(i) \(3^{-2} = ({1\over 3})^2\) = \(1\over9\)
(ii) \(-4^{-2}={1\over -4}\times {1\over -4}\) = \(1\over 16\)
(iii) \({({1\over 2})}^{-5}\) = \({({2\over 1})}^{5}\)
\(= 2\times 2\times 2\times 2\times 2\times 2=32\)
Question 2
Simplify and express the result in power notation with positive exponent.
(i) \((-4)^5 ÷(-4)^8 =(-4)^{5-8}=(-4)^{-3}={1\over (-4)^3}\)
Solution:
\((-4)^5 \div (-4)^8 = (-4)^{5 - 8} \quad \left( a^m \div a^n = a^{m - n} \right)\)
\(= (-4)^{-3}\)
\(= \dfrac{1}{(-4)^3} \quad \left( a^{-m} = \dfrac{1}{a^m} \right)\)
(ii) \(({1\over2^3})^2={1\over 2^6} \)
Solution:
\(\left( \dfrac{1}{2^3} \right)^2 = \dfrac{1}{(2^3)^2}\)
\(= \frac{1}{2^6} \quad \left( (a^m)^n = a^{mn} \right)\)
(iii) \( (-3)^4×({5\over3})^4 \)
Solution:
\((-3)^4 \times \left( \dfrac{5}{3} \right)^4 = (-1 \times 3)^4 \times \dfrac{5^4}{3^4}\)
\(= (-1)^4 \times 3^4 \times \dfrac{5^4}{3^4} \quad \left( (ab)^n = a^n \times b^n \right)\)
\(= (-1)^4 \times 5^4 \quad \left[ (-1)^4 = 1 \right]\)
\(= 5^4\)
(iv) \((3^{-7}÷3^{-10})×3^{-5} \)
Solution:
\(= (3^{-7 - (-10)}) \times 3^{-5} \quad \left( a^m \div a^n = a^{m - n} \right)\)
\(= 3^3 \times 3^{-5}\)
\(= 3^{3 + (-5)} = 3^{-2} \quad \left( a^m \times a^n = a^{m + n} \right)\)
\(= \frac{1}{3^2} \quad \left( a^{-m} = \frac{1}{a^m} \right)\)
(v) \(2^{-3}×(-7)^{-3}\)
Solution:
\(2^{-3} \times (-7)^{-3} = \dfrac{1}{2^3} \times \dfrac{1}{(-7)^3} \quad \left( a^{-m} = \dfrac{1}{a^m} \right)\)
\(2^{-3} \times (-7)^{-3} = \dfrac{1}{2^3} \times \dfrac{1}{(-7)^3} \quad \left( a^{-m} = \dfrac{1}{a^m} \right)\)
\(= \dfrac{1}{[2 \times (-7)]^3} \quad \left( a^m \times b^m = (ab)^m \right)\)
\(= \dfrac{1}{(-14)^3}\)
Question 3: Find the value of
(i) \(3^0 + 4^{-1}) \times 2^2\)
Solution:
\(= (1 + {1\over 4}) \times 2^2 \ \ \ \ \ \ \ \ [\because a^0 = 1]\)
\(= ({5\over4}) \times 4\)
\(= 5\)
Solution:
\(= (1 + {1\over 4}) \times 2^2 \ \ \ \ \ \ \ \ [\because a^0 = 1]\)
\(= ({5\over4}) \times 4\)
\(= 5\)
(ii) \((2^{-1} \times 4^{-1}) ÷ 2^{-2}\)
Solution:
\(= [2^{-1} \times {(2)^2}^{-1}] ÷ 2^{-2}\)
\(= (2^{-1} \times 2^-2) ÷ 2^-2 \ \ \ \ \ \ \ \ [\because (a^m)^n = a^{mn}]\)
\(= 2^{-1 + (-2)} ÷ 2^-2 \ \ \ \ \ \ \ \ [\because a^m \times a^n = a^{m+n}]\)
\(= 2^{-3} ÷ 2^{-2}\)
\(= 2^{-3 - (-2)} \ \ \ \ \ \ \ \ [\because a^m ÷ a^n = a^{m-n}]\)
\(= 2^{-3 + 2}\)
