2022 AIME II Problems/Problem 4: Difference between revisions
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==Problem== | |||
There is a positive real number <math>x</math> not equal to either <math>\tfrac{1}{20}</math> or <math>\tfrac{1}{2}</math> such that<cmath>\log_{20x} (22x)=\log_{2x} (202x).</cmath>The value <math>\log_{20x} (22x)</math> can be written as <math>\log_{10} (\tfrac{m}{n})</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | |||
==Solution== | |||
==See Also== | |||
{{AIME box|year=2022|n=II|num-b=3|num-a=5}} | |||
{{MAA Notice}} | |||
Revision as of 22:13, 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
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. 
