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#### Question 1:

Evaluate:
(i) 4−3
(ii) ${\left(\frac{1}{2}\right)}^{-5}$
(iii) ${\left(\frac{4}{3}\right)}^{-3}$
(iv) ${\left(-3\right)}^{-4}$
(v) ${\left(\frac{-2}{3}\right)}^{-5}$

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

#### Question 2:

Evaluate:
(i) ${\left(\frac{5}{3}\right)}^{2}×{\left(\frac{5}{3}\right)}^{2}$
(ii) ${\left(\frac{5}{6}\right)}^{6}×{\left(\frac{5}{6}\right)}^{-4}$
(iii) ${\left(\frac{2}{3}\right)}^{-3}×{\left(\frac{2}{3}\right)}^{-2}$
(iv) ${\left(\frac{9}{8}\right)}^{-3}×{\left(\frac{9}{8}\right)}^{2}$

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

#### Question 3:

Evaluate:
(i) ${\left(\frac{5}{9}\right)}^{-2}×{\left(\frac{3}{5}\right)}^{-3}×{\left(\frac{3}{5}\right)}^{0}$
(ii) ${\left(\frac{-3}{5}\right)}^{-4}×{\left(\frac{-2}{5}\right)}^{2}$
(iii) ${\left(\frac{-2}{3}\right)}^{-3}×{\left(\frac{-2}{3}\right)}^{-2}$

(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)

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

#### Question 4:

Evaluate:
(i) ${\left\{{\left(\frac{-2}{3}\right)}^{2}\right\}}^{-2}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^{2}\right\}}^{-2}\right]}^{-1}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^{2}$

(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) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^{2}={\left(\frac{3}{2}\right)}^{-2×2}={\left(\frac{3}{2}\right)}^{-4}={\left(\frac{2}{3}\right)}^{4}=\frac{{2}^{4}}{{3}^{4}}=\frac{16}{81}$

#### Question 5:

Evaluate $\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}.$

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

#### Question 6:

Evaluate ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}.$

#### Question 7:

Evaluate $\left[{\left({5}^{-1}×{3}^{-1}\right)}^{-1}÷{6}^{-1}\right].$

#### Question 8:

Find the value of:
(i) (20 + 3−1) × 32
(ii) (2−1 × 3−1) ÷ 2−3
(iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$

(i)

(ii)

(iii)

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

#### Question 9:

Find the value of x for which ${\left(\frac{5}{3}\right)}^{-4}×{\left(\frac{5}{3}\right)}^{-5}={\left(\frac{5}{3}\right)}^{3x}.$

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:

#### Question 10:

Find the value of x for which ${\left(\frac{4}{9}\right)}^{4}×{\left(\frac{4}{9}\right)}^{-7}={\left(\frac{4}{9}\right)}^{2x-1}.$

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

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

#### Question 11:

By what number should (−6)−1 be multiplied so that the product becomes 9−1?

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.

#### Question 12:

By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient may be ${\left(\frac{4}{27}\right)}^{-2}$?

Let the number be $x$.

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

#### Question 13:

If 52x + 1 ÷ 25 = 125, find the value of x.

Given:
${5}^{2x+1}÷25=125$

$\therefore$ $x=2$

#### Question 1:

Write each of the following numbers in standard form:
(i) 57.36
(ii) 3500000
(iii) 273000
(iv) 168000000
(v) 4630000000000
(vi) 345 × 105

(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) $345×{10}^{5}=3.45×{10}^{7}$

#### Question 2:

Write each of the following numbers in usual form:
(i) 3.74 × 105
(ii) 6.912 × 108
(iii) 4.1253 × 107
(iv) 2.5 × 104
(v) 5.17 × 106
(vi) 1.679 × 109

(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) $1.679×{10}^{9}=\frac{1679}{1000}×{10}^{9}=\frac{1679×{10}^{9}}{{10}^{3}}=1679×{10}^{\left(9-3\right)}=1679×{10}^{6}=1679000000$

#### Question 3:

(i) The height of Mount Everest is 8848 m. Write it in standard form.
(ii) The speed of light is 300000000 m/sec. Express it in standard form.
(iii) The distance from the earth to the sun is 149600000000 m. Write it in standard form.

