Exponents And Powers
Exponents And Powers Introduction
If a is any non-zero integer and n is a positive integer, then
x a (n times) is written as a”,
i.e., a” is the continued product of a multiplied by itself n times.
Here, ‘o’ is called the base, and V is called the ‘exponent’ or ‘index’.
The number a” is read as a raised to the power of or simply as ‘nth power of o’.
The notation a” is called the exponential or power notation.
We can write large numbers more conveniently using exponents.
For example :
10000 = 104; 243 = 35; 128 = 27, etc.
Now, we shall learn about negative exponents
Read and Learn More NCERT Solutions For Class 8 Maths
Powers With Negative Exponents
If a is any non-zero integer and m is a positive integer, then
⇒ \(a^{-m}=\frac{1}{a^m}\)
Note: a m is called the multiplicative inverse of am as a-m x am = 1
Am and a ~m are multiplicative inverses of each other.
Note 2: \(a^m=\frac{1}{a^{-m}}\)
Question: What is 10 ~10 equal to?
Solution: \(10^{-10}=\frac{1}{10^{10}}\)
Question 1. Find the multiplicative inverse of the following:
- 2-4
- 10-5
- 7-2
- 5-3
- 10-100
Solution:
The multiplicative inverse is \(2^{-4}=\left(\frac{1}{2^{-4}}\right) \text { is } 2^4\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
The multiplicative inverse of \(10^{-5}=\left(\frac{1}{10^{-5}}\right) \text { is } 10^5\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
The multiplicative inverse of \(7^{-2}=\left(\frac{1}{7^{-2}}\right) \text { is } 7^2\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
The multiplicative inverse \(5^{-3}=\left(\frac{1}{5^{-3}}\right) \text { is } 5^3\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
The multiplicative inverse\(10^{-100}=\left(\frac{1}{10^{-100}}\right) \text { is } 10^{100}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
Question 2. Expand the following numbers using exponents :
- 1025.63
- 1256.249
Solution:
1025.63
1 x 1000 + 0 x 100 + 2 x 10 + 5 x 1 + 6 x\(\frac{1}{10}+3 \times \frac{1}{100}\)
= 1 x 103 + 0 x 102 + 2 x 101 + 5 x 10° + 6 x 10-1 + 3 x 10-2
1256.249
= 1 x 1000 + 2 x 100 + 5 x 10
+ 6x 1+ 2 x \( \frac{1}{10}+4 \times \frac{1}{100}+9 \times \frac{1}{1000}\)
= 1 X 103 + 2 X 102 + 5 X 101 + 6 x 10° + 2 x 10-1 + 4 x 10-2 + 9 x lO-3
Laws Of Exponents
If a, b are non-zero integers and m, n are any integers, then
- am x an = am+ n
- \(\frac{a^m}{a^n}=a^{m-n}\)
- (am)n = amn
- am x bm = (ab)m
- \(\frac{a^m}{b^m}=\left(\frac{a}{b}\right)^m\)
- a° = 1
- \(\left(\frac{a^{-m}}{b^{-n}}\right)=\frac{b^n}{a^m}\)
- \(\left(\frac{a}{b}\right)^{-m}=\left(\frac{b}{a}\right)^m\)
Remember
an” = 1 = n = 0
1n = 1 where n is any integer.
(- 1)n = 1 where n is any even integer.
(-1)n =-1 where n is any odd integer.
