 225.

Sol:

3^x-1× 5^2y-3 = 225

⇒ ${\displaystyle 3^{x-1}\times 5^{2y-3}=3^2\times 5^2}$

⇒ x - 1 = 2 and 2y - 3 = 2
⇒ x = 3 and  2y = 5
⇒ x = 3 and y = ${\displaystyle \frac{5}{2}}$

⇒ x = 3 and y = ${\displaystyle 2\frac{1}{2}}$

Question 4

If ${\displaystyle {\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷{\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)}^{-5}=a^x.b^y}$ , find x + y.

Sol:

${\displaystyle {\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷{\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)}^{-5}=a^x.b^y}$

⇒ ${\displaystyle {\left(\frac{b^6}{a^3}\right)}^7÷{\left(\frac{a^5}{b^8}\right)}^{-5}=a^x.b^y}$

⇒ ${\displaystyle {\left(\frac{b^6}{a^3}\right)}^7÷{\left(\frac{b^8}{a^5}\right)}^5=a^x.b^y}$

⇒ ${\displaystyle \left(\frac{b^{42}}{a^{21}}\right)÷\left(\frac{b^{40}}{a^{25}}\right)=a^x.b^y}$

⇒ ${\displaystyle \left(\frac{b^{42}}{a^{21}}\right)\times \left(\frac{a^{25}}{b^{40}}\right)=a^x.b^y}$

⇒ b² x a⁴ = a^x x b^y
⇒ x = 4 and y = 2
⇒ x + y = 4 + 2 = 6

Question 5

If 3^{x + 1} = 9^{x - 3} , find the value of 2^{1 + x}.

Sol:

3^{x + 1} = 9^{x - 3}
⇒ 3^x x 3 = ( 3² )^{x - 3} 
⇒ 3^x x 3 = ${\displaystyle \frac{3^{2x}}{3^6}}$

⇒ ${\displaystyle 3^6\times 3=\frac{3^{2x}}{3^x}}$

⇒ ${\displaystyle 3^7=3^x}$
⇒ x = 7
⇒ ${\displaystyle 2^{1+x}=2^{1+7}=2^8=256}$

Question 6

If 2^x = 4^y = 8^z and ${\displaystyle \frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}=4}$ , find the value of x.

Sol:

2^x = 4^y = 8^z and ${\displaystyle \frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}=4}$

2^x = 4^y = 8^z 

⇒ 2^x = 2^2y = 2^3z 

⇒ x = 2y = 3z

⇒ y = ${\displaystyle \frac{x}{2}\text{and}z=\frac{x}{3}}$

Now, ${\displaystyle \frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}=4}$

⇒ ${\displaystyle \frac{1}{2x}+\frac{1}{\frac{4x}{2}}+\frac{1}{\frac{8x}{3}}=4}$

⇒ ${\displaystyle \frac{1}{2x}+\frac{2}{4x}+\frac{3}{8x}=4}$

⇒ ${\displaystyle \frac{1}{2x}+\frac{1}{2x}+\frac{3}{8x}=4}$ 

⇒ ${\displaystyle \frac{4+4+3}{8x}=4}$

⇒ ${\displaystyle \frac{11}{8x}=4}$

⇒ x = ${\displaystyle \frac{11}{32}}$.

Question 7

If ${\displaystyle \frac{9^n.3^2.3^n-{\left(27\right)}^n}{{(3^m.2)}^3}=3^{-3}}$

Show that : m - n = 1.

Sol:

${\displaystyle \frac{9^n.3^2.3^n-{\left(27\right)}^n}{{(3^m.2)}^3}=3^{-3}}$

⇒ ${\displaystyle \frac{3^{2n}.3^2.3^n-{\left(3\right)}^{3n}}{3^{3m}.{\left(2\right)}^3} =\frac{1}{3^3}}$

⇒ ${\displaystyle \frac{3^{3n}.3^2-3^{3n}}{3^{3m}.2^3}=\frac{1}{3^3}}$

⇒ ${\displaystyle \frac{3^{3n}(3^2-1)}{3^{3m}\times 8}=\frac{1}{3^3}}$

⇒ ${\displaystyle \frac{3^{3n}\times 8}{3^{3m}\times 8}=\frac{1}{3^3}}$

⇒ ${\displaystyle \frac{1}{3^{3(m-n)}}=\frac{1}{3^{3\times 1}}}$

⇒ m - n = 1  ( proved )

