Exponents and Powers

EXERCISE-12.1

P1. Evaluate

(i) 3⁻²

(ii) (−4)⁻²

(iii) ${\left(\frac{1}{2}\right)}^{\mathbf{–}\mathbf{5}}$

Sol. (i) $3^{–2}=\frac{1}{3^2}=\frac{1}{9} \left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

(ii) ${\left(–4\right)}^{–2}=\frac{1}{{\left(–4\right)}^2}=\frac{1}{9} \left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

(iii) ${\left(\frac{1}{2}\right)}^{–5}=\frac{1}{(2{)}^{–5}}={\left(2\right)}^5=2\times 2\times 2\times 2\times 2=32$

P2. Simplify and express the result in power notation with positive exponent.

(i) $\mathbf{(}\mathbf{–}\mathbf{4}{\mathbf{)}}^{\mathbf{5}}\mathbf{÷}\mathbf{(}\mathbf{–}\mathbf{4}{\mathbf{)}}^{\mathbf{8}}$

(ii) ${\left(\frac{1}{2^3}\right)}^{\mathbf{2}}$

(iii) ${(–3)}^{\mathbf{4}}\mathbf{\times }{\mathbf{\left(}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{\right)}}^{\mathbf{4}}$

(iv) $(3^{–7}÷3^{–10})\mathbf{\times }{\mathbf{3}}^{\mathbf{–}\mathbf{5}}$

(v) ${\mathbf{2}}^{\mathbf{–}\mathbf{3}}\mathbf{\times }\mathbf{(}\mathbf{–}\mathbf{7}{\mathbf{)}}^{\mathbf{–}\mathbf{3}}$

Sol. (i) (−4)⁵ ÷ (−4)⁸ = (−4)^{5 − 8}          (a^m ÷ aⁿ = a^m − n)

= (− 4)⁻³

$\frac{1}{(–4{)}^3} \left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

(ii) ${\left(\frac{1}{2^3}\right)}^2=\frac{1}{{\left(2^3\right)}^2}=\frac{1}{2^6} \left({\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right)$

(iii) ${\left(–3\right)}^4\times {\left(\frac{5}{3}\right)}^4={\left(–1\times 3\right)}^4\times \frac{5^4}{3^4}$

= ${\left(–1\right)}^4\times 3^4\times \frac{5^4}{3^4} \left[(ab{)}^m=a^m\times b^n\right]$

= ${\left(–1\right)}^4\times 3^4\times \frac{5^4}{3^4} \left[(\mathrm{ab}{)}^{\mathrm{m}}={\mathrm{a}}^{\mathrm{m}}\times {\mathrm{b}}^{\mathrm{m}}\right]$

= $(–1{)}^4\times 5^4$

$=5^4\left[(–1{)}^4=1\right]$

(iv)  (3^{− 7} ÷ 3⁻¹⁰) × 3⁻⁵ = (3^{−7 − (−10)}) × 3⁻⁵     

(a^m ÷ aⁿ = a^m − n)

= 3³ × 3⁻⁵

= 3^{3 + (− 5)}          (a^m × aⁿ = a^m + n)

= 3⁻²

$\frac{1}{3^2} \left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

(v) 2⁻³ × (−7)⁻³ 

= $\frac{1}{2^3}\times \frac{1}{(–7{)}^3} \left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

= $\frac{1}{[2\times (–7){]}^3} \left[{\mathrm{a}}^{\mathrm{m}}\times {\mathrm{b}}^{\mathrm{m}}=(\mathrm{ab}{)}^{\mathrm{m}}\right]$

= $\frac{1}{(–14{)}^3}$

P3. Find the value of.

