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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>.
  
−
== Solution ==
+
== Solution 1==
 
Raise both as [[exponent]]s with base 8:
 
Raise both as [[exponent]]s with base 8:
  
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(\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{027}\\
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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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>.
 +
 +
== 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.
 +
 +
<cmath>
 +
\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)}\\
 +
{\log_2 (\frac{1}{3}y)^3} &= {\log_2 (y)}\\
 +
\end{align*}
 +
</cmath>
 +
Solving, we get <math>y^2 = 27</math>, which is what we want.
 +
<math>\boxed{27}</math>
 +
 +
 +
----
 +
Just a quick note-
 +
In this solution, we used 2 important rules of logarithm:
 +
1) <math>\log_a b^n=n\log_a b</math>.
 +
2) <math>\log_{a^n} b=\frac{1}{n}\log_a b</math>.
 +
 +
== Solution 3 ==
 +
 +
First we have
 +
<cmath>\begin{align*}
 +
\log_2(\log_8x)&=\log_8(\log_2x)\\
 +
\frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1
 +
\end{align*}</cmath>
 +
Changing the base in the numerator yields
 +
<cmath>\begin{align*}
 +
\frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\
 +
\frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\
 +
\end{align*}</cmath>
 +
Using the property <math>\frac{\log_ab}{\log_ac}=\log_cb</math> yields
 +
<cmath>\begin{align*}
 +
\log_{\log_2x}(\log_8x)&=\frac{1}{3}\\
 +
(\log_2x)^\frac{1}{3}&=\log_8x\\
 +
\sqrt[3]{\log_2x}&=\frac{\log_2x}{3}
 +
\end{align*}</cmath>
 +
Now setting <math>y=\log_2x</math>, we have
 +
<cmath>\sqrt[3]{y}=\frac{y}{3}</cmath>
 +
Solving gets <math>y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}</math>.
 +
 +
~ Nafer
 +
 +
== Solution 4 ==
 +
Say that <math>\log_{2^3}x=a</math> and <math>\log_2x=b</math> so we have <math>\log_2a=\log_{2^3}b</math>. And we want <math>b^2</math>.
 +
 +
<math>\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.</math>
 +
 +
Because <math>3a=b</math> (as <math>2^{3a}=x</math> and <math>2^b=x</math> from our setup), we have that
 +
 +
<math>b^{1/3}=\frac{b}{3}\\ b^{-2/3}=\frac13\\ b=3^{3/2}\\ \\b^2=3^3=\boxed{27}</math>
 +
 +
~thedodecagon
 +
------
 +
Note that we use the property <math>\log_{x^n}y=\frac1n\log_xy</math> in step 1 and <math>\frac{\log_wx}{\log_wy}=\log_yx</math> in step 2 in this solution.
  
 
== See also ==
 
== See also ==

Latest revision as of 20:52, 18 November 2020

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


Just a quick note- In this solution, we used 2 important rules of logarithm: 1) $\log_a b^n=n\log_a b$. 2) $\log_{a^n} b=\frac{1}{n}\log_a b$.

Solution 3

First we have \begin{align*} \log_2(\log_8x)&=\log_8(\log_2x)\\ \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 \end{align*} Changing the base in the numerator yields \begin{align*} \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ \end{align*} Using the property $\frac{\log_ab}{\log_ac}=\log_cb$ yields \begin{align*} \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ (\log_2x)^\frac{1}{3}&=\log_8x\\ \sqrt[3]{\log_2x}&=\frac{\log_2x}{3} \end{align*} Now setting $y=\log_2x$, we have \[\sqrt[3]{y}=\frac{y}{3}\] Solving gets $y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}$.

~ Nafer

Solution 4

Say that $\log_{2^3}x=a$ and $\log_2x=b$ so we have $\log_2a=\log_{2^3}b$. And we want $b^2$.

$\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.$

Because $3a=b$ (as $2^{3a}=x$ and $2^b=x$ from our setup), we have that

$b^{1/3}=\frac{b}{3}\\ b^{-2/3}=\frac13\\ b=3^{3/2}\\ \\b^2=3^3=\boxed{27}$

~thedodecagon

Note that we use the property $\log_{x^n}y=\frac1n\log_xy$ in step 1 and $\frac{\log_wx}{\log_wy}=\log_yx$ in step 2 in this solution.

See also

1988 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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