Exercise 3.1 - Chapter 3 Algebra Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 3.1
Question 1 .
Fill in the blanks.
1. The exponential form $14^{9}$ should be read as
2. The expanded form of $\mathrm{p}^{3} \mathrm{q}^{2}$ is
3. When base is 12 and exponent is 17 , its exponential form is
4. The value of $(14 \times 21)^{0}$ is

Answers:
1. 14 Power 9
2. $p \times p \times p \times q \times q$
3. $12^{17}$
4. 1

Question 2 .
Say True or False.
1. $2^{3} \times 3^{2}=65$
2. $2^{9} \times 3^{2}=(2 \times 3)^{9 \times 2}$
3. $3^{4} \times 3^{7}=3^{11}$
4. $2^{0} \times 1000^{0}$
5. $2^{3}<3^{2}$
Answers:
1. False
2. False
3. True

4. True
5. True

Question $3 .$
Find the value of the following.
1. $2^{6}$
2. $11^{2}$
3. $5^{4}$
4. $9^{3}$
Solution:
1. $2^{6}=2 \times 2 \times 2 \times 2 \times 2 \times 2=64$
2. $11^{2}=11 \times 11=121$
3. $5^{4}=5 \times 5 \times 5 \times 5=625$
4. $9^{3}=9 \times 9 \times 9=729$

Question $4 .$
Express the following in exponential form.
1. $6 \times 6 \times 6 \times 6$
2. $\mathrm{t} \times \mathrm{t}$
3. $5 \times 5 \times 7 \times 7 \times 7$
4. $2 \times 2 \times \mathrm{a} \times \mathrm{a}$
Solution:

1. $6 \times 6 \times 6 \times 6=6^{1+1+1+1}=64\left[\right.$ Since $\left.\mathrm{a}^{\mathrm{m}} \times \mathrm{a}^{\mathrm{n}}=\mathrm{a}^{\mathrm{m}+\mathrm{n}}\right]$
2. $\mathrm{t} \times \mathrm{t}=\mathrm{t}^{1+1}=\mathrm{t}^{2}$
$3.5 \times 5 \times 7 \times 7 \times 7=5^{1+1} \times 7^{1+1+1}=5^{2} \times 7^{3}$
$4.2 \times 2 \times a \times a=2^{1+1} \times a^{1+1}=2^{2} \times a^{2}=(2 a)^{2}$

Question $5 .$
Express each of the following numbers using exponential form,
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
(i) 512

$512=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=2^{9}$ [Using product rule]
(ii) 343

$343=7 \times 7 \times 7=7^{1+1+1}$
$=7^{3}$ [Using product rule]
(iii) 729

$\begin{aligned}
&729=3 \times 3 \times 3 \times 3 \times 3 \times 3 \\
&=3^{6} \text { [Using product rule] }
\end{aligned}$
$729=3 \times 3$ $=3^{6}[\mathrm{U} \sin$

(iv) 3125

\begin{aligned}
&3125=5 \times 5 \times 5 \times 5 \times 5 \\
&=5^{5} \text { [Using product rule] }
\end{aligned}

Question $6 .$
Identify the greater number in each of the following.
(i) $6^{3}$ or $3^{6}$
(ii) $5^{3}$ or $3^{5}$
(iii) $2^{8}$ or $8^{2}$
Solution:
(i) $6^{3}$ or $3^{6}$
$\begin{aligned}
&6^{3}=6 \times 6 \times 6=36 \times 6=216 \\
&3^{6}=3 \times 3 \times 3 \times 3 \times 3 \times 3=729 \\
&729>216 \text { gives } 3^{6}>6^{3} \\
&\therefore 36 \text { is greater. }
\end{aligned}$

(ii) $5^{3}$ or $3^{5}$
$\begin{aligned}
&5^{3}=5 \times 5 \times 5=125 \\
&3^{5}=3 \times 3 \times 3 \times 3 \times 3=243 \\
&243>125 \text { gives } 3^{5}>5^{3} \\
&\therefore 3^{5} \text { is greater. }
\end{aligned}$
(iii) $2^{8}$ or $8^{2}$
$\begin{aligned}
&2^{8}=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=256 \\
&8^{2}=8 \times 8=64
\end{aligned}$
$\begin{aligned}
&256>64 \text { gives } 2^{8}>8^{2} \\
&\therefore 2^{8} \text { is greater. }
\end{aligned}$

