2022 AIME II Problems/Problem 4
Contents
Problem
There is a positive real number not equal to either
or
such that
The value
can be written as
, where
and
are relatively prime positive integers. Find
.
Solution 1
We could assume a variable which equals to both
and
.
So that
and
Express as:
Substitute to
:
Thus, , where
and
.
Therefore, .
~DSAERF-CALMIT (https://binaryphi.site)
Solution 2
We have \begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}
We have \begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}
Because , we get
\[
\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20}
= \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .
\]
We denote this common value as .
By solving the equality , we get
.
By solving the equality , we get
.
By equating these two equations, we get
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.