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 18:32, 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