Question 8

Solve for x : (13)^√x = 4⁴ - 3⁴ - 6

Sol:

(13)^√x = 4⁴ - 3⁴ - 6
⇒ (13)^√x = 256 - 81 - 6
⇒ (13)^√x  = 169
⇒ (13)^√x  = 13²
⇒ √x = 2
⇒ x = 4

Question 9

If 3^4x = ( 81 )^-1 and ${\displaystyle {10}^{\frac{1}{y}}=0.0001,\text{Find the value of}{2}^{-x}\times {16}^y}$

Sol:

3^4x = ( 81 )^-1 and ${\displaystyle {10}^{\frac{1}{y}}}$ = 0.0001

⇒ 3^4x = ${\displaystyle {\left(3^4\right)}^{-1}\text{and}{10}^{\frac{1}{y}}=\frac{1}{10000}}$

⇒ 3^4x = ${\displaystyle 3^{-4}\text{and}{10}^{\frac{1}{y}}=\frac{1}{{10}^4}}$

⇒ 4x = - 4 and ${\displaystyle {10}^{\frac{1}{y}}={10}^{-4}}$

⇒ x = - 1 and ${\displaystyle \frac{1}{y}}$ = - 4

⇒ x = - 1 and y = ${\displaystyle -\frac{1}{4}}$

∴ ${\displaystyle 2^{-x}\times {16}^y=2^{-(-1)}x{16}^{-\frac{1}{4}}}$

= ${\displaystyle 2\times 2^{4\times -\frac{1}{4}}}$

= ${\displaystyle 2\times 2^{-1}}$

= ${\displaystyle 2^{1-1}}$

= ${\displaystyle 2^0}$ 
= 1

Question 10

Solve : 3(2^x + 1) - 2^x+2 + 5 = 0.

Sol:

3(2^x + 1) - 2^x+2 + 5 = 0.
⇒ 3 x 2^x + 3 - 2^x x 2² + 5 = 0
⇒ 2^x ( 3 - 2² ) + 8 = 0
⇒ 2^x ( 3 - 4 ) = - 8
⇒ 2^x x ( - 1 ) = - 8
⇒ 2^x = 8
⇒ 2^x = 2³
⇒ x = 3

Question 11

If (a^m)ⁿ = a^m .aⁿ, find the value of : m(n - 1) - (n - 1)

Sol:

(a^m)ⁿ = a^m .aⁿ
⇒ a^mn = a^{m + n}
⇒ mn = m + n               ....(1)
Now,
m( n - 1 ) - ( n - 1 )
= mn - m - n + 1
= m + n - m - n + 1      ....[ From (1) ]
= 1

Question 12

If m = ${\displaystyle \sqrt[3]{15}\text{and}n=\sqrt[3]{14},\text{find the value of}m-n-\frac{1}{m^2+mn+n^2}}$

Sol:

${\displaystyle \sqrt[3]{15}\text{and}n=\sqrt[3]{14}}$
⇒ m³ = 15 and n³ = 14

∴ m - n - ${\displaystyle \frac{1}{m^2+mn+n^2}}$

= ${\displaystyle \frac{(m^3+m^{2}n+m{n}^2)-(m^{2}n+m{n}^2+n^3)-1}{m^2+mn+n^2}}$

= ${\displaystyle \frac{m^3+m^{2}n+m{n}^2-m^{2}n-m{n}^2-n^3-1}{m^2+mn+n^2}}$

= ${\displaystyle \frac{m^3-n^3-1}{m^2+mn+n^2}}$

= ${\displaystyle \frac{15-14-1}{m^2+mn+n^2}}$

= ${\displaystyle \frac{1-1}{m^2+mn+n^2}}$

= 0

Question 13

Evaluate :
${\displaystyle \frac{2^n\times 6^{m+1}\times {10}^{m-n}\times {15}^{m+n-2}}{4^m\times 3^{2m+n}\times {25}^{m-1}}}$

Sol:

${\displaystyle \frac{2^n\times 6^{m+1}\times {10}^{m-n}\times {15}^{m+n-2}}{4^m\times 3^{2m+n}\times {25}^{m-1}}}$

