Rs Aggarwal 2019 2020 Solutions for Class 8 Maths Chapter 2 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 8 students for Maths Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 8 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 33:

(i) ${4}^{-3}=\frac{1}{{4}^{3}}=\frac{1}{64}$

(ii) ${\left(\frac{1}{2}\right)}^{-5}={2}^{5}=32$

(iii) ${\left(\frac{4}{3}\right)}^{-3}={\left(\frac{3}{4}\right)}^{3}=\frac{{3}^{3}}{{4}^{3}}=\frac{27}{64}$

(iv) ${\left(-3\right)}^{-4}={\left(\frac{-1}{3}\right)}^{4}=\frac{\left(-1{\right)}^{4}}{{3}^{4}}=\frac{1}{81}$

(v) ${\left(\frac{-2}{3}\right)}^{-5}={\left(\frac{-3}{2}\right)}^{5}=\frac{{\left(-3\right)}^{5}}{{2}^{5}}=\frac{-243}{32}$

Page No 33:

(i)

(ii)

(iii)${\left(\frac{2}{3}\right)}^{-3}×{\left(\frac{2}{3}\right)}^{-2}={\left(\frac{2}{3}\right)}^{\left(-3-2\right)}={\left(\frac{2}{3}\right)}^{-5}={\left(\frac{3}{2}\right)}^{5}=\frac{{3}^{5}}{{2}^{5}}=\frac{243}{32}$

(iv) ${\left(\frac{9}{8}\right)}^{-3}×{\left(\frac{9}{8}\right)}^{2}={\left(\frac{9}{8}\right)}^{\left(-3+2\right)}={\left(\frac{9}{8}\right)}^{-1}=\frac{8}{9}$

Page No 33:

(i)

${\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3}×{\left(\frac{3}{5}\right)}^{0}={\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3+0}\phantom{\rule{0ex}{0ex}}={\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3}={\left(\frac{9}{5}\right)}^{2}×{\left(\frac{5}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{{9}^{2}}{{5}^{2}}×\frac{{5}^{3}}{{3}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({3}^{2}\right)}^{2}}{{5}^{2}}×\frac{{5}^{3}}{{3}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{4}}{{5}^{2}}×\frac{{5}^{3}}{{3}^{3}}=\left({3}^{\left(4-3\right)}\right)×\left({5}^{\left(3-2\right)}\right)=3×5=15$

(ii)

${\left(\frac{-3}{5}\right)}^{-4}×{\left(\frac{-2}{5}\right)}^{2}={\left(\frac{5}{-3}\right)}^{4}×{\left(\frac{-2}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{4}}{-{3}^{4}}×\frac{-{2}^{2}}{{5}^{2}}={5}^{\left(4-2\right)}×\frac{-{2}^{2}}{-{3}^{4}}={5}^{2}×\frac{-{2}^{2}}{-{3}^{4}}\phantom{\rule{0ex}{0ex}}=25×\frac{4}{81}=\frac{100}{81}$

(iii)

${\left(\frac{-2}{3}\right)}^{-3}×{\left(\frac{-2}{3}\right)}^{-2}={\left(\frac{3}{-2}\right)}^{3}×{\left(\frac{3}{-2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{3}}{-{2}^{3}}×\frac{{3}^{2}}{-{2}^{2}}=\frac{{3}^{\left(3+2\right)}}{-{2}^{\left(3+2\right)}}=\frac{{3}^{5}}{-{2}^{5}}=\frac{-243}{32}$

Page No 33:

(i) ${\left\{{\left(\frac{-2}{3}\right)}^{2}\right\}}^{-2}={\left(\frac{-2}{3}\right)}^{2×\left(-2\right)}={\left(\frac{-2}{3}\right)}^{-4}={\left(\frac{3}{-2}\right)}^{4}=\frac{{3}^{4}}{\left(-2{\right)}^{4}}=\frac{{3}^{4}}{{2}^{4}}=\frac{81}{16}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^{2}\right\}}^{-2}\right]}^{-1}={\left[{\left(\frac{-1}{3}\right)}^{2×\left(-2\right)}\right]}^{-1}={\left[{\left(\frac{-1}{3}\right)}^{-4}\right]}^{-1}={\left(\frac{-1}{3}\right)}^{-4×-1}={\left(\frac{-1}{3}\right)}^{4}=\frac{-{1}^{4}}{{3}^{4}}=\frac{{1}^{4}}{{3}^{4}}=\frac{1}{81}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^{2}={\left(\frac{3}{2}\right)}^{-2×2}={\left(\frac{3}{2}\right)}^{-4}={\left(\frac{2}{3}\right)}^{4}=\frac{{2}^{4}}{{3}^{4}}=\frac{16}{81}$