\(= 2^{-1} \)
Solution:
\(= [2^{-1} \times {(2)^2}^{-1}] ÷ 2^{-2}\)
\(= (2^{-1} \times 2^-2) ÷ 2^-2 \ \ \ \ \ \ \ \ [\because (a^m)^n = a^{mn}]\)
\(= 2^{-1 + (-2)} ÷ 2^-2 \ \ \ \ \ \ \ \ [\because a^m \times a^n = a^{m+n}]\)
\(= 2^{-3} ÷ 2^{-2}\)
\(= 2^{-3 - (-2)} \ \ \ \ \ \ \ \ [\because a^m ÷ a^n = a^{m-n}]\)
\(= 2^{-3 + 2}\)
\(= 2^{-1} \)
\(= {1\over 2}\)
(iii) \(({1\over 2})^{-2} + ({1\over 3})^{-2} + ({1\over 4})^{-2}\)
Solution:
\(= ({2\over 1})^2 + ({2\over 1})^2 + ({2\over 1})^2\)
\(= 2^2 + 3^2 + 4^2\)
\(= 4 + 9 + 16\)
\(= 29\)
(iii) \(({1\over 2})^{-2} + ({1\over 3})^{-2} + ({1\over 4})^{-2}\)
Solution:
\(= ({2\over 1})^2 + ({2\over 1})^2 + ({2\over 1})^2\)
\(= 2^2 + 3^2 + 4^2\)
\(= 4 + 9 + 16\)
\(= 29\)
(iv) \((3^{-1} + 4^{-1} + 5^{-1})^0\)
Solution:
\(= ({1\over 3} + {1\over 4}+ {1\over 5})^0\)
\(= 1 \ \ \ \ \ \ (\because (a^0 = 1)\)
(v) \(\left({({-2\over 3})^{-2}}\right)^2\)
Solution:
\(= \left(({3\over-2})^2\right)^2 \ \ \ \ \ \ \ \ [a^{-m} = {1\over a^m}]\)
\(= \left({3^2 \over (-2)^2}\right)^2\)
Solution:
\(= \left(({3\over-2})^2\right)^2 \ \ \ \ \ \ \ \ [a^{-m} = {1\over a^m}]\)
\(= \left({3^2 \over (-2)^2}\right)^2\)
\(= ({9\over 4})^2\)
\(= {81\over 16}\)
Answer:
\(\dfrac{8^{-1} \times 5^3}{2^{-4}}= \dfrac{2^4 \times 5^3}{8^1} = \dfrac{2^4 \times 5^3}{2^3} \quad \left[ a^{-m} = \frac{1}{a^m} \right]\)
\(= 2^{4-3} \times 5^3 \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(= 2 \times 125 = 250\)
\(= {81\over 16}\)
Question 4: Evaluate
(i) \(\dfrac{8^{-1} \times 5^3}{2^{-4}}\)Answer:
\(\dfrac{8^{-1} \times 5^3}{2^{-4}}= \dfrac{2^4 \times 5^3}{8^1} = \dfrac{2^4 \times 5^3}{2^3} \quad \left[ a^{-m} = \frac{1}{a^m} \right]\)
\(= 2^{4-3} \times 5^3 \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(= 2 \times 125 = 250\)
(ii) \((5^{-1} \times 2^{-1}) \times 6^{-1}\)
Answer:
\(= \left( \dfrac{1}{5} \times \dfrac{1}{2} \right) \times \dfrac{1}{6} \quad \left[ a^{-m} = \dfrac{1}{a^m} \right]\)
\(= \dfrac{1}{10} \times \dfrac{1}{6} = \dfrac{1}{60}\)
Question 5: Find the value of \(m\) for which \(5^m \div 5^{-3} = 5^5\)
Answer:
\(5^m \div 5^{-3} = 5^5\)
\(5^{m - (-3)} = 5^5 \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(5^{m+3} = 5^5\)
Since the powers have same bases on both sides, their respective exponents must be equal.
