Difference between revisions of "2020 AIME II Problems/Problem 3"
Topnotchmath (talk | contribs) |
|||
Line 6: | Line 6: | ||
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>. | 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 | ~rayfish | ||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=lPr4fYEoXi0 ~ CNCM | ||
==See Also== | ==See Also== |
Revision as of 18:04, 7 June 2020
Contents
[hide]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
Video Solution
https://www.youtube.com/watch?v=lPr4fYEoXi0 ~ CNCM