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

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Find <math>(\log_2 x)^2</math> if <math>\log_2 (\log_8 x) = \log_8 (\log_2 x)</math>.
 
Find <math>(\log_2 x)^2</math> if <math>\log_2 (\log_8 x) = \log_8 (\log_2 x)</math>.
  
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== Solution ==
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== Solution 1==
 
Raise both as [[exponent]]s with base 8:
 
Raise both as [[exponent]]s with base 8:
  
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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>.
 
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>.
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== Solution 2: Substitution ==
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We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2.
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<cmath>
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\begin{align*}
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{\log_2 (\frac{1}{3}\log_2 x)} &= \frac{1}{3}{\log_2 (\log_2 x)}\\
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{\log_2 x = y}
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{\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\
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{3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\
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\end{align*}
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</cmath>
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Solving, we get <math>y^2 = 27</math>, which is what we want.
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<math>= \boxed{27}</math>
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----
  
 
== See also ==
 
== See also ==

Revision as of 14:13, 14 July 2018

Problem

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

Solution 1

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 $k$.

Solution 2: Substitution

We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2.

\begin{align*} {\log_2 (\frac{1}{3}\log_2 x)} &= \frac{1}{3}{\log_2 (\log_2 x)}\\ {\log_2 x = y} {\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ {3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\ \end{align*} Solving, we get $y^2 = 27$, which is what we want. $= \boxed{27}$


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

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