Page No 33:

$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=\left\{{3}^{3}-{2}^{3}\right\}÷{4}^{3}=\left\{27-8\right\}÷64=\frac{19}{64}$

Page No 33:

(i)

(ii)

(iii)

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

Page No 34:

Consider the left side:
${\left(\frac{5}{3}\right)}^{-4}×{\left(\frac{5}{3}\right)}^{-5}={\left(\frac{5}{3}\right)}^{\left(-4+\left(-5\right)\right)}={\left(\frac{5}{3}\right)}^{-9}$

Given:
${\left(\frac{5}{3}\right)}^{-9}={\left(\frac{5}{3}\right)}^{3x}$
Comparing the powers:

Page No 34:

Given:
${\left(\frac{4}{9}\right)}^{4}×{\left(\frac{4}{9}\right)}^{-7}={\left(\frac{4}{9}\right)}^{2x-1}$

$\therefore$ ${\left(\frac{4}{9}\right)}^{\left(4-7\right)}={\left(\frac{4}{9}\right)}^{-3}={\left(\frac{4}{9}\right)}^{2x-1}$

Page No 34:

Let the required number be $x$.

$\therefore$ $x×\left(-6{\right)}^{-1}={9}^{-1}$

The greatest common divisor for the numerator and the denominator is 3.

Page No 34:

Let the number be $x$.

$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{27}\right)}^{-2}$

Page No 34:

Given:
${5}^{2x+1}÷25=125$

$\therefore$ $x=2$

Page No 36:

(i) $57.36=5.736×10$
(ii) $3500000=3.5×{10}^{6}$
(iii) $273000=2.73×{10}^{5}$
(iv) $168000000=1.68×{10}^{8}$
(v) $4630000000000=4.63×{10}^{12}$
(vi) $345×{10}^{5}=3.45×{10}^{7}$

Page No 36:

(i) $3.74×{10}^{5}=\frac{374}{100}×{10}^{5}=\frac{374×{10}^{5}}{{10}^{2}}=374×{10}^{\left(5-2\right)}=374×{10}^{3}=374000$
(ii) $6.912×{10}^{8}=\frac{6912}{1000}×{10}^{8}=\frac{6912×{10}^{8}}{{10}^{3}}=6912×{10}^{\left(8-3\right)}=6912×{10}^{5}=691200000$
(iii) $4.1253×{10}^{7}=\frac{41253}{10000}×{10}^{7}=\frac{41253×{10}^{7}}{{10}^{4}}=41253×{10}^{\left(7-4\right)}=41253×{10}^{3}=41253000$
(iv) $2.5×{10}^{4}=\frac{25}{10}×{10}^{4}=\frac{25×{10}^{4}}{10}=25×{10}^{\left(4-1\right)}=25×{10}^{3}=25000$
(v) $5.17×{10}^{6}=\frac{517}{100}×{10}^{6}=\frac{517×{10}^{6}}{{10}^{2}}=517×{10}^{\left(6-2\right)}=517×{10}^{4}=5170000$
(vi) $1.679×{10}^{9}=\frac{1679}{1000}×{10}^{9}=\frac{1679×{10}^{9}}{{10}^{3}}=1679×{10}^{\left(9-3\right)}=1679×{10}^{6}=1679000000$

Page No 36:

(i) The height of the Mount Everest is 8848 m.
In standard form, we have:
.
(ii) The speed of light is 300000000 m/s.
In standard form, we have:

(iii) The Sun$-$Earth distance is 149600000000 m.
In standard form, we have: $149600000000=1496×100000000=1.496×1000×100000000=1.496×{10}^{3}×{10}^{8}=1.496×{10}^{11}\mathrm{m}$

Page No 36:

Mass of the Earth =
Now, $5.97×{10}^{24}=5.97×{10}^{\left(2+22\right)}=5.97×{10}^{2}×{10}^{22}=597×{10}^{22}$
So, the mass of the Earth can also be written as .