Q. Simplify and write in exponential form:
(-2)-3X (-2)-4
p3 x p -I0
32 x 3-5 x 36
Solution:
(-2)~3 x (-2)-4 = (- 2)(“3) + (-4>
am x an = am +n
= (-2)-7 = {(-1) x 2} -7
⇒ \(\frac{1}{\{(-1) \times 2\}^7}= \frac{1}{(-1)^7 \times(2)^7}\)
(ab)m = am bm
⇒ \(\frac{1}{(-1) \times 2^7}=-\frac{1}{2^7}\)
⇒ \(\mid(-1)^{\text {odd integer }}=-1\)
⇒ \(p^3 \times p^{-10}=p^{3+(-10)}=p^{-7}=\frac{1}{p^7}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
32 x 3-5 x 36 = 32 + (-5) + 6 = 33
Exponents And Powers Exercise 10.1
Question 1. Evaluate:
- 3-2
- (-4)-2
- \(\left(\frac{1}{2}\right)^{-5}\)
Solution:
⇒ \(3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \((-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(\left(\frac{1}{2}\right)^{-5}=\left(\frac{2}{1}\right)^5\)
⇒ \( a^{-m}=\frac{1}{a^m}\)
⇒ \(\frac{2^5}{1^5}\)
⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
⇒ \(\frac{32}{1}=32\)
1n = 1 where n is an integer
Question 2. Simplify and express the result in power notation with a positive exponent
- \((-4)^5 \div(-4)^8\)
- \(\left(\frac{1}{2^3}\right)^2\)
- \((-3)^4 \times\left(\frac{5}{3}\right)^4\)
- (3-7 – 3-10) x 3-5
- 2-3 x (-7) -3
Solution:
1. (-4)6 ÷ (-4)8
⇒ \(\frac{(-4)^5}{(-4)^8}\)
⇒ \((-4)^{5-8}\)
⇒ \((-4)^{-3}\frac{a^m}{a^n}=a^{m-n}\)
⇒ \(=\frac{1}{(-4)^3} a^{-m}=\frac{1}{a^m}\)
which is the required form.
2. \(\left(\frac{1}{2^3}\right)^2 =\frac{1^2}{\left(2^3\right)^2}\)
⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
⇒ \(\frac{1}{2^{3 \times 2}} \)
⇒ \( \mid\left(a^m\right)^n=a^m\)
⇒ \(\frac{1}{2^6}\)
which is the required form
3. \((-3)^4 \times\left(\frac{5}{3}\right)^4=\{(-1) \times 3\}^4 \times \frac{5^4}{3^4} \)
⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
⇒ \((-1)^4(3)^4 \times \frac{5^4}{3^4}\)
I (ab)m = am bm
= (- 1)4 x 54
= 1 x 54
(-1) even integer = 1
= 54
which is the required form.
4. (3-7 / 3-10) x 3-5
⇒ \(=\frac{3^{-7}}{3^{-10}} \times \frac{1}{3^5}\)
⇒ \( a^{-m}=\frac{1}{a^m}\)
⇒ \(3^{-7-(-10)} \times \frac{1}{3^5}\)
⇒ \(\frac{a^m}{a^n}=a^{m-n}\)
⇒ \(3^3 \times \frac{1}{3^5} \)
⇒ \(\frac{a^m}{a^n}=a^{m-n}\)
⇒ \(3^{3-5} \)
⇒ \(3^{-2} \)
⇒ \(a^{-m}=\frac{1}{a^m}\)
which is the required form.
5. \(2^{-3} \times(-7)^{-3}\)
⇒ \(\frac{1}{2^3} \times \frac{1}{(-7)^3}\)
⇒ \(\frac{1}{[2 \times(-7)]^3}=\frac{1}{a^m}\)
(ab)m = amb
⇒ \(\frac{1}{(-14)^3}\)
which is the required form.
