Difference between revisions of "1983 AHSME Problems/Problem 25"

(Solution)
(Solution 2)
Line 17: Line 17:
 
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
 
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*}
+
<cmath>12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} }</cmath>
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.
 
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.

Revision as of 14:37, 17 April 2021

Problem 25

If $60^a=3$ and $60^b=5$, then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$

Solution

We have that $12=\frac{60}{5}$. We can substitute our value for 5, to get \[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\] Hence \[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\] Since $4=\frac{60}{3\cdot 5}$, we have \[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\] Therefore, we have \[60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}\]

Solution 2

This problem is easier if we turn the first two equations into logs: $a = \log_{60} 3$, $b = \log_{60} 5$. Then

\[12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} }\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)

Using the fact that $1 = \log_{60} 60$, 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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS