Difference between revisions of "1988 AIME Problems/Problem 3"
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Raise both as [[exponent]]s with base 8: | Raise both as [[exponent]]s with base 8: | ||
| − | < | + | <cmath> |
| − | + | \begin{align*} | |
| − | \begin{ | + | 8^{\log_2 (\log_8 x)} &= 8^{\log_8 (\log_2 x)}\\ |
| − | 8^{\log_2 (\log_8 x)} &= | + | 2^{3 \log_2(\log_8x)} &= \log_2x\\ |
| − | 2^{3 \log_2(\log_8x) | + | (\log_8x)^3 &= \log_2x\\ |
| − | (\log_8x)^3 &= | + | \left(\frac{\log_2x}{\log_28}\right)^3 &= \log_2x\\ |
| − | \left(\frac{\log_2x}{\log_28}\right)^3 &= | + | (\log_2x)^2 &= (\log_28)^3 = \boxed{27}\\ |
| − | (\log_2x)^2 &= | + | \end{align*} |
| − | \end{ | + | </cmath> |
| − | </ | ||
---- | ---- | ||
Revision as of 17:32, 10 March 2015
Problem
Find 
if 
.
Solution
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 
.
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. 