Question 3. Find the value of:
(3° + 4 -1) x 22
(2-1 x 4-1)/2 -2
⇒ \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)
⇒ \(\left(3^{-1}+4^{-1}+5^{-1}\right)^0\)
⇒ \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2.\)
Solution:
1. (3° + 4-1) x 22
⇒ \(\left(1+\frac{1}{4}\right) \times 4\)
⇒ \(a^{-m}=\frac{1}{a^m}, \quad a^0=1\)
⇒ \(\frac{5}{4} \times 4=5\)
(2-1 x 4-1) + 2″2
= {2 -l x (22)-1} + 2 -2
= {2-1 x 22x(-1>} -f2-2
(am)n = amn
= (2-’X2-2)T2-2 = 2(-1)+(-2) 2-2
am x a” = am+”
= 2-3 4- 2″2
⇒ \(\frac{2^{-3}}{2^{-2}}=2^{-3-(-2)} \)
⇒ \(\frac{a^m}{a^n}=a^{m-n}\)
⇒ \(2^{-1}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(\frac{1}{2}\)
3.\(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)
⇒ \(\frac{1^{-2}}{2^{-2}}+\frac{1^{-2}}{3^{-2}}+\frac{1^{-2}}{4^{-2}}\)
⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
⇒ \(\frac{2^2}{1^2}+\frac{3^2}{1^2}+\frac{4^2}{1^2}\)
⇒ \(\frac{4}{1}+\frac{9}{1}+\frac{16}{1}\)
= 4 + 9 + 16 = 29
4. [3-1 + 4-1 + 5-1]0
⇒ \({\left[\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right]^0 }\)
⇒ \(\left(\frac{20+15+12}{60}\right)^0\)
⇒ \(\quad \text { | LCM }(3,4,5)=60\)
⇒ \(\left(\frac{47}{60}\right)^0=1\)
\(a^0=1\)Aliter
(3-1 + 4-1 +5-1)° = 1
a°= 1
5. \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2\)
⇒ \(\left(\frac{-2}{3}\right)^{(-2) \times 2}\)
⇒ \(\mid\left(a^m\right)^n=a^{m n}\)
⇒ \(\left(\frac{-2}{3}\right)^{-4}\)
⇒ \(\left(\frac{3}{-2}\right)^4 \)
⇒ \( \left(\frac{a}{b}\right)^{-m}=\left(\frac{b}{a}\right)^m\)
⇒ \(\frac{3^4}{(-2)^4}=\frac{3^4}{(-1 \times 2)^4}\)
⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
⇒ \(\frac{3^4}{(-1)^4(2)^4}\)
⇒ \((a b)^m=a^m b^m\)
⇒ \(\frac{3^4}{1 \times 2^4} \)
⇒ \((-1)^{\text {even integer }}=1\)
⇒ \(\frac{3^4}{2^4}\)
⇒ \(\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2}=\frac{81}{16}\)
Question 4. Evaluate:
- \(\frac{8^{-1} \times 5^3}{2^{-4}}\)
- \(\left(5^{-1} \times 2^{-I}\right) \times 6^{-1}\)
Solution:
1. \(\frac{8^{-1} \times 5^3}{2^{-4}}\)
⇒ \(\frac{2^4 \times 5^3}{8^1}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(\frac{16 \times 125}{8}=250\)
2. \(\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}=\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(\frac{1}{10} \times \frac{1}{6}=\frac{1}{60}\)
Question 5. Find the value of m for which 5m + 5-3 = 55
Solution:
5m + 5-3 = 55
⇒ \(\frac{5^m}{5^{-3}} =5^5\)
⇒ \(5^{m-(-3)} =5^5\)
⇒ \(\frac{a^m}{a^n}=a^{m-n}\)
⇒ \(5^{m+3} =5^5\)
bases are equal exponents are equal
m + 3 = 5
m = 5-3
m = 2
Question 6. Evaluate:
- \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
- \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}.\)
Solution:
⇒ \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
⇒ \(\left(\frac{1^{-1}}{3^{-1}}-\frac{1^{-1}}{4^{-1}}\right)^{-1}\)
⇒ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
⇒ \(\left\{\frac{\frac{1}{1^1}}{\frac{1}{3^1}}-\frac{\frac{1}{1^1}}{\frac{1}{4^1}}\right\}^{-1}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(\left(\frac{3^1}{1^1}-\frac{4^1}{1^1}\right)^{-1}\)
⇒ \(a^1=a\)
⇒ \(\left(\frac{3}{1}-\frac{4}{1}\right)^{-1}\)
⇒ \( (3-4)^{-1}\)
⇒ \((-1)^{-1}=\frac{1}{(-1)^1}\)
⇒ \(| a^{-m}=\frac{1}{a^m}\)
⇒ \(\frac{1}{(-1)}\)
\((-1)^{\text {odd integer }}=-1\)= -1
⇒ \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}=\frac{5^{-7}}{8^{-7}} \times \frac{8^{-4}}{5^{-4}}\)
⇒ \(\frac{5^{-7}}{5^{-4}} \times \frac{8^{-4}}{8^{-7}}\)
⇒ \(=5^{(-7)-(-4)} \times 8^m=\frac{a^m}{b^m}\)
⇒ \(\frac{a^m}{a^n}=a^{m-n}\)
⇒ \(5^{-7+4} \times 8^{-4}+7 \)
⇒ \(\frac{1}{5^3} \times 8^3=\frac{8^3}{5^3}\)
⇒ \(\frac{8 \times 8 \times 8}{5 \times 5 \times 5}=\frac{512}{125}\)
Question 7. Simplify:
- \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \quad(t \neq 0)\)
