A sum of money at 20% rate of compound interest per annum becomes more than 100 times in n years. What is the <strong>LEAST</strong> value of n? (Use \log_{10}2=0.301, \log_{10}3=0.477)

Explanation

We require (1.2)^n > 100. Taking \log_{10} on both sides: n(\log_{10} 12 - 1) > 2. We know \log_{10} 12 = 2\log_{10} 2 + \log_{10} 3 = 2(0.301) + 0.477 = 1.079. Thus, n(1.079 - 1) > 2 \implies 0.079n > 2 \implies n > \frac{2}{0.079} \approx 25.3. The least integer value is 26.

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