Difference between revisions of "1988 AIME Problems/Problem 3"
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\log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ | \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ | ||
(\log_2x)^\frac{1}{3}&=\log_8x\\ | (\log_2x)^\frac{1}{3}&=\log_8x\\ | ||
− | \sqrt[3]{\log_2x}&=\frac{\ | + | \sqrt[3]{\log_2x}&=\frac{\log_2x}{3} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Now setting <math>y=\log_2x</math>, we have | ||
+ | <cmath>\sqrt[3]y=\frac{y}{3}</cmath> | ||
+ | Solving gets $y=\log_2x=3 | ||
== See also == | == See also == |
Revision as of 18:01, 26 August 2019
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: 1) . 2) .
Solution 3
First we have Changing the base in the numerator yields Using the property yields Now setting , we have Solving gets $y=\log_2x=3
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 |
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