Difference between revisions of "1988 AIME Problems/Problem 3"
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{\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ | {\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ | ||
{3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\ | {3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\ | ||
+ | {\log_2 (\frac{1}{3}y)^3} &= {\log_2 (y)}\\ | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 35: | Line 36: | ||
---- | ---- | ||
+ | Just a quick note- | ||
+ | In this solution, we used 2 important rules of logarithm: | ||
+ | a) <math>\log_a b^n=n\log_a b</math> | ||
+ | b) <math>\log_{a^n} b=\frac{1}{n}\log_a b</math> | ||
== See also == | == See also == |
Revision as of 14:16, 14 July 2018
Problem
Find if .
Solution 1
Raise both as exponents with base 8:
A quick explanation of the steps: On the 1st step, we use the property of logarithms that . On the 2nd step, we use the fact that . On the 3rd step, we use the change of base formula, which states for arbitrary .
Solution 2: Substitution
We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2.
Solving, we get , which is what we want.
Just a quick note- In this solution, we used 2 important rules of logarithm: a) b)
See also
1988 AIME (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.