Question 1

Evaluate.

(i) $ 3^{-2} $

(ii) $ (-4)^{-2} $

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

Accepted Answer

Solution:
(i) $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

(ii) $(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$

(iii) $\left( \frac{1}{2} \right)^{-5} = 2^5 = 32$ — Using the property of negative exponents $a^{-n} = \frac{1}{a^n}$, we evaluate each term step-by-step.

Question 2

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

(i) $ (-4)^5 \div (-4)^8 $

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

(iii) $ (-3)^4 	imes \left( \frac{5}{3} ight)^4 $

(iv) $ (3^{-7} \div 3^{-10}) 	imes 3^{-5} $

(v) $ 2^{-3} 	imes (-7)^{-3} $

Accepted Answer

Solution:
(i) $(-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64}$

(ii) $\left( \frac{1}{2^3} ight)^2 = \left( 2^{-3} ight)^2 = 2^{-6} = \frac{1}{2^6} = \frac{1}{64}$

(iii) $(-3)^4 	imes \left( \frac{5}{3} ight)^4 = \left( -3 	imes \frac{5}{3} ight)^4 = (-5)^4 = 5^4 = 625$

(iv) $(3^{-7} \div 3^{-10}) 	imes 3^{-5} = 3^{-7 - (-10)} 	imes 3^{-5} = 3^{3} 	imes 3^{-5} = 3^{3-5} = 3^{-2} = \frac{1}{9}$

(v) $2^{-3} 	imes (-7)^{-3} = \frac{1}{2^3} 	imes \frac{1}{(-7)^3} = \frac{1}{8} 	imes \frac{1}{-343} = -\frac{1}{2744}$ — Use laws of exponents: $a^m \div a^n = a^{m-n}$, $(a^m)^n = a^{mn}$, and $a^m 	imes a^n = a^{m+n}$. Also, negative exponents denote reciprocals.

Question 3

Find the value of.

(i) $ (3^0 + 4^{-1}) 	imes 2^2 $

(ii) $ (2^{-1} 	imes 4^{-1}) \div 2^{-2} $

(iii) $ \left( \frac{1}{2} ight)^{-2} + \left( \frac{1}{3} ight)^{-2} + \left( \frac{1}{4} ight)^{-2} $

(iv) $ (3^{-1} + 4^{-1} + 5^{-1})^0 $

(v) $ \left\{ \left( \frac{-2}{3} ight)^{-2} ight\}^2 $

Accepted Answer

Solution:
(i) $3^0 = 1$, $4^{-1} = \frac{1}{4}$, so $(1 + \frac{1}{4}) 	imes 2^2 = \frac{5}{4} 	imes 4 = 5$

(ii) $(2^{-1} 	imes 4^{-1}) \div 2^{-2} = \left( \frac{1}{2} 	imes \frac{1}{4} ight) \div \frac{1}{2^2} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} 	imes 4 = \frac{1}{2}$

(iii) $\left( \frac{1}{2} ight)^{-2} = 2^2 = 4$, $\left( \frac{1}{3} ight)^{-2} = 3^2 = 9$, $\left( \frac{1}{4} ight)^{-2} = 4^2 = 16$, sum = $4 + 9 + 16 = 29$

(iv) Any number to the power 0 is 1, so $(3^{-1} + 4^{-1} + 5^{-1})^0 = 1$

(v) $\left\{ \left( \frac{-2}{3} ight)^{-2} ight\}^2 = \left( \left( \frac{-2}{3} ight)^{-2} ight)^2 = \left( \frac{3}{-2} ight)^4 = \left( -\frac{3}{2} ight)^4 = \left( \frac{3}{2} ight)^4 = \frac{81}{16}$ — Apply exponent rules and simplify stepwise. Remember $a^0 = 1$ and negative exponents invert the base.

Question 4

Evaluate (i)  $\frac{8^{-1} 	imes 5^3}{2^{-4}}$

(ii)  $(5^{-1} 	imes 2^{-1}) 	imes 6^{-1}$

Accepted Answer

Solution:
(i) $\frac{8^{-1} 	imes 5^3}{2^{-4}} = \frac{\frac{1}{8} 	imes 125}{\frac{1}{2^{-4}}} = \frac{\frac{125}{8}}{2^{-4}} = \frac{125}{8} 	imes 2^{4} = \frac{125}{8} 	imes 16 = 125 	imes 2 = 250$

(ii) $(5^{-1} 	imes 2^{-1}) 	imes 6^{-1} = \frac{1}{5} 	imes \frac{1}{2} 	imes \frac{1}{6} = \frac{1}{60}$ — Use negative exponent rule and simplify numerator and denominator carefully. Recall $a^{-n} = \frac{1}{a^n}$.

