Question 21 of 28intermediate🔧 ApplyShort Answer5 marks

Write equivalent forms of the following. (i) 2−42^{-4} (ii) 10−510^{-5} (iii) (−7)−2(-7)^{-2} (iv) (−5)−3(-5)^{-3} (v) 10−10010^{-100}

(i)
Answer

124=116\frac{1}{2^4} = \frac{1}{16}

(ii)
Answer

1105=1100000\frac{1}{10^5} = \frac{1}{100000}

(iii)
Answer

1(−7)2=149\frac{1}{(-7)^2} = \frac{1}{49}

(iv)
Answer

1(−5)3=1−125=−1125\frac{1}{(-5)^3} = \frac{1}{-125} = -\frac{1}{125}

(v)
Answer

110100\frac{1}{10^{100}}

Explanation

(i) Using n−a=1nan^{-a} = \frac{1}{n^a}: 2−4=124=1162^{-4} = \frac{1}{2^4} = \frac{1}{16} (ii) Using n−a=1nan^{-a} = \frac{1}{n^a}: 10−5=1105=110000010^{-5} = \frac{1}{10^5} = \frac{1}{100000} (iii) Using n−a=1nan^{-a} = \frac{1}{n^a}: (−7)−2=1(−7)2=149(-7)^{-2} = \frac{1}{(-7)^2} = \frac{1}{49} (positive because even power) (iv) Using n−a=1nan^{-a} = \frac{1}{n^a}: (−5)−3=1(−5)3=1−125=−1125(-5)^{-3} = \frac{1}{(-5)^3} = \frac{1}{-125} = -\frac{1}{125} (v) Using n−a=1nan^{-a} = \frac{1}{n^a}: 10−100=11010010^{-100} = \frac{1}{10^{100}}

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