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NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

NCERT solutions for class 8 maths chapter 12 topic 12.2 powers with negative exponents

Question:(i) Find the multiplicative inverse of the following.

2^{-4}

Answer:

The detailed explanation for the question is written below,

The multiplicative inverse is \frac{1}{a^{m}}

So, the multiplicative inverse of 2^{-4} is 2^{4}

Question:(ii) Find the multiplicative inverse of the following.

10^{-5}

Answer:

Here is the detailed solution for the above question,

As we know,

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, the multiplicative inverse of 10^{-5} is 10^{5}

Question:(iii) Find the multiplicative inverse of the following.

7^{-2}

Answer:

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, the multiplicative inverse of 7^{-2} is 7^{2}

Question:(iv) Find the multiplicative inverse of the following.

5^{-3}

Answer:

we know,

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, for 5^{-3} multiplicative inverse is 5^{3}

Question:(v) Find the multiplicative inverse of the following.

10^{-100}

Answer:

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, the multiplicative inverse of 10^{-100} is 10^{100}

NCERT solutions for class 8 maths chapter 12 exponents and powers topic 12.2 powers with negative exponents

Question:(i) Expand the following numbers using exponents.

1025.63

Answer:

1025.63 = 1\times 10^{3}+o\times 10^{2}+2\times 10^{1}+5\times 10^{0}+6\times 10^{-1}+3\times 10^{-2}

Question:(ii) Expand the following numbers using exponents.

1256.249

Answer:

1256.249

= 1\times 10^{3}+2\times 10^{2}+5\times 10^{1}+6\times 10^{0}+2\times 10^{-1}+4\times 10^{-2}+9\times 10^{-3}

NCERT solutions for class 8 maths chapter 12 exponents and powers topic 12.3 laws of exponents

Question:1(i) Simplify and write in exponential form.

(-2)^{-3}\times (-2)^{-4}

Answer:

this is simplified as follows

(-2)^{-3}\times (-2)^{-4}

= \frac{1}{(-2)^{3}}\times \frac{1}{(-2)^{4}}

= \frac{1}{(-2)^{3+4}}=\frac{1}{(-2)^{7}}

= (-2)^{-7}

Question:1(ii) Simplify and write in exponential form .

p^3\times p^{-10}

Answer:

this is simplified as follows

p^3\times p^{-10}

p^{3-10} ............. [a^{m}\times a^{n}=a^{m+n}]

=p^{-7}

Question:1(iii) Simplify and write in exponential form.

3^2 \times 3^{-4}\times 3^6

Answer:

this can be simplified as follows

3^2 \times 3^{-4}\times 3^6

= 3^{2+(-4)+6} ............. [a^{m}\times a^{n}\times a^{o}=a^{m+n+o}]

= 3^{4}= 81

NCERT solutions for class 8 maths chapter 12 exponents and powers-Exercise: 12.1

Question:1 (i) Evaluate.

3^{-2}

Answer:

The detailed explanation for the above-written question is as follows,

We know that,

a^{-m}=\frac{1}{a^{m}}

So, here m =2 and a = 3

3^{-2}=\frac{1}{3^{2}} = \frac{1}{3}\times \frac{1}{3} = \frac{1}{9}

Question: 1(ii) Evaluate.

(-4)^{-2}

Answer:

The detailed explanation for the above-written question is as follows

We know that,

a^{-m}= \frac{1}{a^{m}}

So, here (a = -4) and (m = 2)

Then according to the law of exponent

(-4)^{-2}= \frac{1}{(-4)^{2}} = \frac{1}{(-4)}\times \frac{1}{(-4)} = \frac{1}{16} [ negative \times negative = positive]

Question: 1(iii) Evaluate.

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

Answer:

The detailed solution for the above-written question is as follows

We know that,

(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}\&a^{-m} = \frac{1}{a^{m}}

So, here

a = 1 and b = 2 and m =-5

According to the law of exponent

(\frac{1}{2})^{-5}=\frac{1^{-5}}{2^{-5}} = \frac{-1}{2^{-5}}

=(-1)\times -2\times -2\times -2\times -2\times -2 = 32

Question: 2(i) Simplify and express the result in power notation with a positive exponent.