Question 5

Find the value of $m$ for which $5^m \div 5^{-3} = 5^5$.

Accepted Answer

Solution:
Using the law of exponents:
$5^m \div 5^{-3} = 5^{m - (-3)} = 5^{m+3}$
Given that $5^{m+3} = 5^5$
Therefore, $m + 3 = 5 \Rightarrow m = 2$ — Equate the exponents since bases are same and solve for $m$.

Question 6

Evaluate (i)  $\left\{\left(\frac{1}{3}ight)^{-1} - \left(\frac{1}{4}ight)^{-1}ight\}^{-1}$  (ii)  $\left(\frac{5}{8}ight)^{-7} 	imes \left(\frac{8}{5}ight)^{-4}$

Accepted Answer

Solution:
(i) $\left\{ \left( \frac{1}{3} ight)^{-1} - \left( \frac{1}{4} ight)^{-1} ight\}^{-1} = \left\{ 3 - 4 ight\}^{-1} = (-1)^{-1} = -1$

(ii) $\left( \frac{5}{8} ight)^{-7} 	imes \left( \frac{8}{5} ight)^{-4} = \left( \frac{8}{5} ight)^7 	imes \left( \frac{5}{8} ight)^4 = \frac{8^7}{5^7} 	imes \frac{5^4}{8^4} = \frac{8^{7-4}}{5^{7-4}} = \frac{8^3}{5^3} = \left( \frac{8}{5} ight)^3 = \frac{512}{125}$ — Use negative exponent rule to invert bases and simplify powers by subtracting exponents when multiplying same bases.

Question 7

Simplify.

(i) $\frac{25 	imes t^{-4}}{5^{-3} 	imes 10 	imes t^{-8}} \quad (t 
eq 0)$

(ii) $\frac{3^{-5} 	imes 10^{-5} 	imes 125}{5^{-7} 	imes 6^{-5}}$

Accepted Answer

Solution:
(i) $\frac{25 	imes t^{-4}}{5^{-3} 	imes 10 	imes t^{-8}} = \frac{25}{10} 	imes \frac{t^{-4}}{t^{-8}} 	imes 5^{3} = \frac{5^2}{2 	imes 5} 	imes t^{-4 - (-8)} 	imes 5^{3} = \frac{5}{2} 	imes t^{4} 	imes 5^{3} = \frac{5}{2} 	imes t^{4} 	imes 125 = \frac{5 	imes 125}{2} t^{4} = \frac{625}{2} t^{4}$

(ii) $\frac{3^{-5} 	imes 10^{-5} 	imes 125}{5^{-7} 	imes 6^{-5}} = \frac{3^{-5} 	imes (2 	imes 5)^{-5} 	imes 5^{3}}{5^{-7} 	imes 6^{-5}} = \frac{3^{-5} 	imes 2^{-5} 	imes 5^{-5} 	imes 5^{3}}{5^{-7} 	imes 6^{-5}} = \frac{3^{-5} 	imes 2^{-5} 	imes 5^{-2}}{5^{-7} 	imes 6^{-5}} = 3^{-5} 	imes 2^{-5} 	imes 5^{-2 - (-7)} 	imes 6^{5} = 3^{-5} 	imes 2^{-5} 	imes 5^{5} 	imes 6^{5}$

Since $6 = 2 	imes 3$, $6^{5} = 2^{5} 	imes 3^{5}$

So expression becomes:
$3^{-5} 	imes 2^{-5} 	imes 5^{5} 	imes 2^{5} 	imes 3^{5} = 5^{5} 	imes (3^{-5} 	imes 3^{5}) 	imes (2^{-5} 	imes 2^{5}) = 5^{5} 	imes 1 	imes 1 = 5^{5} = 3125$ — Apply laws of exponents carefully, factorize numbers to prime bases to simplify powers, and combine like bases.