(i) (3⁰ + 4⁻¹) × 2²

(ii) (2⁻¹ × 4⁻¹) ÷2⁻²

(iii) ${\left(\frac{1}{2}\right)}^{\mathbf{–}\mathbf{2}}\mathbf{+}{\left(\frac{1}{3}\right)}^{\mathbf{–}\mathbf{2}}\mathbf{+}{\left(\frac{1}{4}\right)}^{\mathbf{–}\mathbf{2}}$

(iv) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰

(v) ${\left\{{\left(\frac{–2}{3}\right)}^{–2}\right\}}^{\mathbf{2}}$

Sol. (i) $\left(3^0+4^{–1}\right)\times 2^2=\left(1+\frac{1}{4}\right)\times 2^2$       $\left({\mathrm{a}}^0=1\text{ and }{\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

$\frac{5}{4}\times 4=5$

(ii) (2⁻¹ × 4⁻¹) ÷ 2^{− 2} = [2⁻¹ × {(2)²}^{− 1}] ÷ 2^{− 2}

= (2^{− 1} × 2^{− 2}) ÷ 2^{− 2}           $\left({\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right)$

= 2^{−1+ (−2)} ÷ 2⁻²               (a^m × aⁿ = a^m + n)

= 2⁻³ ÷ 2⁻²

= 2^{−3 − (−2)}                          (a^m ÷ aⁿ = a^m − n)

= 2^{−3 + 2} = 2 ⁻¹

$=\frac{1}{2} \left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

(iii) ${\left(\frac{1}{2}\right)}^{–2}+{\left(\frac{1}{3}\right)}^{–2}+{\left(\frac{1}{4}\right)}^{–2}$

${\left(\frac{2}{1}\right)}^2+{\left(\frac{3}{1}\right)}^2+{\left(\frac{4}{1}\right)}^2 \left(\therefore {\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

= $2^2+3^2+4^2=4+9+16=29$

(iv) (3⁻¹ +4⁻¹ +5⁻¹)⁰ 

${\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)}^0 \left(a^{–m}=\frac{1}{a^m}\right)$

= 1 (a⁰ = 1)

(v) ${\left\{{\left(\frac{–2}{3}\right)}^{–2}\right\}}^2={\left\{{\left(\frac{3}{–2}\right)}^2\right\}}^2$

= ${\left\{\frac{3^2}{(–2{)}^2}\right\}}^2$$={\left(\frac{9}{4}\right)}^2=\frac{81}{16}$

P4. Evaluate

(i) $\frac{{\mathbf{8}}^{\mathbf{–}\mathbf{1}}\mathbf{\times }{\mathbf{5}}^{\mathbf{3}}}{{\mathbf{2}}^{\mathbf{–}\mathbf{4}}}$

(ii) $(5^{–1}\times 2^{–1})\mathbf{\times }{\mathbf{6}}^{\mathbf{–}\mathbf{1}}$

Sol. (i) $\frac{8^{–1}\times 5^3}{2^{–4}}=\frac{2^4\times 5^3}{8^1}$

$\frac{2^4\times 5^3}{2^3}=2^{4–3}\times 5^3$

$2\times 125=250$

(ii) $\left(5^{–1}\times 2^{–1}\right)\times 6^{–1}=\left(\frac{1}{5}\times \frac{1}{2}\right)\times \frac{1}{6}$

= $\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}$

P5. Find the value of m for which  5^m ÷5⁻³ = 5⁵.

Sol. 5^m ÷ 5⁻³ = 5⁵

5^{m − (− 3)} = 5⁵              (a^m ÷ aⁿ = a^m − n)

5^{m + 3} = 5⁵

Since the powers have same bases on both sides, their respective exponents must be equal.

m + 3 = 5

m = 5 − 3

m = 2

P6. Evaluate

(i) ${\{{\left(\frac{1}{3}\right)}^{–1}–{\left(\frac{1}{4}\right)}^{–1}\}}^{\mathbf{–}\mathbf{1}}$

(ii) ${\left(\frac{5}{8}\right)}^{\mathbf{–}\mathbf{7}}\mathbf{\times }{\left(\frac{8}{5}\right)}^{\mathbf{–}\mathbf{4}}$

Sol. (i) ${\left\{{\left(\frac{1}{3}\right)}^{–1}–{\left(\frac{1}{4}\right)}^{–1}\right\}}^{–1}={\left\{{\left(\frac{3}{1}\right)}^{–1}–{\left(\frac{4}{1}\right)}^{–1}\right\}}^{–1}$

= $\{3–4{\}}^{–1}=\{–1{\}}^{–1}=\frac{1}{–1}=–1$

(ii) ${\left(\frac{5}{8}\right)}^{–7}\times {\left(\frac{8}{5}\right)}^{–4}=\frac{5^{–7}}{8^{–7}}\times \frac{8^{–4}}{5^{–4}}$

$=\frac{8^7}{5^7}\times \frac{5^4}{8^4}\left({\mathrm{a}}^{–\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)$

= $\frac{8^{7–4}}{5^{7–4}}\left(a^m÷a^n=a^{m–n}\right)$

= $\frac{8^3}{5^3}=\frac{512}{125}$

Q7. Simplify.

