Chapter 8: Exponents and Powers

Ans: $\dfrac{1}{6}$

As, ${\left( {{2^{ - 1}} \times {3^{ - 1}}} \right)^{ - 1}} = {\left( {\dfrac{1}{2} \times \dfrac{1}{3}} \right)^{ - 1}} = {\left( {\dfrac{1}{6}} \right)^{ - 1}} = 6$

Ans: 0

$\left( {{6^0} - {7^0}} \right) \times \left( {{6^0} + {7^0}} \right) = \left( {1 - 1} \right) \times \left( {1 - 1} \right) = 0 \times 0 = 0$

Ans: $\dfrac{1}{{{3^{ - 5}}}}$

Expression can be written as ${3^5} = \dfrac{1}{{{3^{ - 5}}}}$

Ans: 1

$\dfrac{{{{\left( { - 7} \right)}^6}}}{{{{\left( 7 \right)}^6}}} = \dfrac{{{{\left( { - 1 \times 7} \right)}^6}}}{{{{\left( 7 \right)}^6}}} = \dfrac{{{{\left( { - 1} \right)}^6} \times {{\left( 7 \right)}^6}}}{{{{\left( 7 \right)}^6}}} = {\left( { - 1} \right)^6} = 1$

Ans: $\dfrac{{36}}{1} + \dfrac{{36}}{4} + \dfrac{{36}}{9}$

$\left( {{1^{ - 2}} + {2^{ - 2}} + {3^{ - 2}}} \right) \times {6^2} = \left( {\dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}}} \right) \times 36$

= $\left( {\dfrac{1}{1} + \dfrac{1}{4} + \dfrac{1}{9}} \right) \times 36$

=$\dfrac{{36}}{1} + \dfrac{{36}}{4} + \dfrac{{36}}{9}$

=$36+9+4=49$

In questions 66 to 90, state whether the given statements are true (T) or false (F).

Ans: False.

LHS = ${\left( { - 4} \right)^{ - 2}} = \dfrac{1}{{{{\left( { - 4} \right)}^2}}} = \dfrac{1}{{16}}$

RHS = ${\left( 4 \right)^{ - 2}} = \dfrac{1}{{{{\left( 4 \right)}^2}}} = \dfrac{1}{{16}}$

LHS = RHS. Hence, these numbers are the same and not multiplicative.

Ans: True.

${\left( {\dfrac{3}{2}} \right)^2} \times {\left( {\dfrac{2}{3}} \right)^{ - 2}} = {\left( {\dfrac{3}{2}} \right)^2} \times {\left( {\dfrac{3}{2}} \right)^2}$

Ans: True.

Use ${x^{ - m}} = \dfrac{1}{{{x^m}}}$.

${10^{ - 2}} = \dfrac{1}{{{{10}^2}}} = \dfrac{1}{{100}}$

Ans: False

R H S = 2 x 10+ 4 x 1+ 5 x 10+ 8 x 100 = 20 + 4 + 50 + 800 = 874 

So, L H S ≠ R H S

Ans: True

RHS = 3 x 102 + 2 x 101 + 9 x 100 + 2 x 10-1 + 5 x 10-2

= 3 x 10 + 2 x 10 + 9 x 10 + $\dfrac{2}{{10}} + \dfrac{5}{{10 \times 10}}$

Therefore, LHS = RHS.