Difference between revisions of "1988 AIME Problems/Problem 3"

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== Problem ==
 
== Problem ==
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Find <math>(\log_2 x)^2</math> if <math>\log_2 (\log_8 x) = \log_8 (\log_2 x)</math>.
  
 
== Solution ==
 
== Solution ==
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Raise both as [[exponent]]s with base 8:
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 +
<div style="text-align:center;">
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<math>
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\begin{eqnarray*}
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8^{\log_2 (\log_8 x)} &=& 8^{\log_8 (\log_2 x)}\\
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2^{3 \log_2(\log_8x)}} &=& \log_2x\\
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(\log_8x)^3 &=& \log_2x\\
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\left(\frac{\log_2x}{\log_28}\right)^3 &=& \log_2x\\
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(\log_2x)^2 &=& (\log_28)^3 = 27\\
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\end{eqnarray*}
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</math></div>
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----
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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 ==
* [[1988 AIME Problems]]
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{{AIME box|year=1988|num-b=2|num-a=4}}
  
{{AIME box|year=1988|num-b=2|num-a=4}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 17:23, 26 September 2007

Problem

Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$.

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 $a^{\log_a x} = x$. On the 2nd step, we use the fact that $k \log_a x = \log_a x^k$. On the 3rd step, we use the change of base formula, which states $\log_a b = \frac{\log_k b}{\log_k a}$ for arbitrary $k$.

See also

1988 AIME (ProblemsAnswer KeyResources)
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