(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: $149600000000=1496×100000000=1.496×1000×100000000=1.496×{10}^{3}×{10}^{8}=1.496×{10}^{11}\mathrm{m}$

#### Question 4:

Mass of earth is (5.97 × 1024) kg and mass of moon is (7.35 × 1022) kg. What is the total mass of the two?

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:

#### Question 5:

Write each of the following numbers in standard form:
(i) 0.0006
(ii) 0.00000083
(iii) 0.0000000534
(iv) 0.0027
(v) 0.00000165
(vi) 0.00000000689

(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

#### Question 6:

(i) 1 micron = $\frac{1}{1000000}$ m. Express it in standard form.
(ii) Size of a bacteria = 0.0000004 m. Express it in standard form.
(iii) Thickness of a paper = 0.03 mm. Express it in standard form.

(i)

(ii)
(iii)

#### Question 7:

Write each of the following numbers in usual form:
(i) 2.06 × 10−5
(ii) 5 × 10−7
(iii) 6.82 × 10−6
(iv) 5.673 × 10−4
(v) 1.8 × 10−2
(vi) 4.129 × 10−3

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

#### Question 1:

The value of ${\left(\frac{2}{5}\right)}^{-3}$ is
(a) $-\frac{8}{125}$
(b) $\frac{25}{4}$
(c) $\frac{125}{8}$
(d) $-\frac{2}{5}$

(c) $\frac{125}{8}$

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

#### Question 2:

The value of (−3)−4 is
(a) 12
(b) 81
(c) $-\frac{1}{12}$
(d) $\frac{1}{81}$

(d) $\frac{1}{81}$

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

#### Question 3:

The value of (−2)−5 is
(a) −32
(b) $\frac{-1}{32}$
(c) 32
(d) $\frac{1}{32}$

(b) $\frac{-1}{32}$

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

#### Question 4:

(2−5 ÷ 2−2) = ?
(a) $\frac{1}{128}$
(b) $\frac{-1}{128}$
(c) $-\frac{1}{8}$
(d) $\frac{1}{8}$

(d) $\frac{1}{8}$

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

#### Question 5:

The value of (3−1 + 4−1)−1 ÷ 5−1 is
(a) $\frac{7}{10}$
(b) $\frac{60}{7}$
(c) $\frac{7}{5}$
(d) $\frac{7}{15}$

(b) $\frac{60}{7}$

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

#### Question 6:

${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}=?$
(a) $\frac{61}{144}$
(b) $\frac{144}{61}$
(c) 29
(d) $\frac{1}{29}$

(c) 29

#### Question 7:

$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=?$
(a) $\frac{19}{64}$
(b) $\frac{27}{16}$
(c) $\frac{64}{19}$
(d) $\frac{16}{25}$

(a) $\frac{19}{64}$

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

#### Question 8:

${\left[{\left\{{\left(-\frac{1}{2}\right)}^{2}\right\}}^{-2}\right]}^{-1}=?$
(a) $\frac{1}{16}$
(b) 16
(c) $-\frac{1}{16}$
(d) −16

(a) $\frac{1}{16}$

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

#### Question 9:

The value of x for which ${\left(\frac{7}{12}\right)}^{-4}×{\left(\frac{7}{12}\right)}^{3x}={\left(\frac{7}{12}\right)}^{5},$ is
(a) −1
(b) 1
(c) 2
(d) 3

(d) 3

#### Question 10:

If (23x − 1 + 10) ÷ 7 = 6, then x is equal to
(a) −2
(b) 0
(c) 1
(d) 2

(d) 2

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

#### Question 11:

${\left(\frac{2}{3}\right)}^{0}=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 1
(d) 0

(c) 1

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

#### Question 12:

${\left(\frac{-5}{3}\right)}^{-1}=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) $\frac{-3}{5}$
(d) none of these

