2022 AIME II Problems/Problem 4: Difference between revisions
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==Solution== | ==Solution== | ||
We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>. | |||
So that <math>(20x)^v=22x </math> (Eq1) | |||
and <math>(2x)^v=202x </math> (Eq2) | |||
Express Eq1 as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x </math> (Eq3) | |||
Substitute Eq2 to Eq3: <math>202x \cdot (10^v)=22x</math> | |||
Thus, <math>v=\log_{10} (\tfrac{22x}{202x})= \log_{10} (\tfrac{11}{101})</math>, where <math>m=11</math> and <math>n=101</math>. | |||
Therefore, <math>m+n = \boxed{112}</math>. | |||
~DSAERF-CALMIT | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=3|num-a=5}} | {{AIME box|year=2022|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:33, 17 February 2022
Problem
There is a positive real number 
not equal to either 
or 
such that![\[\log_{20x} (22x)=\log_{2x} (202x).\]](http://latex.artofproblemsolving.com/8/b/2/8b2d644c3c35696b97f664a77cacdd39b6995abf.png)
The value 
can be written as 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
We could assume a variable 
which equals to both and 
.
So that 
(Eq1)
and 
(Eq2)
Express Eq1 as: 
(Eq3)
Substitute Eq2 to Eq3: 
Thus, 
, where 
and 
.
Therefore, 
.
~DSAERF-CALMIT
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
