Difference between revisions of "1958 AHSME Problems/Problem 12"

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==Problem==
 
==Problem==
  
If <math> P \equal{} \frac{s}{(1 \plus{} k)^n}</math> then <math> n</math> equals:
+
If <math> P = \frac{s}{(1 + k)^n}</math> then <math> n</math> equals:
  
<math> \textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 \plus{} k)}}\qquad  
+
<math> \textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad  
\textbf{(B)}\ \log{\left(\frac{s}{P(1 \plus{} k)}\right)}\qquad  
+
\textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad  
\textbf{(C)}\ \log{\left(\frac{s \minus{} P}{1 \plus{} k}\right)}\qquad \\
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\textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\
\textbf{(D)}\ \log{\left(\frac{s}{P}\right)} \plus{} \log{(1 \plus{} k)}\qquad  
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\textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad  
\textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 \plus{} k))}}</math>
+
\textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}</math>
  
 
==Solution==
 
==Solution==

Latest revision as of 23:08, 13 March 2015

Problem

If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:

$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad  \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad  \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad  \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}$

Solution

\[P=\frac{s}{(1+k)^n}\]

\[(1+k)^n=\frac{s}{P}\]

Take the $\log$ of each side.

\[n \log(1+k) = \log\left(\frac{s}{P}\right)\]

\[n = \frac{\log\left(\frac{s}{P}\right)}{\log(1+k)} \to \boxed{\text{(A)}}\]


See also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AHSME Problems and Solutions

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