Question 8

TRY THESE

1. Write the following numbers in standard form.

(i) 0.000000564 (ii) 0.0000021 (iii) 21600000 (iv) 15240000

Accepted Answer

Solution:
(i) 0.000000564 = 5.64 	imes 10^{-7}

(ii) 0.0000021 = 2.1 	imes 10^{-6}

(iii) 21600000 = 2.16 	imes 10^{7}

(iv) 15240000 = 1.524 	imes 10^{7} — To write in standard form, move decimal point to get a number between 1 and 10 and multiply by 10 raised to the number of places moved (negative if moved right, positive if moved left).

Question 9

Write all the facts given in the standard form.

Accepted Answer

Solution:
Convert each given fact's number into standard form by expressing as a product of a number between 1 and 10 and a power of 10.

For example:
- Distance from Earth to Sun = 149,600,000,000 m = 1.496 	imes 10^{11} m
- Speed of light = 300,000,000 m/s = 3.0 	imes 10^{8} m/s
- Thickness of Class VII Mathematics book = 20 mm = 2.0 	imes 10^{1} mm
- Diameter of Red Blood Cell = 0.000007 mm = 7.0 	imes 10^{-6} mm
... and so on for all facts. — Apply the method of expressing numbers in standard form by shifting decimal and counting places.

Question 10

Express the following numbers in standard form.

(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

Accepted Answer

Solution:
(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 — To express a number in standard form, write it as a number between 1 and 10 multiplied by a power of 10.

(i) Move decimal point 12 places to the right: 8.5 × 10⁻¹²
(ii) Move decimal point 12 places to the right: 9.42 × 10⁻¹²
(iii) Move decimal point 15 places to the left: 6.02 × 10¹⁵
(iv) Move decimal point 9 places to the right: 8.37 × 10⁻⁹
(v) Move decimal point 10 places to the left: 3.186 × 10¹⁰

Question 11

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

Accepted Answer

Solution:
(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 — To express a number in usual form, move the decimal point according to the power of 10.

(i) 10⁻⁶ means move decimal 6 places left: 0.00000302
(ii) 10⁴ means move decimal 4 places right: 45000
(iii) 10⁻⁸ means move decimal 8 places left: 0.00000003
(iv) 10⁹ means move decimal 9 places right: 1000100000
(v) 10¹² means move decimal 12 places right: 5800000000000
(vi) 10⁶ means move decimal 6 places right: 3614920

Question 12

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

(i) 1 micron is equal to 1/1000000 m.

(ii) Charge of an electron is 0.00000000000000000016 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

Accepted Answer

Solution:
(i) 1 micron = 1/1000000 m = 1 × 10⁻⁶ m
(ii) Charge of electron = 0.00000000000000000016 = 1.6 × 10⁻¹⁹ coulomb
(iii) Size of bacteria = 0.0000005 = 5 × 10⁻⁷ m
(iv) Size of plant cell = 0.00001275 = 1.275 × 10⁻⁵ m
(v) Thickness of paper = 0.07 mm = 7 × 10⁻² mm — Convert each number to standard form by expressing it as a number between 1 and 10 multiplied by a power of 10.

(i) 1/1000000 = 1 × 10⁻⁶
(ii) Count decimal places: 1.6 × 10⁻¹⁹
(iii) Move decimal 7 places right: 5 × 10⁻⁷
(iv) Move decimal 5 places right: 1.275 × 10⁻⁵
(v) 0.07 = 7 × 10⁻²

Question 13

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.

Accepted Answer

Solution:
Thickness of 5 books = 5 × 20 mm = 100 mm
Thickness of 5 sheets = 5 × 0.016 mm = 0.08 mm
Total thickness = 100 mm + 0.08 mm = 100.08 mm — Multiply the number of books and sheets by their respective thicknesses and add the results to get total thickness.

5 books × 20 mm = 100 mm
5 sheets × 0.016 mm = 0.08 mm
Total thickness = 100 + 0.08 = 100.08 mm

Question 14

What is the value of $2^{5}$ expressed as a product of 2's?

Accepted Answer

2 × 2 × 2 × 2 × 2 — $2^{5}$ means multiplying 2 by itself 5 times, which is 2 × 2 × 2 × 2 × 2.

Question 15

Express $2^{m}$ in terms of multiplication for any positive integer $m$.

Accepted Answer

2 × 2 × 2 × ... × 2 (m times) / 2 multiplied by itself m times — By definition, $2^{m}$ means 2 multiplied by itself m times.

Chapter 10

Chapter 10 — Study Notes

10.1 Introduction

10.2 Powers with Negative Exponents

10.3 Laws of Exponents

Practice Questions — Chapter 10

All 13 Chapters in Mathematics