Question $7 .$
Simplify the following
(i) $7^{2} \times 3^{4}$
(ii) $3^{2} \times 2^{4}$
(iii) $5^{2} \times 10^{4}$
Solution:

(i) $7^{2} \times 3^{4}=(7 \times 7) \times(3 \times 3 \times 3 \times 3)$
$=49 \times 81=3969$
(ii) $3^{2} \times 2^{4}=(3 \times 3) \times(2 \times 2 \times 2 \times 2)$
$=9 \times 16=144$
(iii) $5^{2} \times 10^{4}=(5 \times 5) \times(10 \times 10 \times 10 \times 10)$
$=25 \times 10000=2,50,000$

Question $8 .$
Find the value of the following.
(1) $(-4)^{2}$
(ii) $(-3) \times(-2)^{3}$
(iii) $(-2)^{3} \times(-10)^{3}$
Solution:
(i) $(-4)^{2}=(-1)^{2} \times(4)^{2}\left[\right.$ since $\left.a^{m} \times b^{m}=(a \times b)^{m}\right]$
$=1 \times 16=16\left[\right.$ since $(-1)^{\mathrm{n}}=1$ if $\mathrm{n}$ is even $]$
(ii) $(-3) \times(-2)^{3}=(-1) \times(-3) \times(-1)^{3} \times(-2)^{3}$
$=(-1)^{4} \times 24$ [Grouping the terms of same base] $=24$
(iii) $(-2)^{3} \times(-10)^{3}=(-1)^{3} \times(-2)^{3} \times(-1)^{3} \times(-10)^{3}$ $=(-1)^{3+3} \times 2^{3} \times 10^{3}$ [Grouping the terms of same base] $=(-1)^{6} \times(2 \times 10)^{3}$
$\left[\because a^{m} \times b^{m}=(a \times b)^{m}\right]$
$=1 \times 20^{3}\left[\right.$ since $(-1)^{\mathrm{n}}=1$ if $\mathrm{n}$ is even $]$
$=8000$

Question $9 .$
Simplify using laws of exponents.
(i) $3^{5} \times 3^{8}$
(ii) $a^{4} \times a^{10}$
(iii) $7 \mathrm{x} \times 7^{2}$
(iv) $2^{5} \div 2^{3}$
(v) $18^{8} \div 18^{4}$
(vi) $\left(6^{4}\right)^{3}$
(vii) $\left(\mathrm{x}^{\mathrm{m}}\right)^{0}$
(viii) $9^{5} \times 3^{5}$
(ix) $3^{\mathrm{y}} \times 12^{\mathrm{y}}$
(x) $25^{6} \times 5^{6}$
Solution:
(i) $3^{5} \times 3^{8}=3^{5+8}=3^{13}$
[since $a^{m} \times a^{n}=a^{m+n}$ ]
(ii) $a^{4} \times a^{10}=a^{4+10}=a^{14}$
(iii) $7^{x} \times 7^{2}=7^{x+2}$
(iv) $2^{5} \div 2^{3}=2^{5-3}=2^{2}$
[since $\frac{a^{m}}{a^{n}}=a^{m-n}$ ]
(v) $18^{8} \div 18^{4}=18^{8-4}=18^{4}$
[since $\left.\left(a^{m}\right)^{n}=a^{m \times n}\right]$
(vi) $\left(6^{4}\right)_{0}^{3}=6^{4 \times 3}=6^{12}$
[since $\left(a^{m}\right)^{n}=a^{m \times n} ; a^{0}=1$ ]
(vii) $\left(x^{m}\right)^{0}=x^{m \times 0}=x^{0}=1$
[since $a^{m} \times b^{m}=(a \times b)^{m}$ ]
(viii) $9^{5} \times 3^{5}=(9 \times 3)^{5}=27^{5}$
(ix) $3^{y} \times 12^{y}=(3 \times 12)^{y}=36^{y}$
(x) $25^{6} \times 5^{6}=(25 \times 5)^{6}=125^{6}$

