Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 12 Exponents and Powers Ex 12.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1
1. Evaluate:
Question (i)
3-2
Solution:
= \frac{1}{3^{2}}
= \frac{1}{3 \times 3}
= \frac {1}{9}
Question (ii)
(-4)-2
Solution:
= \frac{1}{(-4)^{2}}
= \frac{1}{(-4) \times(-4)}
= \frac {1}{9}
Question (iii)
(\frac {1}{2})-5
Solution:
= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}
= \frac{1}{\frac{1}{32}}
= 32
2. Simplify and express the result in power notation with positive exponent:
Question (i)
(-4)5 ÷ (-4)8
Solution:
= (-4)5 – 8 (∵ am ÷ an = am-n)
= (-4)– 3
= \frac{1}{(-4)^{3}}
Question (ii)
\left(\frac{1}{2^{3}}\right)2
Solution:
= \frac{(1)^{2}}{\left(2^{3}\right)^{2}} [∵ \left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}]
= \frac{1}{2^{3 \times 2}} [∵ (am)n = amn
= \frac{1}{2^{6}}
Question (iii)
(-3)4 × (\frac {5}{3})4
Solution:
= [(-3) × \frac {5}{3}]4 [∵ am × bm = (ab)m]
= [(-1) × 5]4
= (-1)4 × 54
= 1 × 54
= 54
Question (iv)
(3-7 ÷ 3-10) × 3-5
Solution:
= (3(-7)-(-10)) × 3-5 (∵ am ÷ an = am-n)
= (3-7+10 × 3-5
= 33 × 3-5
= 33 + (-5) (∵ am × an = am+n)
= 3-2
= \frac{1}{3^{2}}
Question (v)
2-3 × (-7)-3
Solution:
= [2 × (-7)]-3 [∵ am × bm = (ab)m
= (-14)-3
= \frac{1}{(-14)^{3}}
3. Find the value of:
Question (i)
(30 + 4-1) × 22
Solution:
= (1 + \frac {1}{4}) × 22
= (1 \frac{1}{4}) × 22
= (\frac {5}{4}) × 4
= 5
Question (ii)
(2– 1 × 4-1) ÷ 22
Solution:
= (\frac {1}{2} × \frac {1}{4}) ÷ \frac{1}{2^{2}}
= (\frac {1}{8}) ÷ \frac {1}{4}
= \frac {1}{8} × \frac {4}{1}
= \frac {1}{2}
Question (iii)
(\frac {1}{2})– 2 + (\frac {1}{3})– 2 + (\frac {1}{4})– 2
Solution:
=\frac{1}{\left(\frac{1}{2}\right)^{2}}+\frac{1}{\left(\frac{1}{3}\right)^{2}}+\frac{1}{\left(\frac{1}{4}\right)^{2}}
= \frac{1}{\frac{1}{4}}+\frac{1}{\frac{1}{9}}+\frac{1}{\frac{1}{16}}
= 4 + 9 + 16
= 29
Question (iv)
(3– 1 + 4– 1 + 5– 1)0
Solution:
(3-1 + 4-1 + 5-1)0
∴ [3-1 + 4-1 + 5-1]0 = 1
[∵ a0 = 1]
Question (v)
\left\{\left(\frac{-2}{3}\right)^{-2}\right\}2
Solution:
= \left(\frac{-2}{3}\right)^{(-2) \times 2} [∵ (am)m = amn]
= (\frac {-2}{3})-4
= \frac{(-2)^{-4}}{(3)^{-4}} [∵ \left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}]
= \frac{3^{4}}{(-2)^{4}}
= \frac{3 \times 3 \times 3 \times 3}{(-2) \times(-2) \times(-2) \times(-2)}
= \frac {81}{16}
4. Evaluate:
Question (i)
\frac{8^{-1} \times 5^{3}}{2^{-4}}
Solution:
= \frac{2^{4} \times 5^{3}}{8}
= \frac{2^{4} \times 5^{3}}{2^{3}}
= 24-3 × 53
= 2 × 125
= 250
Question (ii)
(5-1 × 2-1) × 6-1
Solution:
= (\frac {1}{5} × \frac {1}{2}) × \frac {1}{6}
= (\frac {1}{10}) × \frac {1}{6}
= \frac {1}{60}
Another method:
= 5-1 × 2-1 × 6-1
= (5 × 2 × 6)-1
= (60)-1
= \frac {1}{60}
5. Find the value of m for which 5m ÷ 5-3 = 55.
Solution:
∴ 5m-(-3) = 55.
∴ 5m + 3 = 55
∴ m + 3 = 5 (∵ am = an then m = n)
∴ m = 5 – 3
∴ m = 2
Thus, value of m is 2.
6. Evaluate:
Question (i)
\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}
Solution:
= {\frac{3}{1}-\frac{4}{1}}-1 (∵ a-m = \(\frac {1}{am}\))
= {3 – 4}-1
= {-1}-1
= \frac {1}{-1}
= -1
Question (ii)
(\frac {5}{8})-7 × (\frac {8}{5})-4
Solution:
(\frac {5}{8})-7 × (\frac {5}{8})4
(∵ a-m = \frac{1}{a^{m}})
= (\frac {5}{8})-7+4 (∵ am × an = am+n)
= (\frac {5}{8})-3
= (\frac {8}{5})3
= \frac{8 \times 8 \times 8}{5 \times 5 \times 5}
= \frac {512}{125}
7. Simplify:
Question (i)
\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t ≠ 0)
Solution:
Question (ii)
\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}
Solution: