AIME 1988 · 第 3 题

Solution 1

Raise both as exponents with base 8:

$$\begin{aligned} 8^{\log_2 (\log_8 x)} &= 8^{\log_8 (\log_2 x)}\\ 2^{3 \log_2(\log_8x)} &= \log_2x\\ (\log_8x)^3 &= \log_2x\\ \left(\frac{\log_2x}{\log_28}\right)^3 &= \log_2x\\ (\log_2x)^2 &= (\log_28)^3 = \boxed{027}\\ \end{aligned}$$

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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.

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

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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

$$\begin{aligned} \log_2(\log_8x)&=\log_8(\log_2x)\\ \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 \end{aligned}$$ Changing the base in the numerator yields

$$\begin{aligned} \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ \end{aligned}$$ Using the property $\frac{\log_ab}{\log_ac}=\log_cb$ yields

$$\begin{aligned} \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ (\log_2x)^\frac{1}{3}&=\log_8x\\ \sqrt[3]{\log_2x}&=\frac{\log_2x}{3} \end{aligned}$$ 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

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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.