1983 AIME Problems/Problem 1
Problem
Let 
, 
, and 
all exceed 
, and let 
be a positive number such that 
, 
, and 
. Find 
.
Solutions
Solution 1
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential expressions.
, 
, and 
. If we now convert everything to a power of 
, it will be easy to isolate and .
, 
, and 
.
With some substitution, we get 
and 
.
Solution 2
First we'll convert everything to exponential form.
, , and . The only expression with z is . It now becomes clear one way to find is to find what 
and 
are in terms of .
Taking the square root of the equation results in 
. Taking the 
root of equates to 
.
Going back to , we can substitute the and with 
and 
, respectively. We now have **
. Simplify we get 
.
So our answer is \boxed{060}$=== Solution 3 ===
Applying the change of base formula,
<cmath>\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\
\log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\
\log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}</cmath>
Therefore,$ (Error compiling LaTeX. Unknown error_msg) \frac {\log z}{\log
w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$.
Hence,$ (Error compiling LaTeX. Unknown error_msg) \log_z w = \boxed{060}$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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