Difference between revisions of "1988 AIME Problems/Problem 3"
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== Problem == | == Problem == | ||
+ | Find <math>(\log_2 x)^2</math> if <math>\log_2 (\log_8 x) = \log_8 (\log_2 x)</math>. | ||
== Solution == | == Solution == | ||
+ | Raise both as [[exponent]]s with base 8: | ||
+ | |||
+ | <div style="text-align:center;"> | ||
+ | <math> | ||
+ | \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 = 27\\ | ||
+ | \end{eqnarray*} | ||
+ | </math></div> | ||
+ | |||
+ | ---- | ||
+ | |||
+ | A quick explanation of the steps: On the 1st step, we use the property of [[logarithm]]s that <math>a^{\log_a x} = x</math>. On the 2nd step, we use the fact that <math>k \log_a x = \log_a x^k</math>. On the 3rd step, we use the [[change of base formula]], which states <math>\log_a b = \frac{\log_k b}{\log_k a}</math> for arbitrary <math>k</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1988|num-b=2|num-a=4}} | |
− | + | [[Category:Intermediate Algebra Problems]] |
Revision as of 17:23, 26 September 2007
Problem
Find if .
Solution
Raise both as exponents with base 8:
$\begin{eqnarray*} 8^{\log_2 (\log_8 x)} &=& 8^{\log_8 (\log_2 x)}\\ 2^{3 \log_2(\log_8x)}} &=& \log_2x\\ (\log_8x)^3 &=& \log_2x\\ \left(\frac{\log_2x}{\log_28}\right)^3 &=& \log_2x\\ (\log_2x)^2 &=& (\log_28)^3 = 27\\
\end{eqnarray*}$ (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 |