Exercise 11.2 (Revised) - Chapter 13 - Exponents & Powers - Ncert Solutions class 7 - Maths

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Chapter 11 - Exponents & Powers - NCERT Solutions Class 7 Maths

Ex 11.2 Question 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) $\left(5^2\right)^2 \div 5^3$
(vi) $2^5 \times 5^5$
(vii) $a^4 \times b^4$
(viii) $\left(3^4\right)^3$
(ix) $\left(2^{20} \div 2^{15}\right) \times 2^3$
(x) $8^t \div 8^2$

Answer:

(i) $3^2 \times 3^4 \times 3^8=3^{(2+4+8)}=3^{14}\left[\because a^m \times a^n=a^{m+n}\right]$
(ii) $6^{15} \div 6^{10}=6^{15-10}=6^5\left[\because a^m \div a^n=a^{m-n}\right]$
(iii) $a^3 \times a^2=a^{3+2}=a^5\left[\because a^m \times a^n=a^{m+n}\right]$
(iv) $7^x \times 7^2=7^{x+2}\left[\because a^m \times a^n=a^{m+n}\right]$

(v) $\left(5^2\right)^3 \div 5^3=5^{2 \times 3} \div 5^3=5^6 \div 5^3\left[\because\left(a^m\right)^n=a^{m \times n}\right]=5$
(vi) $2^5 \times 5^5=(2 \times 5)^5=10^5\left[\because a^m \times b^m=(a \times b)^m\right]$
(vii) $a^4 \times b^4=(a \times b)^4\left[\because a^m \times b^m=(a \times b)^m\right]$

(viii) $\left(3^4\right)^3=3^{4 \times 3}=3^{12}\left[\because\left(a^m\right)^n=a^{m \times n}\right]$
(ix) $\left(2^{20} \div 2^{15}\right) \times 2^3=\left(2^{20-15}\right) \times 2^3\left[\because a^m \div a^n=a^{m-n}\right]$
$=2^5 \times 2^3=2^{5+3}\left[\because a^m \times b^m=(a \times b)^m\right]=2^8$
(x) $8^t \div 8^2=8^{t-2}\left[\because a^m \div a^n=a^{m-n}\right]$

Ex 11.2 Question 2.

Simplify and express each of the following in exponential form:
(i) $\frac{2^3 \times 3^4 \times 4}{3 \times 32}$
(ii) $\left[\left(5^2\right)^3 \times 5^4\right] \div 5^7$
(iii) $25^4 \div 5^3$
(iv) $\frac{3 \times 7^2 \times 11^8}{21 \times 11}$
(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) $\left(3^0+2^0\right) \times 5^0$
(ix) $\frac{2^8 \times a^5}{4^3 \times a^3}$

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

Answer:

(i) $\frac{2^3 \times 3^4 \times 4}{3 \times 32}=\frac{2^3 \times 3^4 \times 2^2}{3 \times 2^5}=\frac{2^{3+2} \times 3^4}{3 \times 2^5}\left[\because a^m \times a^n=a^{m+n}\right]$
$
\begin{aligned}
& =\frac{2^5 \times 3^4}{3 \times 2^5}=2^{5-5} \times 3^{4-3}\left[\because a^m \div a^n=a^{m-n}\right] \\
& =2^0 \times 3^3=1 \times 3^3=3^3
\end{aligned}
$

$
\begin{aligned}
& \text { (ii) }\left[\left(5^2\right)^3 \times 5^4\right] \div 5^7=\left[5^6 \times 5^4\right] \div 5^7\left[\because\left(a^m\right)^n=a^{m \times n}\right] \\
& =\left[5^{6+4}\right] \div 5^7=5^{10} \div 5^7\left[\because a^m \times a^n=a^{m+n}\right] \\
& =5^{10-7}=5^3\left[\because a^m \div a^n=a^{m-n}\right]
\end{aligned}
$
(iii) $25^4 \div 5^3=\left(5^2\right)^4 \div 5^3=5^8 \div 5^3\left[\because\left(a^m\right)^n=a^{m \times n}\right]$
$
=5^{8-3}=5^3\left[\because a^m \div a^n=a^{m-n}\right]
$
(iv) $\frac{3 \times 7^2 \times 11^8}{21 \times 11^3}=\frac{3 \times 7^2 \times 11^8}{3 \times 7 \times 11^3}=3^{1-1} \times 7^{2-1} \times 11^{8-3}\left[\because a^m \div a^n=a^{m-n}\right]$
$
=3^0 \times 7^1 \times 11^5=7 \times 11^5
$
(v) $\frac{3^7}{3^4 \times 3^3}=\frac{3^7}{3^{4+3}}=\frac{3^7}{3^7}\left[\because a^m \times a^n=a^{m+n}\right]$
$
=3^{7-7}=3^0=1\left[\because a^m \div a^n=a^{m-n}\right]
$
(vi) $2^0+3^0+4^0=1+1+1=3\left[\because a^0=1\right]$
(vii) $2^0 \times 3^0 \times 4^0=1 \times 1 \times 1=1\left[\because a^0=1\right]$
(viii) $\left(3^0+2^0\right) \times 5^0=(1+1) \times 1=2 \times 1=2\left[\because a^0=1\right]$
(ix) $\frac{2^8 \times a^5}{4^3 \times a^3}=\frac{2^8 \times a^5}{\left(2^2\right)^3 \times a^3}=\frac{2^8 \times a^5}{2^6 \times a^3}\left[\because\left(a^m\right)^n=a^{m \times n}\right]$

