Difference between revisions of "2020 AIME II Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | + | Let <math>\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n</math>. Based on the equation, we get <math>(2^x)^n=3^{20}</math> and <math>(2^{x+3})^n=3^{2020}</math>. Expanding the second equation, we get <math>8^n\cdot2^{xn}=3^{2020}</math>. Substituting the first equation in, we get <math>8^n\cdot3^{20}=3^{2020}</math>, so <math>8^n=3^{2000}</math>. Taking the 100th root, we get <math>8^{\frac{n}{100}}=3^{20}</math>. Therefore, <math>(2^{\frac{3}{100}})^n=3^{20}</math>, so <math>n=\frac{3}{100}</math> and the answer is <math>\boxed{103}</math>. | |
| + | ~rayfish | ||
==See Also== | ==See Also== | ||
Revision as of 17:32, 7 June 2020
Problem
The value of 
that satisfies 
can be written as 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Let 
. Based on the equation, we get 
and 
. Expanding the second equation, we get 
. Substituting the first equation in, we get 
, so 
. Taking the 100th root, we get 
. Therefore, 
, so 
and the answer is 
.
~rayfish