\(m + 3 = 5\)
\(m = 5 - 3\)
Answer:
\(= \left( \dfrac{1}{5} \times \dfrac{1}{2} \right) \times \dfrac{1}{6} \quad \left[ a^{-m} = \dfrac{1}{a^m} \right]\)
\(= \dfrac{1}{10} \times \dfrac{1}{6} = \dfrac{1}{60}\)
Question 5: Find the value of \(m\) for which \(5^m \div 5^{-3} = 5^5\)
Answer:
\(5^m \div 5^{-3} = 5^5\)
\(5^{m - (-3)} = 5^5 \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(5^{m+3} = 5^5\)
Since the powers have same bases on both sides, their respective exponents must be equal.
\(m + 3 = 5\)
\(m = 5 - 3\)
\(m = 2\)
Question 6: Evaluate
(i)\( \left\{ \left( \dfrac{1}{3} \right)^{-1} - \left( \dfrac{1}{4} \right)^{-1} \right\}^{-1}\)
Question 6: Evaluate
(i)\( \left\{ \left( \dfrac{1}{3} \right)^{-1} - \left( \dfrac{1}{4} \right)^{-1} \right\}^{-1}\)
Answer:
\(= \left\{ \left( \dfrac{3}{1} \right)^1 - \left( \dfrac{4}{1} \right)^1 \right\}^{-1} \quad \left( a^{-m} = \frac{1}{a^m} \right)\)
\(= \{3 - 4\}^{-1}\)
\(= (-1)^{-1} = \dfrac{1}{-1}\)
\(= -1\)
(ii) \(\left( \dfrac{5}{8} \right)^{-7} \times \left( \dfrac{8}{5} \right)^{-4}\)
Answer:
\(= \dfrac{5^{-7}}{8^{-7}} \times \dfrac{8^{-4}}{5^{-4}} \quad \left[ \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \right]\)
\(= \dfrac{8^7}{5^7} \times \dfrac{5^4}{8^4} \quad \left( a^{-m} = \dfrac{1}{a^m} \right)\)
\(= \dfrac{8^{7-4}}{5^{7-4}} \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(= \dfrac{8^3}{5^3}\)
\(= \dfrac{512}{125}\)
Question 7: Simplify:
(i) \(\dfrac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \quad (t \neq 0)\)
Answer:
\(= \dfrac{5^2 \times t^{-4}}{5^{-3} \times 5 \times 2 \times t^{-8}}\)
\(= \dfrac{5^2 \times t^{-4}}{5^{-3+1} \times 2 \times t^{-8}} \quad \left( a^m \times a^n = a^{m+n} \right)\)
\(= \dfrac{5^2 \times t^{-4}}{5^{-2} \times 2 \times t^{-8}}\)
\(= \dfrac{5^{2 - (-2)} \cdot t^{-4 - (-8)}}{2} \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(= \dfrac{5^4 \cdot t^4}{2} = \frac{625 \cdot t^4}{2}\)
(ii) \(\dfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Answer:
\(= \dfrac{3^{-5} \times (2 \times 5)^{-5} \times 5^3}{5^{-7} \times (2 \times 3)^{-5}}\)
Answer:
\(= \dfrac{3^{-5} \times (2 \times 5)^{-5} \times 5^3}{5^{-7} \times (2 \times 3)^{-5}}\)
\(= \dfrac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}} \quad \left[ (a \times b)^m = a^m \times b^m \right]\)
\(= 3^{-5 - (-5)} \times 2^{-5 - (-5)} \times 5^{-5 + 3 - (-7)} \quad \left( a^m \div a^n = a^{m-n} \right)\)
\(= 3^0 \times 2^0 \times 5^5 \quad \left( a^0 = 1 \right)\)
\(= 5^5\)
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