$
=2^{8-6} x^{5-3}=2^2 x^2
$
(x) $\left(\frac{a^5}{a^3}\right) \times a^8=\left(a^{5-3}\right) \times a^8=a^2 \times a^8\left[\because a^m \div a^n=a^{m-n}\right]$
$
=\mathrm{a}^{2+8}=\mathrm{a}^{10}\left[\because a^m \times a^n=a^{m+n}\right]
$
$
\begin{aligned}
& \text { (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\left[\because a^m \div a^n=a^{m-n}\right] \\
& =1 \times a^3 \times b=a^3 \times \mathrm{b}\left[\because a^0=1\right]
\end{aligned}
$
(xii) $\left(2^3 \times 2\right)^2=\left(2^{3+1}\right)^2=\left(2^4\right)^2\left[\because a^m \times a^n=a^{m+n}\right]$
$
=2^{4 \times 2}=2^8\left[\because\left(a^m\right)^n=a^{m \times n}\right]
$

Ex 11.2 Question 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) $3^0=(1000)^0$

Answer:

(i) $10 \times 10^{11}=100^{11}$
L.H.S. $10^{1+11}=10^{12}$ and R.H.S. $\left(10^2\right)^{11}=10^{22}$

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.

Ex 11.2 Question 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

Answer:

(i) $108 \times 192$
$
\begin{aligned}
& =\left(2^2 \times 3^3\right) \times\left(2^6 \times 3\right) \\
& =2^{2+6} \times 3^{3+1} \\
& =2^8 \times 3^4
\end{aligned}
$
(ii) 270
$
=2 \times 3^3 \times 5
$
(iii) $729 \times 64$
$
=3^6 \times 2^6
$
(iv) 768
$
=2^8 \times 3
$

Ex 11.2 Question 5.

Simplify:
(i) $\frac{\left(2^5\right)^2 \times 7^3}{8^3 \times 7}$
(ii) $\frac{25 \times 5^2 \times t^8}{10^3 \times t^4}$
(iii) $\frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5}$

Answer:

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

$\begin{aligned}
&\begin{aligned}
& =\frac{2^{10} \times 7^3}{2^9 \times 7} \\
& =2^{10-9} \times 7^{3-1}=2 \times 7^2 \\
& =2 \times 49 \\
& =98
\end{aligned}\\
&\begin{aligned}
& \text { (ii) } \frac{25 \times 5^2 \times t^8}{10^3 \times t^4}=\frac{5^2 \times 5^2 \times t^8}{(5 \times 2)^3 \times t^4} \\
& =\frac{5^{2+2} \times t^{8-4}}{2^3 \times 5^3} \\
& =\frac{5^4 \times t^4}{2^3 \times 5^3} \\
& =\frac{5^{4-3} \times t^4}{2^3} \\
& =\frac{5 t^4}{8}
\end{aligned}\\
&\begin{aligned}
& \text { (iii) } \frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5}=\frac{3^5 \times(2 \times 5)^5 \times 5^2}{5^7 \times(2 \times 3)^5} \\
& =\frac{3^5 \times 2^5 \times 5^5 \times 5^2}{5^7 \times 2^5 \times 3^5} \\
& =\frac{3^5 \times 2^5 \times 5^{5+2}}{5^7 \times 2^5 \times 3^5} \\
& =\frac{3^5 \times 2^5 \times 5^7}{5^7 \times 2^5 \times 3^5}
\end{aligned}
\end{aligned}$

$\begin{aligned}
& =2^{5-5} \times 3^{5-5} \times 5^{7-7} \\
& =2^0 \times 3^0 \times 5^0 \\
& =1 \times 1 \times 1 \\
& =1
\end{aligned}$