(-4)^{5}\div (-4)^{8}

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

\frac{a^{m}}{a_{n}} = a^{m-n} and a^{-m}= \frac{1}{a^{m}}

So according to this

a = -4, m = 5 and n = 8

\frac{4^{5}}{4_{8}} = -4^{5-8} = -4^{-3}

= (-\frac{1}{4})\times -(\frac{1}{4})\times(-\frac{1}{4}) = -\frac{1}{64}

Question: 2(ii) Simplify and express the result in power notation with positive exponent.

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

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

\frac{a^{m}}{b^{m}} = (\frac{a}{b})^{m} and a^{-m}= \frac{1}{a^{m}} and (a^{m})^{n} = a^{mn}

So, we have given

a = 1, b=2

By using above exponential law,

\frac{a^{m}}{b^{m}} = \frac{1}{(2^{3})^{2}} = \frac{1}{2^{6}}

Question: 2(iii) Simplify and express the result in power notation with a positive exponent.

(-3)^4\times \left(\frac{5}{3} \right )^{4}

Answer:

The detailed solution for the above-written question is as follows,

We know the exponential formula

(\frac{a}{b})^{m}= \frac{a^{m}}{b^{m}}

So, (-3)^{4}\times (\frac{5}{3})^{4}= (-3)^{4}\times \frac{5^{4}}{3^{4}} = \frac{625}{1}

Question: 2(iv) Simplify and express the result in power notation with a positive exponent.

(3^{-7}\div 3^{-10})\times 3^{-5}

Answer:

The detailed explanation for the above-written question is as follows

As we know the exponential form

\frac{a^{m}}{b^{n}}= a^{m-n} \& (a^{m}\times a^{n})=a^{m+n}

By using these two form we get,

\frac{3^{-7}}{3^{-10}}\times (3)^{-5}=3^{-7-(-10)} \times (3)^{-5}

=3^{3} \times (3)^{-5}= 3^{-2}

=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}

Question:2(v) Simplify and express the result in power notation with positive exponent.

2^{-3} \times (-7)^{-3}

Answer:

The detailed solution for the above-written question is as follows,

we know the exponential forms

a^{-m}=\frac{1}{a^{m}} & a^{m}\times b^{m}= (ab)^{m}

So, according to our data,

here initially we use first forms and then the second one.

2^{-3}\times (-7)^{-3}= \frac{1}{2^{3}}\times \frac{1}{(-7)^{3}}

= \frac{1}{(2\times -7)^{3}}= \frac{1}{(-14)^{3}}

Question:3(i) Find the value of.

(3^0 + 4^{-1})\times 2^2

Answer:

The detailed explanation for the above-written question is as follows,

As we know that a^{0}=1

So, 3^{0}=1

now,

=(1+\frac{1}{4})\times 2^{2}

=\frac{5}{4}\times 2^{2}\Rightarrow \frac{5}{2^{2}}\times 2^{2}=5/1

Question: 3(ii) Find the value of.

(2^{-1}\times 4^{-1})\div 2^{-2}

Answer:

The detailed explanation for the above-written question is as follows

Rewrite the equation

(2^{-1}\times 4^{-1})\div 2^{-2}= (2^{-1}\times 2^{-2})\div 2^{-2}

=(2^{-1+(-2)})\div 2^{-2} ................................. a^{m}\times a^{n}= a^{(m+n)}

=(2^{-3})\div 2^{-2} = 2^{-3-(-2)} ........................ a^{m}\div a^{n}= a^{(m-n)}

= 2^{-1}= \frac{1}{2}

Question: 3(iii) Find the value of.

\left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}

Answer:

The detailed explanation for the above-written question is as follows,

This is the exponential form

(a/b)^{m}= \frac{a^{m}}{b^{m}}

So, \left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}

=\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\frac{1}{4^{-2}} .......................using this form a^{m}= \frac{1}{a^{m}}

=2^{2}+3^{2}+4^{2}

= 4+9+16

= 29

Question: 3(iv) Find the value of.

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

Answer:

since we know that

a^{0}=1

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

Question: 3(v) Find the value of.

