Difference between revisions of "1981 AHSME Problems/Problem 13"
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| − | 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}>\ | + | 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> |
-edited by Maxxie | -edited by Maxxie | ||
Latest revision as of 00:12, 8 September 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 
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
