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

m (Solution)
Line 5: Line 5:
 
Raise both as [[exponent]]s with base 8:
 
Raise both as [[exponent]]s with base 8:
  
<div style="text-align:center;">
+
<cmath>
<math>
+
\begin{align*}
\begin{eqnarray*}
+
8^{\log_2 (\log_8 x)} &= 8^{\log_8 (\log_2 x)}\\
8^{\log_2 (\log_8 x)} &=& 8^{\log_8 (\log_2 x)}\\
+
2^{3 \log_2(\log_8x)} &= \log_2x\\
2^{3 \log_2(\log_8x)}} &=& \log_2x\\
+
(\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 = \boxed{27}\\
(\log_2x)^2 &=& (\log_28)^3 = \boxed{27}\\
+
\end{align*}
\end{eqnarray*}
+
</cmath>
</math></div>
 
  
 
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Revision as of 18:32, 10 March 2015

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{align*} 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{align*}


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

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