Exponents and Powers — Questions and Answers
Welcome to HSLC Guru. This page gives the complete solutions of ASSEB (Assam State School Education Board) Class 8 General Mathematics Chapter 12, Exponents and Powers, covering every question of Exercise 12.1 and Exercise 12.2 with full working, plus extra practice questions.
Summary
In this chapter we extend exponents to negative indices. For any non-zero integer $a$ and positive integer $m$, the number $a^{-m}$ is the multiplicative inverse of $a^{m}$, that is $a^{-m}=\frac{1}{a^{m}}$ and $a^{m}=\frac{1}{a^{-m}}$. For example $2^{-5}=\frac{1}{2^{5}}=\frac{1}{32}$.
For any non-zero rational numbers $a,\ b$ and integers $m,\ n$, the laws of indices hold:
- $a^{m}\times a^{n}=a^{m+n}$
- $a^{m}\div a^{n}=a^{m-n}$
- $(a^{m})^{n}=a^{mn}$
- $a^{m}\times b^{m}=(a\times b)^{m}$
- $\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}$
- $a^{0}=1$
Using exponents, very large and very small numbers are written in standard form (scientific notation) $K\times 10^{n}$, where $1 \le K < 10$ and $n$ is an integer. Moving the decimal point $n$ places to the left gives a positive power of ten, and moving it to the right gives a negative power. For example $5800 = 5.8\times 10^{3}$ and $0.00004 = 4\times 10^{-5}$.
Summary: This ASSEB Class 8 General Mathematics Chapter 12 Exponents and Powers solution explains negative exponents ($a^{-m}=\tfrac{1}{a^{m}}$), the laws of indices, expanded form using powers of ten, and how to write very large and very small numbers in standard (scientific) form $K\times 10^{n}$, with every question of Exercise 12.1 and Exercise 12.2 solved step by step.
Textbook Questions and Answers
Exercise 12.1
1. Evaluate: (i) $5^{-3}$ (ii) $(-4)^{-2}$ (iii) $(-4)^{-3}$ (iv) $\left(-\frac{5}{7}\right)^{5}$ (v) $\left(-\frac{5}{7}\right)^{-5}$ (vi) $\left(-\frac{1}{3}\right)^{8}$
Answer: (i) $5^{-3}=\frac{1}{5^{3}}=\frac{1}{125}$. (ii) $(-4)^{-2}=\frac{1}{(-4)^{2}}=\frac{1}{16}$. (iii) $(-4)^{-3}=\frac{1}{(-4)^{3}}=\frac{1}{-64}=-\frac{1}{64}$. (iv) $\left(-\frac{5}{7}\right)^{5}=\frac{(-5)^{5}}{7^{5}}=-\frac{3125}{16807}$. (v) $\left(-\frac{5}{7}\right)^{-5}=\left(-\frac{7}{5}\right)^{5}=\frac{(-7)^{5}}{5^{5}}=-\frac{16807}{3125}$. (vi) $\left(-\frac{1}{3}\right)^{8}=\frac{(-1)^{8}}{3^{8}}=\frac{1}{6561}$.
2. Express in exponential form: (i) $\frac{343}{125}$ (ii) $\frac{1}{288}$ (iii) $-\frac{27}{343}$ (iv) $-\frac{125}{216}$ (v) $-\frac{27}{16\times 49}$ (vi) $\frac{128}{81}$
Answer: (i) $\frac{343}{125}=\frac{7^{3}}{5^{3}}=\left(\frac{7}{5}\right)^{3}$. (ii) $\frac{1}{288}=\frac{1}{2^{5}\times 3^{2}}=2^{-5}\times 3^{-2}$. (iii) $-\frac{27}{343}=-\frac{3^{3}}{7^{3}}=\left(-\frac{3}{7}\right)^{3}$. (iv) $-\frac{125}{216}=-\frac{5^{3}}{6^{3}}=\left(-\frac{5}{6}\right)^{3}$. (v) $-\frac{27}{16\times 49}=-\frac{3^{3}}{2^{4}\times 7^{2}}$. (vi) $\frac{128}{81}=\frac{2^{7}}{3^{4}}$.
3. Simplify and express the result in power notation with positive exponent: (i) $(-2)^{4}\times\left(\frac{3}{2}\right)^{4}$ (ii) $\left(-\frac{2}{3}\right)^{4}\times\left(-\frac{3}{4}\right)^{3}$ (iii) $5^{-7}\times\left(\frac{1}{5}\right)^{3}$ (iv) $3^{-5}\times(-2)^{-5}\times(-4)^{-5}$
Answer: (i) $(-2)^{4}\times\left(\frac{3}{2}\right)^{4}=2^{4}\times\frac{3^{4}}{2^{4}}=3^{4}$. (ii) $\left(-\frac{2}{3}\right)^{4}\times\left(-\frac{3}{4}\right)^{3}=\frac{2^{4}}{3^{4}}\times\left(-\frac{3^{3}}{2^{6}}\right)=-\frac{1}{2^{2}\times 3}=-\frac{1}{12}$. (iii) $5^{-7}\times\left(\frac{1}{5}\right)^{3}=5^{-7}\times 5^{-3}=5^{-10}=\frac{1}{5^{10}}$. (iv) $3^{-5}\times(-2)^{-5}\times(-4)^{-5}=\left[3\times(-2)\times(-4)\right]^{-5}=24^{-5}=\frac{1}{24^{5}}$.
4. Find the value: (i) $\left(\frac{1}{3}\right)^{4}\times\left(-\frac{3}{5}\right)^{3}\times\left(-\frac{7}{9}\right)^{2}$ (ii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}+\left(\frac{1}{5}\right)^{-2}$ (iii) $\left(\frac{1}{3}\right)^{0}+\left(\frac{1}{3}\right)^{-1}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{3}\right)^{-3}$ (iv) $(5^{-1}+3^{-1}+7^{-2})^{0}$ (v) $(5^{-1}\times 2^{-1})\times 6^{-1}$ (vi) $(3^{-2})^{-3}$
Answer: (i) $\left(\frac{1}{3}\right)^{4}\times\left(-\frac{3}{5}\right)^{3}\times\left(-\frac{7}{9}\right)^{2}=3^{-4}\times\left(-\frac{3^{3}}{5^{3}}\right)\times\frac{7^{2}}{3^{4}}=-\frac{7^{2}}{3^{5}\times 5^{3}}=-\frac{49}{30375}$. (ii) $=2^{2}+3^{2}+4^{2}+5^{2}=4+9+16+25=54$. (iii) $=1+3+9+27=40$. (iv) Any non-zero number raised to power $0$ is $1$, so $(5^{-1}+3^{-1}+7^{-2})^{0}=1$. (v) $(5^{-1}\times 2^{-1})\times 6^{-1}=10^{-1}\times 6^{-1}=\frac{1}{60}$. (vi) $(3^{-2})^{-3}=3^{(-2)\times(-3)}=3^{6}=729$.
5. Write the multiplicative inverse of: (i) $3^{4}$ (ii) $\left(\frac{2}{3}\right)^{6}$ (iii) $\left(-\frac{4}{9}\right)^{50}$ (iv) $\left(\frac{3}{4}\right)^{-5}$ (v) $\left(-\frac{2}{3}\right)^{-7}$ (vi) $\left(\frac{3}{8}\right)^{-4}$
Answer: The multiplicative inverse of $a^{m}$ is $a^{-m}$. (i) $3^{4}\Rightarrow 3^{-4}=\frac{1}{81}$. (ii) $\left(\frac{2}{3}\right)^{6}\Rightarrow\left(\frac{3}{2}\right)^{6}=\frac{729}{64}$. (iii) $\left(-\frac{4}{9}\right)^{50}\Rightarrow\left(-\frac{9}{4}\right)^{50}=\left(\frac{9}{4}\right)^{50}$. (iv) $\left(\frac{3}{4}\right)^{-5}\Rightarrow\left(\frac{3}{4}\right)^{5}=\frac{243}{1024}$. (v) $\left(-\frac{2}{3}\right)^{-7}\Rightarrow\left(-\frac{2}{3}\right)^{7}=-\frac{128}{2187}$. (vi) $\left(\frac{3}{8}\right)^{-4}\Rightarrow\left(\frac{3}{8}\right)^{4}=\frac{81}{4096}$.