- \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Solution:
1. \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\)
⇒ \(\frac{25 \times 5^3}{10} \times \frac{t^8}{t^4}\)
⇒ \(\frac{625}{2} \times \frac{t^8}{t^4}\)
⇒ \(\frac{625 t^4}{2}\)
⇒ \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
⇒ \(\frac{3^{-5} \times(2 \times 5)^{-5} \times(5 \times 5 \times 5)}{5^{-7} \times(2 \times 3)^{-5}}\)
⇒ \(\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}} \)
⇒ \(\quad(a b)^m=a^m b^m\)
⇒ \(\frac{5^{-5} \times 5^3}{5^{-7}}=\frac{5^{(-5)+3}}{5^{-7}}\)
⇒ \(\mid a^m \times a^n=a^{m+n}\)
⇒ \(\frac{5^{-2}}{5^{-7}}=5^{(-2)-(-7)}\)
⇒ \(5^{-2+7}=5^5\)
Use Of Exponents To Express Small Numbers In Standard Form
A number is said to be in standard form if expressed in the form K x 10″ where 1 < K < 10 and n is an integer. A number written in standard form is said to be expressed in scientific notation.
Tiny numbers can be expressed in standard form using negative exponents.
1. To express a large number, we move the decimal point to the left such that only one digit is left to the left side of the decimal point and multiply the resulting number by 10n where n is the number of places to which the decimal point has been moved to the left.
For example : 270,000,000,000 = 2.7 x 1011
(Decimal point has have been moved to the left for11 places)
2. To express a number (< 1), we move the decimal point to the right such that only one digit is left to the left side of the decimal point and multiply the resulting number by 10~’1 where n is the number of places to which the decimal point has been shifted to the right.
For example : 0.000 0009 = 9 x 10-7
(Decimal point has have been shifted to the right for 7 places.)
Question 1. Identify huge and very small numbers from the above facts and write them in the following table:
Solution:
Question 2. Write the following numbers in standard form:
- 0.000000564
- 0.0000021
- 21600000
- 15240000
Solution:
1. 0.000000564
= 5.64 x 10-7
Moving decimal 7 places to the right
2. 0.0000021
0.0000021 = 2.1 x 10-6
Moving decimal 6 places to the right
3. 21600000
21600000 = 2.16 x 107
Moving decimal 7 places to the left
4. 15240000
15240000 = 1.524 x 107
Moving decimal 7 places to the left
Question 2. Write all the facts given in the standard form.
Solution:
(1) The distance from the Earth to the Sun is 1.496 x 1011 m
149, 600,000,000 = 1.496 x 1011.
Moving the decimal 11 places to the left
(2) The speed of light is 3 x 108 m/sec.
300, 000, 000 = 3 x108.
Moving decimal 8 places to the left
The thickness of the Class VII Mathematics book is 2 x 101 mm.
20 = 2 x 10 = 2 x 101
The average diameter of a Red Blood Cell is 7 x 10 6 mm.
0.00 0007 = 7 x 10-6
Moving decimal 6 places to the right
The thickness of human hair is in the range of 5 x 10 -3 cm to 1 x I0″2cm
⇒ \(0.005=\frac{5}{1000}=\frac{5}{10^3}=5 \times 10^{-3}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(0.01=\frac{1}{100}=\frac{1}{10^2}=1 \times 10^{-2}\)
The distance of the moon from the Earth is 3.84467 x 108m
384,467,000 = 3.84467 x 108
I Moved the decimal 8 places to the left
(7) The size of a plant cell is 1.275 x 10-5m
0.00001275 = 1.275 x 105
Moving decimal 5 places to the right
The average radius of the Sun is 6.95 x105km
695000 = 695 x 1000 = 695 x 103 = 6.95 x 105
(9) Mass of fuel in a space shuttle
solid rocket booster is 5.036 x 105 kg
503600
= 5036 x 100 = 5036 x 102
= 5.036 x 103 x 102
= 5.036 x 103+2
am x an = am+n
= 5.036 x 105
(10) Thickness of a piece of paper is 1.6 x10-3 cm
0.0016 = 1.6 x 10-3
Moving decimal 3 places to the right
The diameter of wire on a computer chip is 3 x 10-6 m
is 3 x 10-6
0.000003 = 3 x 013
Moving decimal 6 places to the right
(12) The height of Mount Everest is 8.848 x 103 m.
