Ncert Solutions For Class 7 Maths Ex 13.2

Q.1.Using laws of exponents, simplify and write the answer in exponential form:

(i)$3^{2}\times 3^{4}\times 3^{8}$

(ii)$6^{15}\div 6^{10}$

(iii)$a^{3}\times a^{2}$

(iv)$7^{x}\times 7^{2}$

(v)$(5^{2})^{2} \div 5^{3}$

(vi)$2^{5} \times 5^{5}$

(vii)$a^{4} \times b^{4}$

(viii)$(3^{4} )^{3}$

(ix)$(2^{20}\div 2^{15})\times 2^{3}$

(x)$8^{t}\div 8^{2}$

Solution:

(i)$3^{2}\times 3^{4}\times 3^{8}=3^{(2+4+8)}=3^{14}$

(ii)$6^{15}\div 6^{10}=6^{15-10}=6^{5}$

(iii)$a^{3}\times a^{2}=a^{3+2}=a^{5}$

(iv)$7^{x}\times 7^{2}=7^{x+2}$

(v)$(5^{2})^{2} \div 5^{3}=5^{2\times 3}\div 5^{3}=5^{6}\div 5^{3}=5^{6-3}=5^{3}$

(vi)$2^{5} \times 5^{5}=(2\times 5)^{5}=10^{5}$

(vii)$a^{4}\times b^{4}=(a\times b)^{4}$

(viii)$(3^{4})^{3}=3^{4\times 3}=3^{12}$

(ix)$(2^{20}\div 2^{15})\times 2^{3}=(2^{20-15})\times 2^{3}=2^{5}\times 2^{3}=2^{5+3}=2^{8}$

(x) $8^{t}\div 8^{2}=8^{t-2}$

Formulae:

(i)$a^{m}\times a^{n}=a^{m+n}$

(ii)$a^{m}\div a^{n}=a^{m-n}$

(iii)$(a^{m})^{n}=a^{m\times n}$

(iv)$a^{m}\times b^{m}=(a\times b)^{m}$

 

Q.2.(i)$\frac{2^{3}\times 3^{4}\times 4}{3\times 32}$

(ii)$[(5^{2})^{3}\times 5^{4}]\div 5^{7}$

(iii)$25^{4}\div 5^{3}$

(iv) $\frac{3\times 7^{2}\times 11^{8}}{21\times 11^{3}}$

(v)$\frac{3^{7}}{3^{4}\times 3^{3}}$

(vi)$2^{0}+3^{0}+4^{0}$

(vii)$2^{0}\times 3^{0}\times 4^{0}$

(viii)$3^{0}+2^{0}\times 5^{0}$

(ix)$\frac{2^{8}\times a^{5}}{4^{3}\times a^{3}}$

(x)$(\frac{a^{5}}{a^{3}})\times a^{8}$

(xi)$\frac{4^{5}\times a^{8}b^{3}}{4^{5}\times a^{5}b^{2}}$

(xii)$(2^{3}\times2)^{2}$

Solution:

(i)$\frac{2^{3}\times 3^{4}\times 4}{3\times 32}=\frac{2^{3}\times 3^{4}\times 2^{2}}{3\times 2^{5}}=\frac{2^{3+2}\times 3^{4}}{3\times 2^{5}}\\ =\frac{2^{5}\times 3^{4}}{3\times 2^{5}}=2^{5-5}\times 3^{4-3}\\ =2^{0}\times 3^{3}=1\times 3^{3}=3^{3}$

 

(ii)$[(5^{2})^{3}\times 5^{4}]\div 5^{7}=[5^{6}\times 5^{4}]\div 5^{7}\\ =[5^{6+4}]\div 5^{7}\\ =5^{10}\div 5^{7}\\ =5^{10-7}=5^{3}$

 

(iii)$25^{4}\div 5^{3}=(5^{2})^{4}\div 5^{3}\\ =5^{8}\div 5^{3}\\ =5^{8-3}=5^{5}$

 

(iv) $\frac{3\times 7^{2}\times 11^{8}}{21\times 11^{3}}=\frac{3\times 7^{2}\times 11^{8}}{3\times 7\times 11^{3}}\\ =3^{1-1}\times 7^{2-1}\times 11^{8-3}\\ =3^{0}\times 7^{1}\times 11^{5}=7\times 11^{5}$

 

(v)$\frac{3^{7}}{3^{4}\times 3^{3}}=\frac{3^{7}}{3^{4+3}}=\frac{3^{7}}{3^{7}}\\ =3^{7-7}=3^{0}=1$

