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 $log_{10}^{3}=0.466$ , thus by log rules we have $2log_{10}^{3}=log_{10}^{9}=2*0.477=0.954$ , thus $n=\frac{1}{.046} > 21$ , and our answer is $\boxed{(B) 22}$

-edited by Maxxie

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1981 AHSME Problems/Problem 13

Problem

Solution