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

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.

@@ Line 11: / Line 11: @@
 (\log_8x)^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*}
 </cmath>
@@ Line 64: / Line 64: @@
 ~ 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 ==

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