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

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==Problem==
 
==Problem==
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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 <math>n</math> such that after <math>n</math> years, the money will have lost at least <math>90\%</math> of its value (To the nearest thousandth <math>log_{10}^{3}=0.477</math>).
+
Suppose that at the end of any year, a unit of money has lost <math>10\%</math> of the value it had at the beginning of that year. Find the smallest integer <math>n</math> such that after <math>n</math> years, the money will have lost at least <math>90\%</math> of its value (To the nearest thousandth <math>\log_{10}{3}=0.477</math>).
  
 
<math>\textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22</math>
 
<math>\textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22</math>
  
 
==Solution==
 
==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>2\log_{10}{3}=\log_{10}{9}=2*0.477=0.954</math>, thus <math>n=\frac{1}{.046} \approx 21.7</math>, and our answer is <math>\boxed{(\text{E}) 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}=0.477</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} \approx 21.7</math>, and our answer is <math>\boxed{(\text{E}) 22}</math>.
  
 
-edited by Maxxie and maxamc
 
-edited by Maxxie and maxamc

Latest revision as of 13:46, 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 $\log_{10}{3}=0.477$, 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

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