CMIC Contest Question

Algebra Level 3

$\large \log_{2}x +\log_{8}x+\log_{64}x = \log_{x}2+\log_{x}16+\log_{x}128$

Let $x$ be a real number satisfying the equation above.

If the value of $\log_{2}x + \log_{x}2$ can be expressed as $\dfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are coprime positive integers and $b$ is square-free, compute $abc.$

The answer is 72.

1 solution

Sabhrant Sachan
Jun 2, 2016

Relevant wiki: Logarithms

$\log_{2}{x}+\log_{8}{x}+\log_{64}{x}=\log_{x}{2}+\log_{x}{16}+\log_{x}{128} \\ \log_{2}{x}+\dfrac{1}{3}\log_{2}{x}+\dfrac{1}{6}\log_{2}{x}=\log_{x}{2}+4\log_{x}{2}+7\log_{x}{2} \\ \dfrac{3}{2}\log_{2}{x}=12\log_{x}{2} \\ \log_{2}{x}=2\sqrt{2} \quad \quad \text{ (Why not } -2\sqrt{2} \text{ ? )}\\ \log_{2}{x}+\log_{x}{2}=2\sqrt{2}+\dfrac1{2\sqrt{2}} \implies \boxed{\dfrac{9\sqrt{2}}{4}} \\ \text{In comparison } a=9,b=2,c=4 \\ abc =\boxed{72}$

Typo: $\dfrac{1}{3} \log_2 x$

Hung Woei Neoh - 5 years ago

thanks :) , edited

Sabhrant Sachan - 5 years ago

When you went from $\frac{3}{2} \log_2 x = 12 \log_x 2$ to $\log_2 x = 2 \sqrt{2}$ , you omitted the other possible solution $\log_2 x = -2 \sqrt{2}$ .

Jon Haussmann - 5 years ago

The question stated that $a$ and $c$ are coprime positive integers

Hung Woei Neoh - 5 years ago

Right, and this means the problem is incomplete. The problem should acknowledge the existence of two different possible values of $x$ .

Jon Haussmann - 5 years ago

CMIC Contest Question

The answer is 72.

1 solution

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