 

(vi)$2^{0}+3^{0}+4^{0}=1+1+1=3$

 

(vii)$2^{0}\times 3^{0}\times 4^{0}=1\times 1\times 1=1$

 

(viii)$3^{0}+2^{0}\times 5^{0}=(1+1)\times 1=2\times 1=2$

 

(ix)$\frac{2^{8}\times a^{5}}{4^{3}\times a^{3}}=\frac{2^{8}\times a^{5}}{(2^{2})^{3}\times a^{3}} =\frac{2^{8}\times a^{5}}{2^{6}\times a^{3}}\\ =2^{8-6}\times a^{5-2}=2^{2}\times a^{2}\\ =(2a)^{2}$

 

(x)$(\frac{a^{5}}{a^{3}})\times a^{8}=(a^{5-3})\times a^{8}=a^{2}\times a^{8}\\ =a^{2+8}=a^{10}$

 

(xi)$\frac{4^{5}\times a^{8}b^{3}}{4^{5}\times a^{5}b^{2}}=4^{5-5}\times a^{8-5}\times b^{3-2}=4^{0}\times a^{3}\times b\\ =1\times a^{3}\times b=a^{3}b$

 

(xii)$(2^{3}\times2)^{2}=(2^{3+1})^{2}=(2^{4})^{2}\\ =2^{4\times 2}=2^{8}$

 

Q.3.Say true or false and justify your answer:

(i)$10 \times 10^{11} = 100^{11}$

(ii)$2^{3} > 5^{2}$

(iii)$2^{3} \times 3^{2} = 6^{5}$

(iv)$3^{0} = (1000)^{0}$

Solution:

(i)$10 \times 10^{11} = 100^{11}$

L.H.S.  $10^{1+11}=10^{12}$

And

R.H.S. $(10^{2})^{11}=10^{22}$

Since, $L.H.S.\neq R.H.S.$

Therefore, it is false.

(ii)$2^{3} > 5^{2}$

L.H.S. $2^{3}=8$

And

R.H.S. $5^{2}=25$

Since, L.H.S. is not greater than R.H.S.

Therefore, it is false.

(iii)$2^{3} \times 3^{2} = 6^{5}$

L.H.S. $2^{3}\times 3^{2}=8\times 9=72$

And

R.H.S. $6^{5}=7,776$

Since, $L.H.S.\neq R.H.S.$

Therefore, it is false.

(iv)$3^{0} = (1000)^{0}$

L.H.S. $3^{0}=1$

And

R.H.S. $(1000)^{0}=1$

Since, L.H.S. = R.H.S.

Therefore, it is true.

 

Q.4. Express each of the following as a product of prime factors only in exponential form:

(i)$108 \times 192$

(ii)270

(iii)$729 \times 64$

(iv)768

Solution:

(i)$108 \times 192$

$108 \times 192 =(2^{2}\times 3^{3})\times (2^{6}\times 3)\\ =2^{2+6}\times 3^{3+1}\\ =2^{8}\times 3^{4}$

 

(ii)270

$270 =2\times 3^{5}\times 5$

 

(iii)$729 \times 64$

$729 \times 64=3^{6}\times 2^{6}$\

 

(iv)768

$768=2^{8}\times 3$

 

Q.5. Simplify:

(i)$\frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}$

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

(iii)$\frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}$

Solution:

(i)$\frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}$

$\frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}\\ \\ =\frac{(2^{5\times 2})\times 7^{3}}{(2^{3})^{3}\times 7}\\ \\ =\frac{2^{10}\times 7^{3}}{2^{9}\times 7}\\ \\ =2^{10-9}\times 7^{3-1}=2\times 7^{2}\\ \\ =2\times 49=98$

 

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

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

 

(iii)$\frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}$

$\frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}=\frac{3^{5}\times (2\times 5)^{5}\times 5^{2}}{5^{7}\times (2\times 3)^{5}}\\ \\ =\frac{3^{5}\times 2^{5}\times 5^{5}\times 5^{2}}{5^{7}\times 2^{5}\times 3^{5}}\\ \\ =\frac{3^{5}\times 2^{5}\times 5^{5+2}}{5^{7}\times 2^{5}\times 3^{5}}\\ \\ =\frac{3^{5}\times 2^{5}\times 5^{7}}{5^{7}\times 2^{5}\times 3^{5}}\\ \\ =2^{5-5}\times 3^{5-5}\times 5^{5-5}\\ =2^{0}\times 3^{0}\times 5^{0}=1\times 1\times 1=1$

 

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