Mass of the Moon =
Sum of the masses of the Earth and the Moon:

Page No 36:

(i) $0.0006=\frac{6}{{10}^{4}}=6×{10}^{-4}$

(ii) $0.00000083=\frac{83}{{10}^{8}}=\frac{8.3×10}{{10}^{8}}=8.3×{10}^{\left(1-8\right)}=8.3×{10}^{-7}$

(iii) $0.0000000534=\frac{534}{{10}^{10}}=\frac{5.34×{10}^{2}}{{10}^{10}}=5.34×{10}^{\left(2-10\right)}=5.34×{10}^{-8}$

(iv) $0.0027=\frac{27}{{10}^{4}}=\frac{2.7×10}{{10}^{4}}=2.7×{10}^{\left(1-4\right)}=2.7×{10}^{-3}$

(v) 0.00000165 = (2-8)  = 1.65×10-6

(vi) 0.00000000689 = (2-11)= 6.89×10-9

(i)

(ii)
(iii)

Page No 36:

(i) $2.06×{10}^{-5}=\frac{206}{100}×\frac{1}{{10}^{5}}=\frac{206}{{10}^{2}×{10}^{5}}=\frac{206}{{10}^{\left(5+2\right)}}=\frac{206}{{10}^{7}}=\frac{206}{10000000}=0.0000206$

(ii) $5×{10}^{-7}=\frac{5}{{10}^{7}}=\frac{5}{10000000}=0.0000005$

(iii) $6.82×{10}^{-6}=\frac{682}{100}×\frac{1}{{10}^{6}}=\frac{682}{{10}^{2}×{10}^{6}}=\frac{682}{{10}^{\left(2+6\right)}}=\frac{682}{{10}^{8}}=\frac{682}{100000000}=0.00000682$

(iv)$5.673×{10}^{-4}=\frac{5673}{1000}×\frac{1}{{10}^{4}}=\frac{5673}{{10}^{3}×{10}^{4}}=\frac{5673}{{10}^{\left(3+4\right)}}=\frac{5673}{{10}^{7}}=\frac{5673}{10000000}=0.0005673$

(v)$1.8×{10}^{-2}=\frac{18}{10}×\frac{1}{{10}^{2}}=\frac{18}{10×{10}^{2}}=\frac{18}{{10}^{\left(1+2\right)}}=\frac{18}{{10}^{3}}=\frac{18}{1000}=0.018$

(vi) $4.129×{10}^{-3}=\frac{4129}{1000}×\frac{1}{{10}^{3}}=\frac{4129}{{10}^{3}×{10}^{3}}=\frac{4129}{{10}^{\left(3+3\right)}}=\frac{4129}{{10}^{6}}=\frac{4129}{1000000}=0.004129$

Page No 37:

(c) $\frac{125}{8}$

${\left(\frac{2}{5}\right)}^{-3}={\left(\frac{5}{2}\right)}^{3}=\frac{{5}^{3}}{{2}^{3}}=\frac{125}{8}$

Page No 37:

(d) $\frac{1}{81}$

${\left(-3\right)}^{-4}=\frac{1}{{\left(-3\right)}^{4}}=\frac{1}{{\left(-1\right)}^{4}×{\left(3\right)}^{4}}=\frac{1}{{\left(3\right)}^{4}}=\frac{1}{81}$

Page No 37:

(b) $\frac{-1}{32}$

${\left(-2\right)}^{-5}=\frac{1}{{\left(-2\right)}^{5}}=\frac{1}{-32}=\frac{1×\left(-1\right)}{-32×\left(-1\right)}=\frac{-1}{32}$

Page No 37:

(d) $\frac{1}{8}$

$\left({2}^{-5}÷{2}^{-2}\right)=\left(\frac{1}{{2}^{5}}÷\frac{1}{{2}^{2}}\right)=\left(\frac{1}{32}÷\frac{1}{4}\right)=\left(\frac{1}{32}×4\right)=\frac{4}{32}=\frac{1}{8}$

