1981 AHSME Problems/Problem 13

Problem

Suppose that at the end of any year, a unit of money has lost $10\%$ of the value it had at the beginning of that year. Find the smallest integer $n$ such that after $n$ years, the money will have lost at least $90\%$ of its value (To the nearest thousandth $\log_{10}{3}=0.477$ ).

$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22$

Solution

What we are trying to solve is $\log_{0.9}{0.1}=n$ . This turns into $\frac{\log{0.1}}{\log{0.9}}=\frac{-1}{\log{9}-1}=n$ We know that $\log_{10}{3}=0.477$ , thus by log rules we have $2\log_{10}{3}=\log_{10}{9}=2*0.477=0.954$ , thus $n=\frac{1}{.046} \approx 21.7$ , and our answer is $\boxed{(\text{E}) 22}$ .

-edited by Maxxie and maxamc

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1981 AHSME Problems/Problem 13

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