(i) $\frac{\mathbf{25}\mathbf{\times }{\mathbf{t}}^{\mathbf{–}\mathbf{4}}}{{\mathbf{5}}^{\mathbf{–}\mathbf{3}}\mathbf{\times }\mathbf{10}\mathbf{\times }{\mathbf{t}}^{\mathbf{–}\mathbf{8}}}(t\ne 0)$

(ii) $\frac{{\mathbf{3}}^{\mathbf{–}\mathbf{5}}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}}\mathbf{\times }\mathbf{125}}{{\mathbf{5}}^{\mathbf{–}\mathbf{7}}\mathbf{\times }{\mathbf{6}}^{\mathbf{–}\mathbf{5}}}$

Sol. (i) $\frac{25\times t^{–4}}{5^{–3}\times 10\times t^{–8}}=\frac{5^2\times t^{–4}}{5^{–3}\times 5\times 2\times t^{–8}}$

= $\frac{5^2\times t^{–4}}{5^{–3+1}\times 2\times t^{–8}} \left(a^m\times a^n=a^{m+n}\right)$

$=\frac{5^2\times t^{–4}}{5^{–2}\times 2\times t^{–8}}$

$=\frac{5^{2+2}\times t^{–4+8}}{2}$

$$\begin{gathered} =\frac{5^4\times {\mathrm{t}}^4}{2} \\ =\frac{625{\mathrm{t}}^4}{2} \end{gathered}$$

(ii) $\frac{3^{–5}\times {10}^{–5}\times 125}{5^{–7}\times 6^{–5}}$

$=\frac{3^{–5}\times (2\times 5{)}^{–5}\times 5^3}{5^{–7}\times (2\times 3{)}^{–5}}$

= $3^{–5–(–5)}\times 2^{–5–(–5)}\times 5^{–5+3(–7)}$

= $3^0\times 2^0\times 5^5 \left({\mathrm{a}}^0=1\right)$

= $5^5$

EXERCISE-12.2

P1. Express the following numbers in standard form.

(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

Sol. (i) 0.0000000000085 = 8.5 × 10⁻¹²

(ii) 0.00000000000942 = 9.42 × 10⁻¹²

(iii) 6020000000000000 = 6.02 × 10¹⁵

(iv) 0.00000000837 = 8.37 × 10⁻⁹

(v)  31860000000 = 3.186 × 10¹⁰

P2. Express the following numbers in usual form.

(i) 3.02 × 10⁻⁶        (ii) 4.5 × 10⁴

(iii) 3 × 10⁻⁸               (iv) 1.0001 × 10⁹

(v) 5.8 × 10¹²             (vi) 3.61492 × 10⁶

Sol. (i) 3.02 × 10⁻⁶ = 0.00000302

(ii) 4.5 × 10⁴ = 45000

(iii) 3 × 10⁻⁸ = 0.00000003

(iv) 1.0001 × 10⁹ = 1000100000

(v) 5.8 × 10¹² = 5800000000000

(vi) 3.61492 × 10⁶ = 3614920

P3. Express the number appearing in the following statements in standard form.

(i)  1 micron is equal to $\frac{1}{1000000} \mathrm{m}$.

(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm

Sol. (i) $\frac{1}{1000000}$= 1 × 10⁻⁶

(ii) 0.000, 000, 000, 000, 000, 000, 16 = 1.6 × 10⁻¹⁹

(iii) 0.0000005 = 5 × 10⁻⁷

(iv) 0.00001275 = 1.275 × 10⁻⁵

(v) 0.07 = 7 × 10⁻²

P4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Sol. Thickness of each book = 20 mm

Hence, thickness of 5 books = (5 × 20) mm = 100 mm

Thickness of each paper sheet = 0.016 mm

Hence, thickness of 5 paper sheets = (5 × 0.016) mm = 0.080 mm

Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets

= (100 + 0.080) mm

= 100.08 mm

= 1.0008 × 10² mm