8848 = 8.848 x 1000 = 8.848 x 103
Exponents And Powers Exercise 10.2
Question 1. Express the following numbers in standard form:
- 0.0000000000085
- 0.00000000000942
- 6020000000000000
- 0.00000000837
- 31860000000.
Solution:
1. 0.0000000000085
0.0000000000085 = 8.5 x 10-12
Moving the decimal 12 places to the right
2. 0.00000000000942
0.00000000000942 = 9.42 x 10-15
Moving the decimal 12 places to the right
3. 6020000000000000
6020000000000000 = 6.02 x 1015
Moving the decimal 15 places to the left
4. 0.00000000837
0.00000000837 = 8.37 x 10-9
Moving decimal 9 places to the right
5. 31860000000
31860000000 = 3.186 x 1010
Moving decimal 10 places to the left
Question 2. Express the following numbers in the usual form:
- 3.02 x 10-6
- 4.5 x 104
- 3 x 10-8
- 1.0001 x 109
- 5.8 x 1012
- 3.61492 x 106
Solution:
⇒ \(3 \times 10^{-8}=\frac{3}{10^8}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(=\frac{3.02}{1000000}\)
⇒ 0.00000302
2. 4.5 x 104 = 4.5 x 10000 = 45000
3. \(3.02 \times 10^{-6}=\frac{3.02}{10^6}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(\frac{3}{100000000}\)
=0.00000003
4. 1.0001 X 109
= 1.0001 X 1000,000,000
= 1,000,100,000
5. 5.8 X 1012
= 5.8 X 1,000,000,000,000
= 5,800,000,000,000
6. 3.61492 x 106
= 3.61492 x 1,000,000
= 3,614,920
Question 3. Express the number appearing in the following statements in standard form:
- 1 micron is equal to \(\frac{1}{1000000} m\)
- The charge of an electron is 0. 000, 000,000,000,000,000,1 6 coulomb.
- The size of bacteria is 0. 0000005 m
- The size of a plant cell is 0. 00001 275 m
- The thickness of thick paper is 0.07 mm.
Solution:
1. \(\frac{1}{1000000} \mathrm{~m}= \frac{1}{10^6}\)
⇒ \(1 \times 10^{-6} \mathrm{~m}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
which is the required standard form.
2. 0.000,000,000,000,000,000,16 coulomb
⇒ \(\frac{16}{100,000,000,000,000,000,000} \text { coulomb }\)
⇒ \(frac{16}{10^{20}} \text { coulomb }\)
⇒ \(\frac{1.6 \times 10}{10^{20}} \text { coulomb }\)
⇒ \(\frac{1.6 \times 10^1}{10^{20}} \text { coulomb } \quad \mid a^1=a\)
⇒ \(1.6 \times 10^{1-20} \text { coulomb } \frac{a^m}{a^n}=a^{m-n}\)
⇒ \(1.6 \times 10^{-19} \text { coulomb }\)
which is the required standard form.
3. 0.0000005 m
⇒ \(\frac{5}{10000000} \mathrm{~m} \)
⇒ \(\frac{5}{10^7} \mathrm{~m}\)
⇒ \(5 \times 10^{-7} \mathrm{~m}\)
which is the required standard form.
4. \(0.00001275 \mathrm{~m} =\frac{1275}{100,000,000} \mathrm{~m}\)
⇒ \(\frac{1275}{10^8} \mathrm{~m}\)
⇒ \(\frac{1275}{10^3 \times 10^5} \mathrm{~m}\)
| am x a” = am + n
⇒ \(\frac{1275}{10^3} \times 10^{-5} \mathrm{~m}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
⇒ \(1.275 \times 10^{-5} \mathrm{~m}\)
which is the required standard form.