6. Simplify using laws of indices: (i) $\left(-\frac{4}{5}\right)^{3}\times\left(-\frac{4}{5}\right)^{2}\times\left(-\frac{4}{5}\right)$ (ii) $\left(\frac{5}{3}\right)^{0}\times\left(\frac{5}{3}\right)^{-3}\times\left(\frac{5}{3}\right)^{-2}$ (iii) $\left\{\left(-\frac{5}{3}\right)^{15}\times\left(-\frac{5}{3}\right)^{-8}\right\}\div\left(-\frac{5}{3}\right)^{6}$ (iv) $\left(-\frac{3}{2}\right)^{-5}\times\left(-\frac{3}{2}\right)^{-7}\times\left(-\frac{2}{3}\right)^{8}\times\left(-\frac{2}{3}\right)^{4}$ (v) $(3^{-4})^{-2}\times(3^{-5})^{2}\div(3^{-2})^{-3}$
Answer: (i) $\left(-\frac{4}{5}\right)^{3+2+1}=\left(-\frac{4}{5}\right)^{6}=\frac{4^{6}}{5^{6}}=\frac{4096}{15625}$. (ii) $\left(\frac{5}{3}\right)^{0+(-3)+(-2)}=\left(\frac{5}{3}\right)^{-5}=\left(\frac{3}{5}\right)^{5}=\frac{243}{3125}$. (iii) $\left(-\frac{5}{3}\right)^{15+(-8)-6}=\left(-\frac{5}{3}\right)^{1}=-\frac{5}{3}$.
(iv) Grouping like bases gives $\left(-\frac{3}{2}\right)^{-12}\times\left(-\frac{2}{3}\right)^{12}$. Since $\left(-\frac{3}{2}\right)^{-12}=\left(-\frac{2}{3}\right)^{12}$, the product is $\left(-\frac{2}{3}\right)^{12}\times\left(-\frac{2}{3}\right)^{12}=\left(-\frac{2}{3}\right)^{24}=\frac{2^{24}}{3^{24}}$. (v) $(3^{-4})^{-2}\times(3^{-5})^{2}\div(3^{-2})^{-3}=3^{8}\times 3^{-10}\div 3^{6}=3^{8-10-6}=3^{-8}=\frac{1}{3^{8}}=\frac{1}{6561}$.
7. (i) If $\left(\frac{5}{7}\right)^{-7}\times\left(\frac{7}{5}\right)^{-9}=\left(\frac{5}{7}\right)^{2m}$, find $m$. (ii) If $\left(\frac{9}{49}\right)^{-5}\times\left(\frac{9}{49}\right)^{7}=\left(\frac{9}{49}\right)^{-6k}$, find $k$. (iii) If $(1.4)^{8}\times(1.4)^{5}=(1.4)^{3}\times(1.4)^{k}$, find $k$. (iv) Find $m$ such that $5^{m}\div 5^{-3}=5^{5}$.
Answer: (i) Since $\left(\frac{7}{5}\right)^{-9}=\left(\frac{5}{7}\right)^{9}$, LHS $=\left(\frac{5}{7}\right)^{-7+9}=\left(\frac{5}{7}\right)^{2}$, so $2=2m\Rightarrow m=1$. (ii) LHS $=\left(\frac{9}{49}\right)^{-5+7}=\left(\frac{9}{49}\right)^{2}$, so $2=-6k\Rightarrow k=-\frac{1}{3}$. (iii) $(1.4)^{13}=(1.4)^{3+k}\Rightarrow 13=3+k\Rightarrow k=10$. (iv) $5^{m-(-3)}=5^{5}\Rightarrow 5^{m+3}=5^{5}\Rightarrow m+3=5\Rightarrow m=2$.
