Difference between revisions of "1981 AHSME Problems/Problem 13"

Revision as of 13:43, 28 October 2024

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

-edited by Maxxie and maxamc

@@ Line 5: / Line 5: @@
 ==Solution==
-What we are trying to solve is <math>log_{0.9}^{0.1}=n</math>. This turns into <math>\frac{log{0.1}}{log{0.9}}=\frac{-1}{log{9}-1}=n</math> We know that <math>log_{10}^{3}=0.466</math>, thus by log rules we have <math>2log_{10}^{3}=log_{10}^{9}=2*0.477=0.954</math>, thus <math>n=\frac{1}{.046} > 21</math>, and our answer is <math>\boxed{(B) 22}</math>
+What we are trying to solve is <math>\log_{0.9}^{0.1}=n</math>. This turns into <math>\frac{\log{0.1}}{\log{0.9}}=\frac{-1}{\log{9}-1}=n</math> We know that <math>\log_{10}^{3}=0.466</math>, thus by log rules we have <math>2\log_{10}^{3}=\log_{10}^{9}=2*0.477=0.954</math>, thus <math>n=\frac{1}{.046} > 21</math>, and our answer is <math>\boxed{(B) 22}</math>
--edited by Maxxie
+-edited by Maxxie and maxamc

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Difference between revisions of "1981 AHSME Problems/Problem 13"

Revision as of 13:43, 28 October 2024

Problem

Solution