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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.1

Question 1.

Fill in the blanks.

- The exponential form 14
^{9}should be read as ______ - The expanded form of p
^{3}q^{2}is ______ - When base is 12 and exponent is 17, its e×ponential form is _____
- The value of (14 × 21)
^{0}is _____

Answers:

- 14 Power 9
- p × p × p × q × q
- 12
^{17} - 1

Question 2.

Say True or False.

- 2
^{3}× 3^{2}= 65 - 2
^{9}× 3^{2}= (2 × 3)^{9×2} - 3
^{4}× 3^{7}= 3^{11} - 2
^{0}× 1000^{0} - 2
^{3}< 3^{2}

Answers:

- False
- False
- True
- True
- True

Question 3.

Find the value of the following.

- 2
^{6} - 11
^{2} - 5
^{4} - 9
^{3}

Solution:

- 2
^{6}= 2 × 2 × 2 × 2 × 2 × 2 = 64 - 11
^{2}= 11 × 11 = 121 - 5
^{4}= 5 × 5 × 5 × 5 = 625 - 9
^{3}= 9 × 9 × 9 = 729

Question 4.

Express the following in e×ponential form.

- 6 × 6 × 6 × 6
- t × t
- 5 × 5 × 7 × 7 × 7
- 2 × 2 × a × a

Solution:

- 6 × 6 × 6 × 6 = 6
^{1+1+1+1}= 64 [Since a^{m}× a^{n}= a^{m+n}] - t × t = t
^{1+1}= t^{2} - 5 × 5 × 7 × 7 × 7 = 5
^{1+1}× 7^{1+1+1}= 5^{2}× 7^{3} - 2 × 2 × a × a = 2
^{1+1}× a^{1+1}= 2^{2}× a^{2}= (2a)^{2}

Question 5.

E×press each of the following numbers using e×ponential form,

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution:

(i) 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2^{9} [Using product rule]

(ii) 343

343 = 7 × 7 × 7 = 7^{1+1+1}

= 7^{3} [Using product rule]

(iii) 729

729 = 3 × 3 × 3 × 3 × 3 × 3

= 3^{6} [Using product rule]

(iv) 3125

3125 = 5 × 5 × 5 × 5 × 5

= 5^{5} [Using product rule]

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}

6^{3} = 6 × 6 × 6 = 36 × 6 = 216

3^{6} = 3 × 3 × 3 × 3 × 3 × 3 = 729

729 > 216 gives 3^{6} > 6^{3}

∴ 36 is greater.

(ii) 5^{3} or 3^{5}

5^{3} = 5 × 5 × 5 = 125

3^{5} = 3 × 3 × 3 × 3 × 3 = 243

243 > 125 gives 3^{5} > 5^{3}

∴ 3^{5} is greater.

(iii) 2^{8} or 8^{2}

2^{8} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8^{2} = 8 × 8 = 64

256 > 64 gives 2^{8} > 8^{2}

∴ 2^{8} is greater.

Question 7.

Simplify the following

(i) 7^{2} × 3^{4}

(ii) 3^{2} × 2^{4}

(iii) 5^{2} × 10^{4}

Solution:

(i) 7^{2} × 3^{4} = (7 × 7) × (3 × 3 × 3 × 3)

= 49 × 81 = 3969

(ii) 3^{2} × 2^{4} = (3 × 3) × (2 × 2 × 2 × 2)

= 9 × 16 = 144

(iii) 5^{2} × 10^{4} = (5 × 5) × (10 × 10 × 10 × 10)

= 25 × 10000 = 2,50,000

Question 8.

Find the value of the following.

(i) (-4)^{2}

(ii) (-3) × (-2)^{3}

(iii) (-2)^{3} × (-10)^{3}

Solution:

(i) (-4)^{2} = (-1)^{2} × (4)^{2} [since a^{m} × b^{m} = (a × b)^{m}]

= 1 × 16 = 16 [since (-1)^{n} = 1 if n is even]

(ii) (-3) × (-2)^{3} = (-1) × (-3) × (-1)^{3} × (-2)^{3}

= (-1)^{4} × 24 [Grouping the terms of same base]

= 24

(iii) (-2)^{3} × (-10)^{3} = (-1)^{3} × (-2)^{3} × (-1)^{3} × (-10)^{3}

= (-1)^{3+3} × 2^{3} × 10^{3} [Grouping the terms of same base]

= (-1)^{6} × (2 × 10)^{3}

[∵ a^{m} × b^{m} = (a × b)^{m}]

= 1 × 20^{3} [since (-1)^{n} = 1 if n is even]

= 8000

Question 9.

Simplify using laws of exponents.

(i) 3^{5} × 3^{8}

(ii) a^{4} × a^{10}

(iii) 7x × 7^{2}

(iv) 2^{5} ÷ 2^{3}

(v) 18^{8} ÷ 18^{4}

(vi) (6^{4})^{3}

(vii) (x^{m})^{0}

(viii) 9^{5} × 3^{5}

(ix) 3^{y} × 12^{y}

(x) 25^{6} × 5^{6}

Solution:

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) (a – b)^{a}

Solution:

(i) a^{b} + b^{a}

a = 3 and b = 2

we get 3^{2} + 2^{3} = (3 × 3) + (2 × 2 × 2) = 9 + 8 = 17

(ii) (a^{a} – b^{b})

Substituting a = 3 and b = 2

we get 3^{2} – 2^{2} = (3 × 3 × 3) – (2 × 2) = 27 – 4 = 23

(iii) (a + b)^{b}

Substituting a = 3 and b = 2

we get (3 + 2)^{2} = 5^{2} = 5 × 5 = 25

(iv) (a – b)^{a}

Substituting a = 3 and b = 2

we get (3 – 2)^{3} = 1^{3} = 1 × 1 × 1 = 1

Question 11.

Simplify and express each of the following in exponential form:

(i) 4^{5} × 4^{2} × 4^{4}

(ii) (3^{2} × 3^{3})^{7}

(iii) (5^{2} × 5^{8}) ÷ 5^{s}

(iv) 2^{0} × 3^{0} × 4^{0}

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

Solution:

Objective Type Questions

Question 12.

a × a × a × a × a equal to

(i) a^{5}

(ii) 5 a

(iii) 5a

(iv) a + 5

Answer:

(i) a^{5}

Question 13.

The exponential form of 72 is

(i) 72

(ii) 27

(iii) 2^{2} × 3^{3}

(iv) 2^{3} × 3^{2}

Answer:

(iv) 2^{3} × 3^{2}

Question 14.

The value of x in the equation a^{13} = x^{3} × a^{10} is

(i) a

(ii) 13

(iii) 3

(iv) 10

Answer:

(i) 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}