1983 AIME Problems/Problem 1
Problem
Let ,
, and
all exceed
, and let
be a positive number such that
,
, and
. Find
.
Solution
Solution 1
[hide="Solution"] 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
.
[/hide]
Solution 2
Applying the change of base formula,
\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*}$ (Error compiling LaTeX. Unknown error_msg)Therefore, .
Hence, .
[/hide]
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 |