Difference between revisions of "2020 AIME II Problems/Problem 3"
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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 | ||
+ | ==Easiest Solution== | ||
+ | Recall the identity <math>\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b </math> (which is easily proven using exponents) | ||
+ | Then this problem turns into <cmath>\frac{30}{x}\log_{2} 3 = \frac{3030}{x+3}\log_{2} 3</cmath> | ||
+ | Divide <math>\log_{2} 3</math> from both sides. And we are left with <math>\frac{30}{x}=\frac{3030}{x+3}</math>.Solving this simple equation we get <cmath>x = \tfrac{3}{100} \Rightarrow \boxed{103}</cmath> | ||
+ | ~mlgjeffdoge21 | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/lPr4fYEoXi0 ~ CNCM | https://youtu.be/lPr4fYEoXi0 ~ CNCM |
Revision as of 21:07, 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
Easiest Solution
Recall the identity (which is easily proven using exponents)
Then this problem turns into
Divide
from both sides. And we are left with
.Solving this simple equation we get
~mlgjeffdoge21
Video Solution
https://youtu.be/lPr4fYEoXi0 ~ CNCM
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.