2022 AIME II Problems/Problem 4
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
Define to be
, what we are looking for. Then, by the definition of logs,
Dividing the first equation by the second equation gives us
, so by the definition of logs,
. This is what the problem asked for, so the fraction
gives us
.
~ihatemath123
Solution 2
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 3
We have
We have
Because , we get
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)
Solution 4
By the change of base rule, we have , or
. We also know that if
, then this also equals
. We use this identity and find that
. The requested sum is
~MathIsFun286
Video Solution
https://www.youtube.com/watch?v=4qJyvyZN630
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 |
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