A sum of money at 20% rate of compound interest per annum becomes more than 100 times in n years. What is the least value of n? (Use log₁₀2 = 0.301, log₁₀3 = 0.477)

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Accepted Answer

26. (1.2)^n > 100. n·log(1.2) > 2. log(1.2) = log(12)−1 = log4+log3−1 = 2·0.301+0.477−1 = 0.602+0.477−1 = 0.079. n > 2/0.079 ≈ 25.3. Least integer = 26.

A sum of money at 20% rate of compound interest per annum becomes more than 100 times in n years. What is the least value of n? (Use log₁₀2 = 0.301, log₁₀3 = 0.477)

Explanation

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