\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}

Answer:

The detailed explanation for the above-written question is as follows

\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}

=(-2/3)^{-2\times 2} .............. BY using these form of exponential (a^{m})^{n}=a^{mn}

(-2/3)^{-4}=(-3/2)^{4} ......... use this a^{-m}=\frac{1}{a^{m}}

=\frac{81}{16}

Question:4(i) Evaluate

\frac{8^{-1}\times 5^{3}}{2^{-4}}

Answer:

The detailed explanation for the above written question is as follows

\frac{8^{-1}\times 5^{3}}{2^{-4}}

after rewriting the above equation we get,

\frac{2^{-3}\times 5^{3}}{2^{-4}}=2^{-3-(-4)}\times 5^{3} ...........as we know that \frac{a^{m}}{a^{n}}= a^{m-n}

\\=2^{1}\times 5^{3}\\ = 2\times 125 = 250

An alternate method,

= \frac{5^{3}}{2^{-4}\times 2^{3}}

here you can use first a^{-m}= \frac{1}{a^{m}} and after that use a^{m}\times a^{n}= a^{m+n}

Question: 4(ii) Evaluate

(5^{-1}\times 2^{-1})\times 6^{-1}

Answer:

The detailed explanation for the above-written question is as follows

We clearly see that this is in the form of a^{-m}=\frac{1}{a^{m}}

So, (5^{-1}\times 2^{-1})\times 6^{-1}= (\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}

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Question: 5 Find the value of m for which

5^m \div 5^{-3} = 5^5

Answer:

We have,

a^{m}\div a^{n}= a^{m-n}

Here a = 5 and n =-3 and m-n = 5

therefore,

5^{m}\div 5^{-3}= 5^{m+3} = 5^{5}

By comparing from both sides we get

m+3 = 5

m= 2

Question: 6(i) Evaluate

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

Answer:

The detailed solution for the above-written question is as follows

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

= [(1\times 3)-(1\times 4)]^{-1} .............by using a^{-m}= \frac{1}{a^{m}}

= [3-4]^{-1}

\\= [-1]^{-1}\\=-1

Question: 6(ii) Evaluate

\left ( \frac{5}{8} \right )^{-7}\times \left (\frac{8}{5}\right)^{-4}

Answer:

The detailed solution for the above-written question is as follows

\left ( \frac{5}{8} \right )^{-7}\times \left (\frac{8}{5}\right)^{-4}

=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}} ................ using the form \frac{a^{m}}{b^{m}}= (a/b)^{m}

=5^{-7+4}\times 8^{-4+7} ............using a^{m} \div a^{n} = a^{m-n}

= 5^{-3}\times 8^{+3}

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

Question: 7(i) Simplify

\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\;\;(t\neq 0)

Answer:

The detailed solution for the above-written question is as follows

\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\;\;(t\neq 0)

we can write 25= 5^{2}

So, after rewriting the equation,

\frac{5^{2}\times t^{-4}}{5^{-3}\times 10\times t^{-8}}

= \frac{5^{2+3}\times t^{-4+8}}{10} .................using the form [a^{m}\div a^{n}= a^{m-n}]

= \frac{5^{5}\times t^{4}}{10} .............(By expanding we have now)

= \frac{625t^{4}}{2}

Question: 7(ii) Simplify.

\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}

Answer:

The detailed solution for the above-written question is

\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}

we can write 125 = 5^{3} and 6^{-5} can be written as (2\times 3)^{-5}

Now, rewriting the equation, we get

=\frac{3^{-5}\times 10^{-5}\times 5^{3}}{5^{-7}\times(2\times 3) ^{-5}}

=\frac{3^{-5}\times 10^{-5}\times 5^{3+7}}{(2\times 3) ^{-5}} .............by using [a^{m}\div a^{n}=a^{m-n}]

=\frac{ 10^{-5}\times 5^{10}}{(2) ^{-5}} .....................Use [a^{m}\div a^{n}=a^{m-n}]

5^{10-5}= 5^{5}=3125 .........................As [10^{-5} = (2\times 5)^{-5}=2^{-5}\times 5^{-5}] . 2^{-5} can be cancelled out with the denominator 2^{-5}

NCERT solutions for class 8 maths chapter 12 exponents and powers topic 12.4 use of exponents to express small numbers in standard form

Question: 1(i) Write the following numbers in standard form.

0.000000564

Answer:

the standard form of 0.000000564 is

\frac{564}{1000000000}=5.64\times10^{-7}

Question: 1(ii) Write the following numbers in standard form.

0.0000021

Answer:

The standard form 0.0000021 is

=\frac{21}{10000000}=2.1\times 10^{-6}

Question: 1(iii) Write the following numbers in standard form.

21600000

Answer:

The standard form 21600000 is

=2.16\times 10^{7}

Question: 1(iv) Write the following numbers in standard form.

