Difference between revisions of "2022 AIME II Problems/Problem 4"
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and <math>(2x)^v=202x \textcircled{2}</math> | and <math>(2x)^v=202x \textcircled{2}</math> | ||
− | Express <math>\textcircled{1}</math> as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x \textcircled{3}</math> | + | Express <math>\textcircled{1}</math> as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}</math> |
− | Substitute <math>\textcircled{2}</math> to <math>\textcircled{3}</math>: <math>202x \cdot (10^v)=22x</math> | + | Substitute <math>\textcircled{{2}}</math> to <math>\textcircled{3}</math>: <math>202x \cdot (10^v)=22x</math> |
− | Thus, <math>v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})</math>, where <math>m=11</math> and <math>n=101</math>. | + | Thus, <math>v=\log_{10} \left(\frac{22x}{202x}\right)= \log_{10} \left(\frac{11}{101}\right)</math>, where <math>m=11</math> and <math>n=101</math>. |
Therefore, <math>m+n = \boxed{112}</math>. | Therefore, <math>m+n = \boxed{112}</math>. | ||
− | |||
− | |||
==Solution 3== | ==Solution 3== | ||
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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>\ | + | 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> | ||
~MathIsFun286 | ~MathIsFun286 | ||
+ | |||
+ | ==Solution 6== | ||
+ | By change of base formula, | ||
+ | <cmath> | ||
+ | \frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1 | ||
+ | </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== | ||
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==Video Solution by Power of Logic== | ==Video Solution by Power of Logic== | ||
− | https://youtu.be/ | + | https://youtu.be/m2Cm9r5_Jvs |
~Hayabusa1 | ~Hayabusa1 |
Latest revision as of 17:14, 30 January 2024
Contents
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, .
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.