Question 10 .
If $a=3$ and $b=2$, then find the value of the following.
(i) $a^{b}+b^{a}$
(ii) $a^{a}-b^{b}$
(iii) $(a+b)^{b}$
(iv) $(\mathrm{a}-\mathrm{b})^{\mathrm{a}}$
Solution:
(i) $a^{b}+b^{a}$
$a=3$ and $b=2$
we get $3^{2}+2^{3}=(3 \times 3)+(2 \times 2 \times 2)=9+8=17$
(ii) $\left(a^{a}-b^{b}\right)$
Substituting $a=3$ and $b=2$
we get $3^{2}-2^{2}=(3 \times 3 \times 3)-(2 \times 2)=27-4=23$
(iii) $(\mathrm{a}+\mathrm{b})^{\mathrm{b}}$
Substituting $a=3$ and $b=2$
we get $(3+2)^{2}=5^{2}=5 \times 5=25$
(iv) $(a-b)^{a}$
Substituting $\mathrm{a}=3$ and $\mathrm{b}=2$
we get $(3-2)^{3}=1^{3}=1 \times 1 \times 1=1$

Question 11 .
Simplify and express each of the following in exponential form:
(i) $4^{5} \times 4^{2} \times 4^{4}$
(ii) $\left(3^{2} \times 3^{3}\right)^{7}$
(iii) $\left(5^{2} \times 5^{8}\right) \div 5^{\mathrm{s}}$
(iv) $2^{0} \times 3^{0} \times 4^{0}$
(v) $\frac{5^{5} \times a^{5} \times b^{3}}{4^{3} \times a^{5} \times b^{2}}$
Solution:
(i) $4^{5} \times 4^{2} \times 4^{4}=4^{5+2+4}=4^{11}$
[Using product rule]
(ii) $\left(3^{2} \times 3^{3}\right)^{7}$
(iii) $\left(5^{2} \times 5^{8}\right) \div 5^{5}=\left(3^{2+3}\right)^{7}=\left(3^{5}\right)^{7}=3^{5 \times 7}=3^{5 \times 7}=3^{35}[$ Using product rule $]$
(iii) $\left(5^{2} \times 5^{8}\right) \div 5^{5}=5^{2+8} \div 5^{5}$
[Using product rule]
$=5^{10} \div 5^{5}=5^{10-5}=5$
$\left[\because a^{0}=1\right]$
(v) $\frac{5^{5} \times a^{8} \times b^{3}}{4^{3} \times a^{5} \times b^{2}}$
[Using quotient rule]
$4^{2} \times a^{3} \times b^{2}=4 \times 4 \times a^{3} \times b=16 a^{3} b$
 

Objective Type Questions

Question $12 .$
$a \times a \times a \times a \times a$ equal to
(i) $\mathrm{a}^{5}$
(ii) $5 \mathrm{a}$
(iii) $5 \mathrm{a}$
(iv) $a+5$
Answer:
(i) $a^{5}$

Question 13.
The exponential form of 72 is
(i) 72
(ii) 27
(iii) $2^{2} \times 3^{3}$
(iv) $2^{3} \times 3^{2}$
Answer:
(iv) $2^{3} \times 3^{2}$

Question $14 .$
The value of $x$ in the equation $a^{13}=x^{3} \times a^{10}$ is
(i) $\mathrm{a}$
(ii) 13
(iii) 3
(iv) 10
Answer:
(i) $\mathrm{a}$

Question $15 .$
How many zeros are there in $100^{10}$ ?
(i) 2
(ii) 3
(iii) 100
(iv) 20
Answer:
(iv) 20

Question $16 .$

$2^{40}+2^{40}$ is equal to
(i) $4^{40}$
(ii) $2^{80}$
(iii) $2^{41}$
(iv) $4^{80}$
Answer:
(iii) $2^{41}$