Difference between revisions of "1981 AHSME Problems/Problem 13"
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==Problem== | ==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 <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} | + | 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> | ||
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| + | ==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}=0.477</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>. | ||
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| + | -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 
of the value it had at the beginning of that year. Find the smallest integer 
such that after years, the money will have lost at least 
of its value (To the nearest thousandth 
).
Solution
What we are trying to solve is 
. This turns into 
We know that , thus by log rules we have 
, thus 
, and our answer is 
.
-edited by Maxxie and maxamc
