Chapter 8: Exponents and Powers

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

So, ${100^{ - 10}} = \dfrac{1}{{{{100}^{10}}}}$

Ans: Using law of exponents, am × an = am+n

${2^{ - 2}} \times {2^{ - 3}} = {2^{ - 2 - 3}} = {2^{ - 5}}$

Ans: Use ${a^m} \div {a^n} = {a^{m - n}}$

${\left( {\dfrac{1}{2}} \right)^{ - 2}} \div {\left( {\dfrac{1}{2}} \right)^{ - 3}} = {\left( {\dfrac{1}{2}} \right)^{ - 2 - \left( { - 3} \right)}} = {\left( {\dfrac{1}{2}} \right)^1} = \dfrac{1}{2}$

Ans: Using law of exponents, am × an = am+n

${3^{ - 5}} \times {3^{ - 4}} = {3^{ - 5 - 4}} = {3^{ - 9}} = \dfrac{1}{{{3^9}}}$

Ans: Since $2 \times 2 \times 2 \times 2 = 16 = {2^4}$,

${16^{ - 2}} = {\left( {{2^4}} \right)^{ - 2}} = {\left( 2 \right)^{4 \times  - 2}} = {2^{ - 8}}$

Ans: 27 = 3 × 3 × 3 = 33

-27 = -3 × -3 × -3 = (-3)3

64 = 4 × 4 × 4 = 43

Now use ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$,

$\dfrac{{27}}{{64}} = \dfrac{{{3^3}}}{{{4^3}}} = {\left( {\dfrac{3}{4}} \right)^3}$

$\Rightarrow \dfrac{{ - 27}}{{64}} = \dfrac{{{{\left( { - 3} \right)}^3}}}{{{4^3}}} = {\left( {\dfrac{{ - 3}}{4}} \right)^3}$

Ans: 16 = 4 × 4 = (4)2

81 = 9 × 9 = 92

Now use ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$,

$\dfrac{{16}}{{81}} = \dfrac{{{4^2}}}{{{9^2}}} = {\left( {\dfrac{4}{9}} \right)^2}$

$\Rightarrow \dfrac{{ - 16}}{{81}} = \dfrac{{ - {{\left( 4 \right)}^2}}}{{{9^2}}} =  - {\left( {\dfrac{4}{9}} \right)^2}$

Ans: Use ${\left( {{a^m}} \right)^n} = {a^{mn}}$ and ${a^{ - m}} = \dfrac{1}{{{a^m}}}$

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

Ans: Use ${a^m} \div {a^n} = {a^{m - n}}$ and ${a^m} \times {a^n} = {a^{m + n}}$

$\left( {{2^5} \div {2^8}} \right) \times {2^{ - 7}} = \left( {{2^{5 - 8}}} \right) \times {2^{ - 7}}$

= ${2^{ - 3}} \times {2^{ - 7}} = {2^{ - 3 - 7}} = {2^{ - 10}}$

Ans: Since, cube of (-2)= (-2)3

And square of (+4)= (+4)2

Therefore, the product= (-2)3 x (4)2 = (-8) x 16 = -128

Ans: Use ${a^{ - m}} = \dfrac{1}{{{a^m}}}$,

${\left( {\dfrac{1}{4}} \right)^{ - 2}} + {\left( {\dfrac{1}{2}} \right)^{ - 2}} + {\left( {\dfrac{1}{3}} \right)^{ - 2}} = {4^2} + {2^2} + {3^2} = 16 + 4 + 9 = 29$

Ans: Use ${a^m} \div {a^n} = {a^{m - n}}$

${\left( {{{\left( {\dfrac{{ - 2}}{3}} \right)}^{ - 2}}} \right)^3} \times {\left( {\dfrac{1}{3}} \right)^{ - 4}} \times {3^{ - 1}} \times \dfrac{1}{6} = {\left( {\dfrac{{ - 2}}{3}} \right)^{ - 2 \times 3}} \times {3^4} \times \dfrac{1}{3} \times \dfrac{1}{6}$

= ${\left( {\dfrac{{ - 2}}{3}} \right)^{ - 6}} \times {3^4} \times \dfrac{1}{3} \times \dfrac{1}{{2 \times 3}}$

= $\dfrac{{{3^6}}}{{{2^6}}} \times {3^4} \times \dfrac{1}{3} \times \dfrac{1}{{2 \times 3}}$

= $\dfrac{{{3^{10 - 2}}}}{{{2^{6 + 1}}}} = \dfrac{{{3^8}}}{{{2^7}}}$

Ans: Use ${a^m} \div {a^n} = {a^{m - n}}$ 

$\dfrac{{49 \times {z^{ - 3}}}}{{{7^{ - 3}} \times 10 \times {z^{ - 5}}}} = \dfrac{{{7^2} \times {z^{ - 3}}}}{{{7^{ - 3}} \times 10 \times {z^{ - 5}}}} = \dfrac{{{7^{2 + 3}} \times {z^{ - 3 + 5}}}}{{10}}$

= $\dfrac{{{7^5} \times {Z^2}}}{{10}} = \dfrac{{{7^{^5}}}}{{10}}{z^2}$

Ans: Use ${a^m} \div {a^n} = {a^{m - n}}$ 

$\left( {{2^5} \div {2^8}} \right) \times {\left( 2 \right)^{ - 7}} = {\left( 2 \right)^{5 - 8}}{\left( 2 \right)^{ - 7}}$

= ${2^{ - 3}} \times {2^{ - 7}} = {2^{ - 3 - 7}} = {2^{ - 10}} = \dfrac{1}{{{2^{10}}}} = \dfrac{1}{{1024}}$

Ans: Use ${a^m} \times {a^n} = {a^{m + n}}$,

${\left( {\dfrac{5}{3}} \right)^{ - 2}} \times {\left( {\dfrac{5}{3}} \right)^{ - 14}} = {\left( {\dfrac{5}{3}} \right)^{8x}}$

$\Rightarrow {\left( {\dfrac{5}{3}} \right)^{ - 2 - 14}} = {\left( {\dfrac{5}{3}} \right)^{8x}}$

$\Rightarrow {\text{Compare }}{\text{both }}{\text{sides,}}$

$\Rightarrow - 16 = 8x$

$\Rightarrow x =  - 2$

Ans: Use, am × an = am+n

Then, (-2)3 x (-2)-6 = (-2)2x-1

(-2)3-6 = (-2)2x-1

(-2)-3 = (-2)2x-1

Compare both the sides and the result is=

$-3 = 2x – 1$

$\Rightarrow 2 x = -3 + 1$

$\Rightarrow 2x = -2$

$\Rightarrow x= -1$

Ans: Use ${a^{ - m}} = \dfrac{1}{{{a^m}}}$,

${\left( {{2^{ - 1}} + {4^{ - 1}} + {6^{ - 1}} + {8^{ - 1}}} \right)^x} = 1$

$\Rightarrow {\left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{8}} \right)^x} = 1$

$\Rightarrow {\left( {\dfrac{{12 + 6 + 4 + 3}}{{24}}} \right)^x} = 1$

$\Rightarrow {\left( {\dfrac{{25}}{{24}}} \right)^x} = 1$

This can be possible only if x = 0 because a0 = 1.

Ans: Use ${a^{ - m}} = \dfrac{1}{{{a^m}}}$,

1000000 = 106

$\Rightarrow \dfrac{{293}}{{{{10}^6}}} = 293 \times {10^{ - 6}}$

$= 2.93 \times {10^{ - 6}} \times {10^2} = 2.93 \times {10^{ - 4}}$