Difference between revisions of "1988 AIME Problems/Problem 3"
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(\log_8x)^3 &=& \log_2x\\ | (\log_8x)^3 &=& \log_2x\\ | ||
\left(\frac{\log_2x}{\log_28}\right)^3 &=& \log_2x\\ | \left(\frac{\log_2x}{\log_28}\right)^3 &=& \log_2x\\ | ||
| − | (\log_2x)^2 &=& (\log_28)^3 = 27\\ | + | (\log_2x)^2 &=& (\log_28)^3 = \boxed{27}\\ |
\end{eqnarray*} | \end{eqnarray*} | ||
</math></div> | </math></div> | ||
Revision as of 20:59, 29 March 2011
Problem
Find 
if 
.
Solution
Raise both as exponents with base 8:
$\begin{eqnarray*} 8^{\log_2 (\log_8 x)} &=& 8^{\log_8 (\log_2 x)}\\ 2^{3 \log_2(\log_8x)}} &=& \log_2x\\ (\log_8x)^3 &=& \log_2x\\ \left(\frac{\log_2x}{\log_28}\right)^3 &=& \log_2x\\ (\log_2x)^2 &=& (\log_28)^3 = \boxed{27}\\
\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)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 | ||
