Difference between revisions of "2022 AIME II Problems/Problem 4"
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10^a &= \dfrac{11}{101}. \ | 10^a &= \dfrac{11}{101}. \ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Thus, we see that <math>\log_10 \left(\dfrac{11}{101}right) = a = \log_{20x} 22x</math>, so the answer is <math>11 + 101 = \boxed{112}</math>. | + | Thus, we see that <math>\log_10 \left(\dfrac{11}{101}\right) = a = \log_{20x} 22x</math>, so the answer is <math>11 + 101 = \boxed{112}</math>. |
− | ==Solution | + | ~A_MatheMagician |
+ | |||
+ | ==Solution 5== | ||
By the change of base rule, we have <math>\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}</math>, or <math>\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k</math>. We also know that if <math>a/b=c/d</math>, then this also equals <math>\frac{a-c}{b-d}</math>. We use this identity and find that <math>k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}</math>. The requested sum is <math>11+101=\boxed{112}.</math> | By the change of base rule, we have <math>\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}</math>, or <math>\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k</math>. We also know that if <math>a/b=c/d</math>, then this also equals <math>\frac{a-c}{b-d}</math>. We use this identity and find that <math>k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}</math>. The requested sum is <math>11+101=\boxed{112}.</math> | ||
Revision as of 10:40, 22 January 2023
Contents
[hide]Problem
There is a positive real number not equal to either or such thatThe 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, .
~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 (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
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.