# NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 8 Maths Chapter 12 Exponents And Powers Exercise 12.1

1.Evaluate:
(i) $$3^{-2}$$

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

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

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i) We have:
$$3^{-2}=\frac{1}{3^2}=\frac{1}9\quad$$[∵$$a^{-n}=\frac{1}{a^n}$$]

(ii)We have:
$$(-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\quad$$[∵$$a^{-n}=\frac{1}{a^n}$$]

(iii)We have:
$$(\frac { 1 }{ 2 })^{-5}=2^5=32\quad$$[∵$$(\frac{a}b)^{-n}=(\frac{b}a)^n$$]

2.Simplify and express the result in power notation with a positive exponent. (i)$$(-4)^5÷(-4)^8$$
(ii)$$(\frac{1}{2^3})^2$$

(iii)$$(-3)^4\times (\frac{5}3)^4$$

(iv)$$(3^{-7}÷3^{-10})\times 3^{-5}$$

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

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:$$(-4)^5÷(-4)^8$$
$$=(-4)^{5-8}=(-4)^{-3}=\frac{1}{(-4)^3}\quad$$[∵$$a^{m}÷a^{n}=a^{m-n}$$]

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

(ii)We have:$$(\frac{1}{2^3})^2$$

$$(\frac{1}{2^3})^2=\frac{(1)^2}{(2^3)^2}=\frac{1}{2^6}=(\frac{1}2)^6$$

(iii)We have:$$(-3)^4\times (\frac{5}3)^4$$

$$=(-3)^4\times (\frac{5}3)^4=(-3)^4\times \frac{(5)^4}{(3)^4}$$

$$=\frac{(3)^4\times(5)^4}{(3)^4}=(5)^4$$

(iv)We have:$$(3^{-7}÷3^{-10})\times 3^{-5}$$

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

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

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

(v)We have:$$2^{-3}\times(-7)^{-3}$$

$$=2^{-3}\times(-7)^{-3}=[2\times(-7)]^{-3}=(-14)^{-3}$$

$$=-(14)^{-3}=-\frac{1}{14^3}=(-\frac{1}{14})^3$$

3.Find the value of
(i)$$3^0+4^{-1}\times 2^2$$

(ii)$$2^{-1}\times4^{-1}÷ 2^{-2}$$

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

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

(v)$$[(\frac{-2}3)^{-2}]^2$$

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:$$3^0+4^{-1}\times 2^2$$

$$=3^0+4^{-1}\times 2^2=(1+\frac{1}4)\times4$$

$$=(\frac{4+1}4)\times4=\frac{5}4\times 4=5$$

(ii)We have:$$2^{-1}\times4^{-1}÷ 2^{-2}$$

$$=2^{-1}\times4^{-1}÷ 2^{-2}=(\frac{1}2\times\frac{1}4)÷\frac{1}{2^2}$$

$$=\frac{1}8÷\frac{1}4=\frac{1}8\times\frac{4}1$$

$$=\frac{1}2$$

(iii)We have:$$(\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}$$

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

$$=4+9+16=29$$

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

$$=(3^{-1}+4^{-1}+5^{-1})^0=1\quad$$[∵$$a^0=1$$]

(v)We have:$$[(\frac{-2}3)^{-2}]^2$$

$$=[(\frac{-2}3)^{-2}]^2=[(-\frac{3}2)^2]^2$$

$$=(\frac{9}4)^2=\frac{81}{16}$$

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

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

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:$$\frac { { 8 }^{ -1 }\times { 5 }^{ 3 } }{ { 2 }^{ -4 } }$$

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

$$=\frac{1}8\times125\times 16=125\times 2=250$$

(ii)We have:$$(5^{-1} \times 2^{-1}) \times 6^{-1}$$

$$=(\frac{1}5\times\frac{1}2)\times\frac{1}6$$

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

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

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

We have given: $$5^m ÷ 5^{-3} = 5^5$$.

$$\Rightarrow 5^{m-(-3)} = 5^5\quad$$ [∵$$a^m ÷ a^n = a^{m-n}$$]

$$\Rightarrow 5^{m+3} = 5^5$$

Now, Comparing the powers of equal bases, we have m + 3 = 5

$$\Rightarrow m = 5 – 3 = 2$$, i.e., m = 2

6.Evaluate:
(i)$$[(\frac{1}3)^{-1}-(\frac{1}4)^{-1}]^{-1}$$

(ii)$$(\frac{5}8)^{-7}\times (\frac{8}5)^{-4}$$

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:$$[(\frac{1}3)^{-1}-(\frac{1}4)^{-1}]^{-1}$$

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

(ii)We have: $$(\frac{5}8)^{-7}\times (\frac{8}5)^{-4}$$
$$=(\frac{8}5)^7\times(\frac{5}8)^4=\frac{8^7}{5^7}\times\frac{5^4}{8^4}$$

$$=\frac{8^{7-4}}{5^{7-4}}=\frac{8^3}{5^3}=\frac{512}{125}$$

7.Simplify :
(i)$$\frac{25\times t^{-4}}{5^{-3}\times10\times t^{-8}}(t\neq 0)$$

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

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:$$\frac{25\times t^{-4}}{5^{-3}\times10\times t^{-8}}(t\neq 0)$$

$$=\frac{25\times 5^3}{10}\times t^{(-4+8)}=\frac{5\times 5^3}2\times t^4$$

$$=\frac{625}2t^4$$
(ii)We have:$$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$$

$$=\frac{5^{7}\times 6^{5}\times 125}{3^{5}\times 10^{5}}$$

$$=\frac{5^{7}\times (3\times 2)^{5}\times 5^3}{3^{5}\times (2\times5)^{5}}=\frac{5^{7}\times 3^{5}\times 2^5\times 5^3}{3^{5}\times 2^5\times5^{5}}=5^{7+3-5}=5^5$$