15240000

Answer:

The standard form 15240000

=1.524\times 10^{7}

Question: 2 Write all the facts given in the standard form .

Answer:

  1. Distance between sun and earth 1.496\times10^{11}m

  2. speed of light is 3\times10^{8}m/s

  3. The avg. diameter of red blood cells is (7\times10^{-6}mm)

  4. the distance of the moon from the earth is (3.84467\times10^{8}m)

  5. size of the plant cell is (1.275\times10^{-5}m)

  6. The diameter of the wire on a computer chip is (3\times10^{-6}m)

  7. the height of the Mount Everest is (8.848\times10^{3}m)

NCERT solutions for class 8 maths chapter 12 exponents and powers-Exercise: 12.2

Question: 1(i) Express the following numbers in standard form .

0.0000000000085

Answer:

The standard form is 8.5\times 10^{-12}

Question: 1(ii) Express the following numbers in standard form.

0.00000000000942

Answer:

The standard form is 9.42\times 10^{-12}

Question: 1(iii) Express the following numbers in standard form.

6020000000000000

Answer:

The standard form is 6.02\times 10^{15}

Question: 1(iv) Express the following numbers in standard form.

0.00000000837

Answer:

The standard form of the given number is 8.37\times10^{-9}

Question: 1(v) Express the following numbers in standard form.

31860000000

Answer:

The standard form is 3.186\times10^{10}

Question: 2(i) Express the following numbers in usual form.

3.02\times 10^{-6}

Answer:

3.02\times 10^{-6}

=\frac{3.02}{1000000}

=0.00000302

this is the usual form

Question: 2(ii) Express the following numbers in usual form.

4.5 \times 10^4

Answer:

4.5 \times 10^4=4.5\times 10000

=45000

this is the usual form

Question:2(iii) Express the following numbers in the usual form.

3\times 10^{-8}

Answer:

3\times 10^{-8}

\frac{3}{100000000} = 0.000000030

this is the usual form

Question: 2(iv) Express the following numbers in usual form.

1.0001\times 10^9

Answer:

1.0001\times 10^9

=1.0001\times 100000000

=1000100000

this is the usual form

Question: 2(v) Express the following numbers in usual form.

5.8\times 10^{12}

Answer:

5.8\times 10^{12}

=5.8\times 100000000000

=5800000000000

this is the usual form

Question:2(vi) Express the following numbers in usual form.

3.61492 \times 10^6

Answer:

3.61492 \times 10^6

= 3.61492\times 1000000

= 3614920

17155

Question:3(i) Express the number appearing in the following statements in standard form.

1 micron is equal to \frac{1}{1000000}m .

Answer:

1 micron is equal to

\frac{1}{1000000}m

= 1\times 10^{-6}

Question:3(ii) Express the number appearing in the following statements in standard form.

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

Answer:

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

=1.6\times 10^{-19} coulomb.

Question:3(iii) Express the number appearing in the following statements in standard form.

Size of a bacteria is 0.0000005 m.

Answer:

Size of a bacteria is 0.0000005 m

\frac{5}{10000000}=5\times 10^{-7}m

Question:3(iv) Express the number appearing in the following statements in standard form.

Size of a plant cell is 0.00001275 m.

Answer:

Size of a plant cell is0.00001275m

=\frac{1275}{10000}=1.275\times10^{-5}m

Question:3(v) Express the number appearing in the following statements in standard form.

Thickness of a thick paper is 0.07 mm

Answer:

The thickness of a thick paper is 0.07

=\frac{7}{100}=7\times 10^{-2}mm

Question:4 In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.

Answer:

the thickness of each book = 20mm

So, the thickness of 5 books = (5\times 20)=100 mm

the thickness of one paper sheet =0.016mm

So, the thickness of 5 paper sheet = (5\times 0.016)=0.08 mm

the total thickness of the stack = (100+0.08)mm

=100.08 mm or

(1.008\times10^{-2}mm)

NCERT Class 8 Mathematics Solutions

Chapter 01 - Rational Numbers

Chapter 02 - Linear Equations in One Variable

Chapter 03 -Understanding Quadrilaterals

Chapter 04 - Practical Geometry

Chapter 05 - Data Handling

Chapter 06 - Squares and Square Roots

Chapter 07 - Cubes and Cube Roots

Chapter 08 - Comparing Quantities

Chapter 09 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Indirect proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

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