5. \( 0.07 \mathrm{~mm} =\frac{7}{100}\)
⇒ \(\frac{7}{10^2}=7 \times 10^{-2} \mathrm{~mm}\)
⇒ \(a^{-m}=\frac{1}{a^m}\)
which is the required standard form
Question 4. In a stack, there are 5 books each of thickness 20 mm, and 5 paper sheets each of thickness 0. 016mm. What is the total thickness of the stack?
Solution:
Total thickness of books = 5 x 20 mm = 100 mm
Total thickness of paper sheets = 5 x 0.016 mm = 0.080 mm
Total thickness of the stack
= Total thickness of books + Total thickness of paper sheets
= 100 mm + 0.080 mm
= (100 + 0.080) mm
= 100.080 mm
= 1.0008 x 102 mm.
Moving decimal 2 places to the left
Hence, the total thickness of the stack is 1.0008 x IQ2 mm.
Exponents And Powers Multiple-Choice Question And Solutions
Question 1. am x an is equal to
- am+n
- am-n
- amn
- an-m
Solution: 1. am+n
Question 2. am ÷ am is equal to
- am+n
- am-n
- amn
- an-m
Solution: 1. am+n
Question 3. (am)is equal to
- am+n
- am-n
- amn
- an-m
Solution: 3. amn
Question 4. Am x is equal to
- (ab)m
- (ab)-m
- amb
- abm
Solution: 1. (ab)m
Question 5. a0 is equal to
- 0
- 1
- -1
- a
Solution: 2. 1
Question 6. \(\frac{a^m}{b^m}\) is equal to
- \(\left(\frac{a}{b}\right)^m\)
- \(\left(\frac{b}{a}\right)^m\)
- \(\frac{a^m}{b}\)
- \(\frac{a}{b^m}\)
Solution: 1. \(\left(\frac{a}{b}\right)^m\)
Question 7. 2 x 2 x 2 x 2 x 2 is equal to
- 24
- 23
- 22
- 25
Solution: 4. 25
Question 8. In 102, the exponents
- l
- 2
- 10
- 1
Solution: 2. 2
Question 9. In 102, the base is
- 1
- 0
- 10
- 100
Solution: 3. \(\frac{1}{10}\)
Question 10. 10-1 is equal to
- 10
- -1
- \(\frac{1}{10}\)
- \(-\frac{1}{10} \text {. }\)
Solution: 3. \(\frac{1}{10}\)
Question 11. The multiplicative inverse of 2-3 is
- 2
- 3
- -3
- 23
Solution: 4. 23
Question 12. The multiplicative inverse of 105 is
- 5
- 10
- 10-5
- 105
Solution: 3. 10-5
Question 13. The multiplicative inverse of \( \frac{1}{2^2}\)
- 2-2
- 22
- 2
- 1
Solution: 2. 22
Question 14. The multiplicative inverse of 10 “10 is
- 10
- \(\frac{1}{10}\)
- 10-10
- 1010
Solution: 4. 1010
Question 15. The multiplicative inverse of am is
- a
- m
- am
- a-m
Solution: 5. a-m
Question 16. 53 x 5-1 is equal to
- 5
- 53
- 5-1
- 52
Solution: 4. 52
Question 17. (-2)5 x (- 2)6 is equal to
- 2
- -2
- -5
- 6
Solution: 2. -2
Question 18. 32 x 3-4 x 35 is equal to
- 3
- 32
- 33
- 35
Solution: 3. 33
19. (- 2) 2 is equal to
- \(\frac{1}{4}\)
- \(\frac{1}{2}\)
- \(-\frac{1}{2}\)
- \(-\frac{1}{4}\)
Solution: 1. \(\frac{1}{4}\)
Question 20. \(\left(\frac{1}{2}\right)^{-4}\) is equal to
- 2
- 24
- 1
- 2-4
Solution: 2. 24
Question 21. (20 + 4-1) x 22 is equal to
- 2
- 3
- 4
- 5
Solution: 4. 5
Question 22. (2-1 + 3-1 + 5-1)0 is equal to
- 2
- 3
- 5
- 1
Solution: 4. 1
Question 23. 3m÷ 3-3 = 35 ⇒ m is equal to