Page No 37:

(b) $\frac{60}{7}$

${\left({3}^{-1}+{4}^{-1}\right)}^{-1}÷{5}^{-1}={\left(\frac{1}{3}+\frac{1}{4}\right)}^{-1}÷\frac{1}{5}={\left(\frac{4+3}{12}\right)}^{-1}÷\frac{1}{5}={\left(\frac{7}{12}\right)}^{-1}÷\frac{1}{5}=\left(\frac{12}{7}\right)÷\frac{1}{5}=\frac{12}{7}×5=\frac{60}{7}$

(c) 29

Page No 37:

(a) $\frac{19}{64}$

$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}\phantom{\rule{0ex}{0ex}}=\left\{{3}^{3}-{2}^{3}\right\}÷{4}^{3}\phantom{\rule{0ex}{0ex}}=\left\{27-8\right\}÷64\phantom{\rule{0ex}{0ex}}=19÷64\phantom{\rule{0ex}{0ex}}=\frac{19}{64}$

Page No 37:

(a) $\frac{1}{16}$

${\left[{\left\{{\left(-\frac{1}{2}\right)}^{2}\right\}}^{-2}\right]}^{-1}\phantom{\rule{0ex}{0ex}}={\left[{\left\{-\frac{1}{2}\right\}}^{-4}\right]}^{-1}\phantom{\rule{0ex}{0ex}}={\left(-\frac{1}{2}\right)}^{\left(-4×-1\right)}\phantom{\rule{0ex}{0ex}}={\left(-\frac{1}{2}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}$

(d) 3

(d) 2

23x-1  = 42 -10
23x-1 = 32
23x-1  = 25
3x-1 = 5
3x = 6
Therefore,
x =  2

Page No 37:

(c) 1

Using the law of exponents ${\left(\frac{a}{b}\right)}^{0}=1$:
$\therefore$ ${\left(\frac{2}{3}\right)}^{0}=1$

Page No 38:

(c) $\frac{-3}{5}$

${\left(\frac{-5}{3}\right)}^{-1}={\left(\frac{3}{-5}\right)}^{1}=\frac{3}{-5}=\frac{3×\left(-1\right)}{-5×\left(-1\right)}=\frac{-3}{5}$

Page No 38:

(d) $\frac{-1}{8}$

${\left(\frac{-1}{2}\right)}^{3}=\frac{-{1}^{3}}{{2}^{3}}=\frac{-1}{8}$

Page No 38:

(b) $\frac{9}{16}$

${\left(\frac{-3}{4}\right)}^{2}=\frac{{\left(-3\right)}^{2}}{{\left(4\right)}^{2}}=\frac{9}{16}$

Page No 38:

(c) $3.67×{10}^{6}$

$3670000=367×{10}^{4}=3.67×100×{10}^{4}=3.67×{10}^{2}×{10}^{4}=3.67×{10}^{\left(2+4\right)}=3.67×{10}^{6}$

Page No 38:

(b) $4.63×{10}^{-5}$

$0.0000463=\frac{463}{{10}^{7}}=\frac{4.63×{10}^{2}}{{10}^{7}}=4.63×{10}^{\left(2-7\right)}=4.63×{10}^{-5}$

Page No 38:

(a) 3.67

$0.000367×{10}^{4}=\frac{367}{{10}^{6}}×{10}^{4}=367×{10}^{\left(4-6\right)}=367×{10}^{-2}=\frac{367}{{10}^{2}}=\frac{367}{100}=3.67$

Page No 39:

(i) ${3}^{-4}=\frac{1}{{3}^{4}}=\frac{1}{81}$

(ii) ${\left(-4\right)}^{3}={\left(-1\right)}^{3}×{\left(4\right)}^{3}=-1×64=-64$

(iii) ${\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^{2}=\frac{{4}^{2}}{{3}^{2}}=\frac{16}{9}$

(iv) ${\left(\frac{-2}{3}\right)}^{-5}={\left(\frac{3}{-2}\right)}^{5}=\frac{{3}^{5}}{-{2}^{5}}=\frac{243}{-32}=\frac{243×-1}{-32×-1}=\frac{-243}{32}$

