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

(Solution)
Line 24: Line 24:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 18:10, 4 July 2013

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 = \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 $a^{\log_a x} = x$. 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png