Difference between revisions of "2022 AIME II Problems/Problem 4"
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==Solution 6== | ==Solution 6== | ||
+ | By change of base formula, | ||
<cmath> | <cmath> | ||
− | \frac{\log_{2x} 22x}{\log_{2x} 20x} | + | \frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1 |
</cmath> | </cmath> | ||
+ | <cmath> | ||
+ | \log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1 | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101 | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \log_{10} {\frac{11}{1010}} = \log_{2x} 101 | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \log_{10} {\frac{11}{1010}} + 1 = \log_{2x} 101 + 1 = \log_{2x} 202x = \log_{20x} {22x} | ||
+ | </cmath> | ||
+ | Thus, | ||
+ | <cmath> | ||
+ | \log_{20x} 22x = \log_{10} \left( \frac{11}{1010} \times 10 \right) = \log_{10} \frac{11}{101} | ||
+ | </cmath> | ||
+ | The requested answer is <math>11 + 101 = \boxed{112}</math>. | ||
+ | |||
+ | ~ adam_zheng | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 18:14, 30 January 2024
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
Define to be
, what we are looking for. Then, by the definition of the logarithm,
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, .
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 (Solution 1 with more reasoning)
Let be the exponent such that
and
. Dividing, we get
Thus, we see that
, so the answer is
.
~A_MatheMagician
Solution 5
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
Solution 6
By change of base formula,
Thus,
The requested answer is
.
~ adam_zheng
Video Solution
https://www.youtube.com/watch?v=4qJyvyZN630
Video Solution by Power of Logic
~Hayabusa1
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.