(c) $\frac{-3}{5}$

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

#### Question 13:

${\left(\frac{-1}{2}\right)}^{3}=?$
(a) $\frac{-1}{6}$
(b) $\frac{1}{6}$
(c) $\frac{1}{8}$
(d) $\frac{-1}{8}$

(d) $\frac{-1}{8}$

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

#### Question 14:

${\left(\frac{-3}{4}\right)}^{2}=?$
(a) $\frac{-9}{16}$
(b) $\frac{9}{16}$
(c) $\frac{16}{9}$
(d) $\frac{-16}{9}$

(b) $\frac{9}{16}$

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

#### Question 15:

3670000 in standard form is
(a) 367 × 104
(b) 36.7 × 105
(c) 3.67 × 106
(d) none of these

(c) $3.67×{10}^{6}$

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

#### Question 16:

0.0000463 in standard form is
(a) 463 × 10−7
(b) 4.63 × 10−5
(c) 4.63 × 10−9
(d) 46.3 × 10−6

(b) $4.63×{10}^{-5}$

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

#### Question 17:

0.000367 × 104 in usual form is
(a) 3.67
(b) 36.7
(c) 0.367
(d) 0.0367

(a) 3.67

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

#### Question 1:

Evaluate:
(i) 3−4
(ii) (−4)3
(iii) ${\left(\frac{3}{4}\right)}^{-2}$
(iv) ${\left(\frac{-2}{3}\right)}^{-5}$
(v) ${\left(\frac{5}{7}\right)}^{0}$

(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:

#### Question 2:

Evaluate: ${\left\{{\left(\frac{-2}{3}\right)}^{3}\right\}}^{-2}.$

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

#### Question 3:

Simplify: $\left({3}^{-1}+{6}^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1}.$

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

#### Question 4:

By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient is ${\left(\frac{4}{9}\right)}^{-2}$?

Let the number be $x$.

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

#### Question 5:

By what number should (−3)−1 be multiplied so that the product becomes 6−1?

Let the number be $x$.

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

#### Question 6:

Express each of the following in standard form:
(i) 345
(ii) 180000
(iii) 0.000003
(iv) 0.000027

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

#### Question 7:

Mark (✓) against the correct answer
The value of (−3)−3 is
(a) −27
(b) 9
(c) $\frac{-1}{27}$
(d) $\frac{1}{27}$

(c) $\frac{-1}{27}$

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

#### Question 8:

Mark (✓) against the correct answer
The value of ${\left(\frac{3}{4}\right)}^{-3}$ is
(a) $\frac{-27}{64}$
(b) $\frac{64}{27}$
(c) $\frac{-9}{4}$
(d) $\frac{27}{64}$

(b) $\frac{64}{27}$

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

#### Question 9:

Mark (✓) against the correct answer
(3−6 ÷ 34) = ?
(a) 3−2
(b) 32
(c) 3−10
(d) 310

(c) ${3}^{-10}$

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

#### Question 10:

Mark (✓) against the correct answer
If
(a) −1
(b) 1
(c) 2
(d) 3

(d) 3

#### Question 11:

Mark (✓) against the correct answer
${\left(\frac{3}{5}\right)}^{0}=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) 1
(d) 0

(c) 1

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

#### Question 12:

Mark (✓) against the correct answer
${\left(\frac{-6}{5}\right)}^{-1}=?$
(a) $\frac{6}{5}$
(b) $\frac{-6}{5}$
(c) $\frac{5}{6}$
(d) $\frac{-5}{6}$

(d) $\left(\frac{-5}{6}\right)$

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

#### Question 13:

Mark (✓) against the correct answer
${\left(\frac{-1}{3}\right)}^{3}=?$
(a) $\frac{-1}{9}$
(b) $\frac{1}{9}$
(c) $\frac{-1}{27}$
(d) $\frac{1}{27}$

(c) $\frac{-1}{27}$

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

#### Question 14:

Fill in the blanks.
(i) 360000 written in standard form is .........
(ii) 0.0000123 written in standard form is .........
(iii) ${\left(\frac{-2}{3}\right)}^{-2}=.........$
(iv) 3 × 10−3 in usual form is .........
(v) 5.32 × 10−4 in usual form is .........

(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)

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