## NCERT solutions for class 8 Maths Chapter 12 Exponents And Powers Exercise 12.2

1.Express the following numbers in standard form:
(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:0.0000000000085
$$=\frac{85}{10000000000000}=\frac{8.5\times 10}{10^{13}}$$

$$=8.5\times 10^{(1-13)}=8.5\times 10^{-12}$$

(ii)We have:0.00000000000942
$$=\frac{942}{100000000000000}=\frac{9.42\times 10^2}{10^{14}}$$

$$=9.42\times 10^{(2-14)}=9.42\times 10^{-12}$$

(iii)We have:6020000000000000=$$6.02\times1000000000000000$$

$$=6.02\times 10^{15}$$

(iv)We have:0.00000000837

$$=\frac{837}{100000000000}=\frac{8.37\times 100}{10^{11}}$$

$$=8.37\times 10^{(2-11)}=8.37\times 10^{-9}$$

(v)We have: 31860000000

$$= 3.186 \times 10000000000$$

$$= 3.186 \times 10^10$$

Hence, the required standard from = $$3.186 × 10^10$$

2.Express the following numbers in usual form.
(i) $$3.02 × 10^{-6}$$

(ii) $$4.5 × 10^4$$

(iii) $$3 × 10^{-8}$$

(iv) $$1.0001 × 10^9$$

(v) $$5.8 × 10^{12}$$

(vi) $$3.61492 \times 10^6$$

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have:$$3.02 × 10^{-6}$$

$$=\frac{302}{100}\times\frac{1}{10^6}=\frac{302}{100000000}$$

$$=302\times 10^{-8}=0.00000302$$

Thus, we have $$3.02 × 10^{-6}=0.00000302$$

(ii)We have:$$4.5 × 10^4$$

$$=\frac{45}{10}\times10^4=45\times 10^3$$

$$=45000$$

Thus, we have $$4.5 × 10^4=45000$$

(iii)We have:$$3 × 10^{-8}$$

$$=\frac{3}{10^8}=\frac{3}{100000000}$$

$$=0.00000003$$

Thus, we have $$3 × 10^{-8}=0.00000003$$

(iv)We have: $$1.0001 × 10^9$$

$$=\frac{10001}{10000}\times10^9=\frac{10001}{10^4}\times10^9$$

$$=10001\times 10^5=1000100000$$

Thus, we have$$1.0001 × 10^9=1000100000$$

(v)We have: $$5.8 × 10^{12}$$

$$=\frac{58}{10}\times10^{12}=58\times 10^{11}$$

$$=5800000000000$$

Thus, we have $$5.8 × 10^{12}=5800000000000$$

(vi)We have: $$3.61492 \times 10^6$$

$$=\frac{361492}{100000}\times10^6=\frac{361492}{10^5}\times10^6$$

$$=361492\times 10=3614920$$

Thus, we have $$3.61492 \times 10^6=3614920$$

3.Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to $$\frac { 1 }{ 1000000 }$$ 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.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

(i)We have 1 micron=$$\frac { 1 }{ 1000000 }$$

$$=\frac{1}{10^6}m=10^{-6}$$

(ii)We know charge of an electron=0.000,000,000,000,000,000,16

$$\frac{16}{1,000,000,000,000,000,000,00}=\frac{1.6\times 10}{1,000,000,000,000,000,000,00}$$

$$=\frac{1.6\times 10}{10^{20}}=1.6\times 10^{1-20}=1.6\times 10^{-19}$$

(iii)We have size of bacteria=0.0000005m

$$=\frac{5}{10000000}m=\frac{0.5\times 10}{10^7}m$$

$$=0.5\times 10^{1-7}m=0.5\times 10^{-6}m$$

$$=5\times 10^{-7}m$$

(iv)We have size of plant cell=0.00001275m

$$=\frac{1275}{100000000}m=\frac{1.275\times 10^3}{10^8}m$$

$$=1.275\times 10^{3-8}m=1.275\times 10^{-5}m$$

$$=1.275\times 10^{-5}m$$

(v)We have thickness of paper=0.07mm

$$=\frac{7}{100}mm=\frac{0.7\times 10}{10^2}mm$$

$$=0.7\times 10^{1-2}mm=0.7\times 10^{-1}mm$$

$$=7\times 10^{-2}mm$$

4.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?

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

We have the thickness of books = $$5 \times 20 = 100 mm$$

Also, the thickness of 5 paper sheets = $$5 \times 0.016 mm = 0.080 mm$$.

Total thickness of the stack = $$100 mm + 0.080 mm = 100.080 mm = 1.0008 \times 10^2 mm$$

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