Difference between revisions of "1983 AHSME Problems/Problem 25"
Sevenoptimus (talk | contribs) m (Fixed formatting) |
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Therefore, we have | Therefore, we have | ||
<cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath> | <cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | This problem is easier if we turn the first two equations into logs: <math>a = \log_{60} 3</math>, <math>b = \log_{60} 5</math>. Then | ||
+ | |||
+ | \begin{align*} | ||
+ | 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} } | ||
+ | \end{align*} | ||
+ | |||
+ | Using the fact that <math>1 = \log_{60} 60</math>, we can combine the linear combination of logs into one log. In fact, adding and subtracting logs in this way is how multiplication and division is done with log tables and slide rules. | ||
+ | |||
+ | \begin{align*} | ||
+ | 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{\log_{60} 60/(3 \cdot 5)}{2(\log_{60} 60/5)} } \\ | ||
+ | &= 12^{ \frac{\log_{60} 4}{2\log_{60} 12} } \\ | ||
+ | &= 12^{\frac{1}{2} \log_{12} 4} \\ | ||
+ | &= 4^{1/2} \\ | ||
+ | &= \boxed{2}. | ||
+ | \end{align*} | ||
==See Also== | ==See Also== |
Revision as of 14:36, 17 April 2021
Contents
Problem 25
If and , then is
Solution
We have that . We can substitute our value for 5, to get Hence Since , we have Therefore, we have
Solution 2
This problem is easier if we turn the first two equations into logs: , . Then
\begin{align*} 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} } \end{align*}
Using the fact that , we can combine the linear combination of logs into one log. In fact, adding and subtracting logs in this way is how multiplication and division is done with log tables and slide rules.
\begin{align*} 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{\log_{60} 60/(3 \cdot 5)}{2(\log_{60} 60/5)} } \\
&= 12^{ \frac{\log_{60} 4}{2\log_{60} 12} } \\ &= 12^{\frac{1}{2} \log_{12} 4} \\ &= 4^{1/2} \\ &= \boxed{2}.
\end{align*}
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.