8. Simplify: (i) $\frac{125\times 3^{-4}\times 2^{2}}{5^{-4}\times 100\times 3^{-7}}$ (ii) $\frac{3^{2k}\times 27\times 9^{-3}}{81^{-2k}\times 3^{-4}\times 3^{5}}$ (iii) $\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$ (iv) $\frac{25\times l^{-4}}{5^{-3}\times 10\times l^{-8}}$ ($l\neq 0$) (v) $\frac{2^{m+2}\times 3^{2m-n}\times 6^{n}}{6^{m}\times 2^{n}\times 4\times 3^{m}}$
Answer (i): Writing $125=5^{3}$ and $100=2^{2}\times 5^{2}$,
$$\frac{5^{3}\times 3^{-4}\times 2^{2}}{5^{-4}\times 2^{2}\times 5^{2}\times 3^{-7}}=5^{3-(-2)}\times 3^{-4-(-7)}\times 2^{0}=5^{5}\times 3^{3}=84375$$
(ii) With $27=3^{3}$, $9^{-3}=3^{-6}$, $81^{-2k}=3^{-8k}$: $=\frac{3^{2k+3-6}}{3^{-8k-4+5}}=\frac{3^{2k-3}}{3^{-8k+1}}=3^{(2k-3)-(-8k+1)}=3^{10k-4}$.
(iii) Using $10^{-5}=2^{-5}\times 5^{-5}$, $125=5^{3}$, $6^{-5}=2^{-5}\times 3^{-5}$: $=\frac{2^{-5}\times 3^{-5}\times 5^{-2}}{5^{-7}\times 2^{-5}\times 3^{-5}}=5^{-2-(-7)}=5^{5}=3125$. (iv) With $25=5^{2}$, $10=2\times 5$: $=\frac{5^{2}\times l^{-4}}{2\times 5^{-2}\times l^{-8}}=\frac{5^{4}}{2}\times l^{4}=\frac{625\,l^{4}}{2}$. (v) Using $6^{n}=2^{n}3^{n}$, $6^{m}=2^{m}3^{m}$, $4=2^{2}$, both numerator and denominator become $2^{m+n+2}\times 3^{2m}$, so the value is $1$.
9. Express in expanded form: (i) $15737.348$ (ii) $35792.39$
Answer (i):
$$15737.348 = 1\times 10^{4}+5\times 10^{3}+7\times 10^{2}+3\times 10^{1}+7\times 10^{0}+3\times 10^{-1}+4\times 10^{-2}+8\times 10^{-3}$$
(ii)
$$35792.39 = 3\times 10^{4}+5\times 10^{3}+7\times 10^{2}+9\times 10^{1}+2\times 10^{0}+3\times 10^{-1}+9\times 10^{-2}$$
Exercise 12.2
1. Express the following in standard form: (i) $35700000$ (ii) $705030000$ (iii) $37800.35$ (iv) $5362.8\times 10^{6}$ (v) $4003.2\times 10^{5}$
Answer: (i) $35700000=3.57\times 10^{7}$. (ii) $705030000=7.0503\times 10^{8}$. (iii) $37800.35=3.780035\times 10^{4}$. (iv) $5362.8\times 10^{6}=5.3628\times 10^{3}\times 10^{6}=5.3628\times 10^{9}$. (v) $4003.2\times 10^{5}=4.0032\times 10^{3}\times 10^{5}=4.0032\times 10^{8}$.
2. Express the following in standard form: (i) $0.0000000382$ (ii) $0.00000009057$ (iii) $0.00000756$ (iv) $0.00023\times 10^{-2}$ (v) $0.000314\times 10^{-3}$
Answer: (i) $0.0000000382=3.82\times 10^{-8}$. (ii) $0.00000009057=9.057\times 10^{-8}$. (iii) $0.00000756=7.56\times 10^{-6}$. (iv) $0.00023\times 10^{-2}=2.3\times 10^{-4}\times 10^{-2}=2.3\times 10^{-6}$. (v) $0.000314\times 10^{-3}=3.14\times 10^{-4}\times 10^{-3}=3.14\times 10^{-7}$.