- 1
- 2
- 3
- 4
Solution: 2. 2
Question 24. (-2)m+1 x (-2)4 = (- 2)6 ⇒ m =
- 0
- 1
- -1
- none of these
Solution: 2. 1
Question 25. (-1)60 is equal to
- -1
- 1
- 50
- -50
Solution: 2. 1
Question 26. (-1)51 is equal to
- -1
- 1
- 51
- -51
Solution: 1. -1
Question 27. 149600000000 is equal to
- 1.496 x 1011
- 1.496 x lO10
- 1.496 x 1012
- 1.496 x 105
Solution: 1. 1.496 x 1011
Question 28. 300000000 is equal to
- 3 x 108
- 3 x 107
- 3 x 106
- 3 x 109
Solution: 1. 3 x 108
Question 29. 0.000007 is equal to
- 7 x 10-6
- 7 x 10-6
- 7 x 10-4
- 7 x 10-3
Solution: 1. 7 x 10-6
Question 30. 384467000 is equal to
- 3.84467 x 1o8
- 3.84467 x 103
- 3.84467 x 107
- 3.84467 x 106
Solution: 1. 3.84467 x 108
Question 31. 0.00001275 is equal to
- 1.275 x 10-6
- 1.275 x 10-3
- 1.275 x 104
- 1.275 x 103
Solution: 1. 1.275 x 10-6
Question 32. 695000 is equal to
- 6.95 x 105
- 6.95 x 103
- 6.95 x 106
- 6.95 x 104
Solution: 1. 6.95 x 105
Question 33. 503600 is equal to
- 5.036 x 105
- 5.036 x 106
- 5.036 x 104
- 5.036 x 107
Solution: 1. 5.036 x 105
Question 34. 0.0016is equal to
- 1.6 x 10 -3
- 1.6 x 10-2
- 1.6 x 10 -4
- 1.6 x lO-5
Solution: 1. 1.6 x 10-3
Question 35. 0.000003 is equal to
- 3 x 10-6
- 3 x 106
- 3 x 105
- 3 x 10-5
Solution: 1. 3 x 10-6
Question 36. 8848 is equal to
- 8.848 x 103
- 8.848 x 102
- 8.848 x 10
- 8.848 x 104
Solution: 1. 8.848 x 103
Question 37. 1.5 x 1011 is equal to
- 150000000000
- 15000000000
- 1500000000
- 1500000000000
Solution: 1. 150000000000
Question 38. 2.1 x 10-6 is equal to
- 0.0000021
- 0.000021
- 0.00021
- 0.0021.
Solution: 1. 0.0000021
Question 39. 2.5 x 104 is equal to
- 25
- 250
- 2500
- 25000
Solution: 4. 25000
Question 40. 0.07 x 1O10is equal to
- 700000000
- 7000000
- 7000
- 7
Solution: 1. 700000000
Exponents And Powers True-False
Write whether the following statements are True or False:
1. The value of \(\left\{(-1)^{-1}\right\}^{-1}\) is 1: False
2. The reciprocal of \(\left(\frac{4}{3}\right)^0\) is 1: True
3. The standard form of \(\frac{1}{1000000}\) is 1.0 x 10 -6: True
4. If 6m + 6 “3 = 66, then the value of m is 3: True
5. 2345.6 = 2 x 1000 + 3 x 100 + 4 x 10 + 5 x 1 + 6 x 10 – 1: True
Exponents And Powers Fill in the Blanks
1. (1000)° = 1
2. The standard form of 1,234,500,000,000 is: 1.2345 x 1012
3. The multiplicative inverse of(-3) ~2 is: (-3)2
4. (- 9)4 -5- (- 9)10 is equal to (-9)-6
5. The value of (2 -1 + 3 -1 + 4 -1)° is : 1
6. Write 1.0002 x 109 in the usual form: 1000200000
7. Write the reciprocal of 10 _1: 10
8. Find the value of*if* “3 = (100)1-4 + (100)°: 100
9. By what number should (-8) -1 be divided, so that the quotient may be equal to (-8) -1: 1
10. If = \(\frac{5^m \times 5^2 \times 5^{-3}}{5^{-5}}=5^4\) then find the value of m: 0