(v) Using the property ${\left(\frac{a}{b}\right)}^{0}=1$, we have:

Page No 39:

${\left\{{\left(\frac{-2}{3}\right)}^{3}\right\}}^{-2}={\left(\frac{-2}{3}\right)}^{-6}={\left(\frac{3}{-2}\right)}^{6}=\frac{{3}^{6}}{-{2}^{6}}=\frac{729}{64}$

Page No 39:

$\left({3}^{-1}+{6}^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{3}+\frac{1}{6}\right)÷{\left(\frac{4}{3}\right)}^{1}\phantom{\rule{0ex}{0ex}}=\left(\left[\frac{1×2}{3×2}\right]+\left[\frac{1×1}{6×1}\right]\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{2+1}{6}\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{3}{6}\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)÷\left(\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)×\left(\frac{3}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{8}$

Page No 39:

Let the number be $x$.

$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{9}\right)}^{-2}$

Page No 39:

Let the number be $x$.

$\therefore$ ${\left(-3\right)}^{-1}×x={\left(6\right)}^{-1}$

Page No 39:

(i) $345=3.45×100=3.45×{10}^{2}$

(ii) $180000=18×10000=18×{10}^{4}=1.8×10×{10}^{4}=1.8×{10}^{\left(1+4\right)}=1.8×{10}^{5}$

(iii) $0.000003=\frac{3}{1000000}=3×{10}^{-6}$

(iv) $0.000027=\frac{27}{100000}=\frac{27}{{10}^{6}}=\frac{2.7×10}{{10}^{6}}=2.7×{10}^{\left(1-6\right)}=2.7×{10}^{-5}$

Page No 39:

(c) $\frac{-1}{27}$

${\left(-3\right)}^{-3}={\left(\frac{1}{-3}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{{1}^{3}}{-{3}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{1}{-27}\phantom{\rule{0ex}{0ex}}=\frac{1×-1}{-27×-1}\phantom{\rule{0ex}{0ex}}=\frac{-1}{27}$

Page No 39:

(b) $\frac{64}{27}$

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

Page No 39:

(c) ${3}^{-10}$

$\left({3}^{-6}÷{3}^{4}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{{3}^{6}}÷{3}^{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{3}^{6}}×\frac{1}{{3}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{3}^{\left(6+4\right)}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{3}^{10}}\phantom{\rule{0ex}{0ex}}={3}^{-10}$

(d) 3

Page No 39:

(c) 1

Using the law of exponents, which says ${\left(\frac{a}{b}\right)}^{0}=1,$ we get:

Page No 39:

(d) $\left(\frac{-5}{6}\right)$

${\left(\frac{-6}{5}\right)}^{-1}={\left(\frac{5}{-6}\right)}^{1}=\frac{5}{-6}=\frac{5×-1}{-6×-1}=\frac{-5}{6}$

Page No 39:

(c) $\frac{-1}{27}$

${\left(\frac{-1}{3}\right)}^{3}=\frac{-{1}^{3}}{{3}^{3}}=\frac{-1}{27}$

Page No 40:

(i) The standard form of 36000 is $3.6×{10}^{5}$
$360000=36×{10}^{4}=3.6×10×{10}^{4}=3.6×{10}^{\left(1+4\right)}=3.6×{10}^{5}$

(ii) The standard form of 0.0000123 is $1.23×{10}^{-5}$.
$0.0000123=\frac{123}{10000000}=\frac{123}{{10}^{7}}=\frac{1.23×100}{{10}^{7}}=\frac{1.23×{10}^{2}}{{10}^{7}}=1.23×{10}^{\left(2-7\right)}=1.23×{10}^{-5}$

(iii) ${\left(\frac{-2}{3}\right)}^{-2}=\frac{9}{4}$
${\left(\frac{-2}{3}\right)}^{-2}={\left(\frac{3}{-2}\right)}^{2}=\frac{{3}^{2}}{-{2}^{2}}=\frac{9}{4}$

(iv)
$3×{10}^{-3}=\frac{3}{{10}^{3}}=\frac{3}{1000}=0.003$

(v)

View NCERT Solutions for all chapters of Class 8