3. Express the following in usual form: (i) $7.02\times 10^{5}$ (ii) $3.972\times 10^{7}$ (iii) $1.001\times 10^{8}$ (iv) $3\times 10^{-8}$ (v) $2.1\times 10^{-6}$ (vi) $3.09\times 10^{-5}$
Answer: (i) $7.02\times 10^{5}=702000$. (ii) $3.972\times 10^{7}=39720000$. (iii) $1.001\times 10^{8}=100100000$. (iv) $3\times 10^{-8}=0.00000003$. (v) $2.1\times 10^{-6}=0.0000021$. (vi) $3.09\times 10^{-5}=0.0000309$.
4. Write the statement in standard form:
- (i) Speed of light is $300000$ km per second $= 3\times 10^{5}$ km/s.
- (ii) Distance between the sun and Saturn is $1{,}433{,}500{,}000{,}000$ m $= 1.4335\times 10^{12}$ m.
- (iii) $18$ g water contains $602{,}300{,}000{,}000{,}000{,}000{,}000{,}000$ molecules $= 6.023\times 10^{23}$.
- (iv) Diameter of a molecule is $0.000000015$ cm $= 1.5\times 10^{-8}$ cm.
- (v) Size of a virus is $0.0000005$ m $= 5\times 10^{-7}$ m.
- (vi) Diameter of a wire is $0.0000032$ m $= 3.2\times 10^{-6}$ m.
- (vii) $1$ micron $=\frac{1}{1000000}$ m $= 1\times 10^{-6}$ m.
5. Express the following in standard form and arrange in ascending order: $925\times 10^{4}$; $94.2\times 10^{5}$; $875\times 10^{5}$; $87.5\times 10^{4}$.
Answer: In standard form: $925\times 10^{4}=9.25\times 10^{6}$; $94.2\times 10^{5}=9.42\times 10^{6}$; $875\times 10^{5}=8.75\times 10^{7}$; $87.5\times 10^{4}=8.75\times 10^{5}$. Arranged in ascending order:
$$8.75\times 10^{5} < 9.25\times 10^{6} < 9.42\times 10^{6} < 8.75\times 10^{7}$$
That is, $87.5\times 10^{4} < 925\times 10^{4} < 94.2\times 10^{5} < 875\times 10^{5}$.
6. Add: (i) $3.04\times 10^{11}+5.02\times 10^{10}$ (ii) $6.03\times 10^{7}+6.03\times 10^{8}$
Answer: (i) Converting to the same exponent, $3.04\times 10^{11}=30.4\times 10^{10}$, so $30.4\times 10^{10}+5.02\times 10^{10}=(30.4+5.02)\times 10^{10}=35.42\times 10^{10}=3.542\times 10^{11}$. (ii) $6.03\times 10^{8}=60.3\times 10^{7}$, so $6.03\times 10^{7}+60.3\times 10^{7}=66.33\times 10^{7}=6.633\times 10^{8}$.
7. Subtract: (i) $6.47\times 10^{8}-3.15\times 10^{6}$ (ii) $3.76\times 10^{7}-3.76\times 10^{5}$
Answer: (i) $6.47\times 10^{8}=647\times 10^{6}$, so $647\times 10^{6}-3.15\times 10^{6}=(647-3.15)\times 10^{6}=643.85\times 10^{6}=6.4385\times 10^{8}$. (ii) $3.76\times 10^{7}=376\times 10^{5}$, so $376\times 10^{5}-3.76\times 10^{5}=(376-3.76)\times 10^{5}=372.24\times 10^{5}=3.7224\times 10^{7}$.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The value of $2^{-3}$ is (a) $8$ (b) $-8$ (c) $\frac{1}{8}$ (d) $-\frac{1}{8}$
Answer: (c) $\frac{1}{8}$.
2. For $a\neq 0$, $a^{0}$ equals (a) $0$ (b) $1$ (c) $a$ (d) undefined
Answer: (b) $1$.