= ${\displaystyle \frac{2^n\times 6^m\times 6\times {10}^m\times {10}^{-n}\times {15}^m\times {15}^n\times {15}^{-2}}{4^m\times {\left(3^2\right)}^m\times 3^n\times {25}^m\times {25}^{-1}}}$

= ${\displaystyle \frac{{(2\times \frac{1}{10}\times 15)}^n\times {(6\times 10\times 15)}^m\times 6\times \frac{1}{{15}^2}}{3^n\times {(4\times 3^2\times 25)}^m\times \frac{1}{25}}}$

= ${\displaystyle \frac{3^n\times {900}^m\times \frac{6}{225}}{3^n\times {900}^m\times \frac{1}{25}}}$

= ${\displaystyle \frac{6}{225}\times \frac{25}{1}}$

= ${\displaystyle \frac{6}{9}}$

= ${\displaystyle \frac{2}{3}}$

Question 14

Evaluate : ${\displaystyle {\left(\frac{x^q}{x^r}\right)}^{\frac{1}{qr}}\times {\left(\frac{x^r}{x^p}\right)}^{\frac{1}{rp}}\times {\left(\frac{x^p}{x^q}\right)}^{\frac{1}{pq}}}$

Sol:

${\displaystyle {\left(\frac{x^q}{x^r}\right)}^{\frac{1}{qr}}\times {\left(\frac{x^r}{x^p}\right)}^{\frac{1}{rp}}\times {\left(\frac{x^p}{x^q}\right)}^{\frac{1}{pq}}}$

= ${\displaystyle \frac{x^{q\times \frac{1}{qr}}}{x^{r\times \frac{1}{qr}}}\times \frac{x^{r\times \frac{1}{rp}}}{x^{p\times \frac{1}{rp}}}\times \frac{x^{p\times \frac{1}{pq}}}{x^{q\times \frac{1}{pq}}}}$

= ${\displaystyle \frac{x^{\frac{1}{r}}}{x^{\frac{1}{q}}}\times \frac{x^{\frac{1}{p}}}{x^{\frac{1}{r}}}\times \frac{x^{\frac{1}{q}}}{x^{\frac{1}{p}}}}$

= 1

Question 15.1

Prove that : ${\displaystyle \frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}=\frac{2}{b^2-a^2}}$

Sol:

${\displaystyle \frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}=\frac{2}{b^2-a^2}}$

L.H.S. = ${\displaystyle \frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}}$

= ${\displaystyle \frac{\frac{1}{a}}{\frac{1}{a}+\frac{1}{b}}+\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}}$

= ${\displaystyle \frac{\frac{1}{a}}{\frac{b+a}{ab}}+\frac{\frac{1}{a}}{\frac{b-a}{ab}}}$

= ${\displaystyle \frac{1}{a}\times \frac{ab}{b+a}+\frac{1}{a}\times \frac{ab}{b-a}}$

= ${\displaystyle \frac{b}{b+a}+\frac{b}{b-a}}$

= ${\displaystyle \frac{b^2-ab+b^2+ab}{b^2-a^2}}$

= ${\displaystyle \frac{2b^2}{b^2-a^2}}$

= R.H.S.

Question 15.2

Prove that : ${\displaystyle \frac{a+b+c}{a^{-1}{b}^{-1}+b^{-1}{c}^{-1}+c^{-1}{a}^{-1}}=abc}$

Sol:

L.H.S. = ${\displaystyle \frac{a+b+c}{a^{-1}{b}^{-1}+b^{-1}{c}^{-1}+c^{-1}{a}^{-1}}}$

= ${\displaystyle \frac{a+b+c}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}}$

= ${\displaystyle \frac{a+b+c}{\frac{c+a+b}{abc}}}$

= ${\displaystyle \frac{(a+b+c)\left(abc\right)}{a+b+c}}$

= abc

= R.H.S.

Question 16

Evaluate : ${\displaystyle \frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}}$

Sol:

${\displaystyle \frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}}$

= ${\displaystyle \frac{4}{{\left(6^3\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(4^4\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(3^5\right)}^{-\frac{1}{5}}}}$

= ${\displaystyle \frac{4}{{\left(6\right)}^{-2}}+\frac{1}{{\left(4\right)}^{-3}}+\frac{2}{{\left(3\right)}^{-1}}}$

= 4 x 6² + 1 x 4³ + 2 x 3

= 4 x 36 + 1 x 64 + 6

= 144 + 64 + 6

= 214