3. $\frac{1}{10^{-3}}$ equals (a) $10^{3}$ (b) $10^{-3}$ (c) $\frac{1}{1000}$ (d) $30$
Answer: (a) $10^{3}$.
4. $a^{m}\times a^{n}$ equals (a) $a^{mn}$ (b) $a^{m+n}$ (c) $a^{m-n}$ (d) $a^{m/n}$
Answer: (b) $a^{m+n}$.
5. The standard form of $57300$ is (a) $5.73\times 10^{3}$ (b) $5.73\times 10^{4}$ (c) $57.3\times 10^{3}$ (d) $0.573\times 10^{5}$
Answer: (b) $5.73\times 10^{4}$.
6. $\left(\frac{2}{3}\right)^{-2}$ equals (a) $\frac{4}{9}$ (b) $\frac{9}{4}$ (c) $-\frac{4}{9}$ (d) $\frac{2}{3}$
Answer: (b) $\frac{9}{4}$.
7. $(-1)^{-7}$ equals (a) $1$ (b) $-1$ (c) $7$ (d) $-7$
Answer: (b) $-1$.
8. The multiplicative inverse of $3^{4}$ is (a) $3^{4}$ (b) $-3^{4}$ (c) $3^{-4}$ (d) $4^{-3}$
Answer: (c) $3^{-4}$.
9. $(2^{2})^{3}$ equals (a) $2^{5}$ (b) $2^{6}$ (c) $2^{8}$ (d) $2^{9}$
Answer: (b) $2^{6}$.
10. The standard form of $0.0004$ is (a) $4\times 10^{-4}$ (b) $4\times 10^{4}$ (c) $4\times 10^{-3}$ (d) $0.4\times 10^{-3}$
Answer: (a) $4\times 10^{-4}$.
Fill in the Blanks
- $a^{-m}=$ ______. Answer: $\frac{1}{a^{m}}$.
- $10^{-2}=$ ______. Answer: $0.01$.
- In standard form $K\times 10^{n}$, $1\le K <$ ______. Answer: $10$.
- $2^{5}\times 2^{-5}=$ ______. Answer: $1$.
- $5^{0}=$ ______. Answer: $1$.
True or False
- $2^{-1}=\frac{1}{2}$. Answer: True.
- $(-3)^{-2}$ is a negative number. Answer: False (it is $\frac{1}{9}$, positive).
- $a^{m}\div a^{n}=a^{m+n}$. Answer: False (it is $a^{m-n}$).
- When a very small number is written in standard form, the power of $10$ is negative. Answer: True.
- $(2\times 3)^{2}=2^{2}\times 3^{2}$. Answer: True.
Short Answer Questions
1. Define $a^{-m}$.
Answer: For a non-zero integer $a$ and positive integer $m$, $a^{-m}=\frac{1}{a^{m}}$; it is the multiplicative inverse of $a^{m}$.
2. Verify the law $a^{m}\times a^{n}=a^{m+n}$ for $a=2,\ m=-3,\ n=5$.
Answer: $2^{-3}\times 2^{5}=\frac{1}{8}\times 32=4$; also $2^{-3+5}=2^{2}=4$. Both are equal, so the law holds.
3. Write $384000000$ in standard form.
Answer: $384000000=3.84\times 10^{8}$.
4. Write $6\times 10^{-4}$ in usual form.
Answer: $6\times 10^{-4}=0.0006$.
Key Terms
| Term | Meaning |
|---|---|
| Exponent / Index | The number showing how many times the base is multiplied by itself |
| Power | The combined form of a base and an exponent, e.g. $2^{5}$ |
| Base | The number that is multiplied repeatedly |
| Negative exponent | An exponent of the form $a^{-m}=\frac{1}{a^{m}}$ |
| Multiplicative inverse | A number whose product with the given number is $1$ |
| Laws of indices | Rules for multiplying, dividing and raising powers |
| Standard form | $K\times 10^{n}$, where $1\le K<10$ |
| Scientific notation | Another name for standard form |
| Expanded form | A number written as the sum of place values as powers of ten |
| Ascending order | Arranging numbers from smallest to largest |