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I was working through an Art of Problem Solving workbook when I encountered a very frustrating problem.

Solve the equation $\log_{2x}216=x$, where $x\in \mathbb{R}$.

I understand how to find the answer by inspection (all you have to do is stare at it for a few seconds), but I'm trying to figure out how to solve it algebraically. Every time I try to manipulate the problem, I either find myself running in circles or I create some convoluted expression that's even harder to deal with than the original problem. Forgive me if this is a very easy question - I'm just completely lost here. Thanks!

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  • $\begingroup$ Typeset $\log_{2x} 216 = x$ as $\log_{2x} 216 = x$. $\endgroup$ – DMcMor Apr 22 '17 at 22:06
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$$ x = \log_{2x}{216} = \frac{\log{216}}{\log{2x}} = \frac{3\log{2}+3\log{3}}{\log{2}+\log{x}} = 3\frac{1+\log_2{3}}{1+\log_2{x}} $$

But then multiplying up gives $$ x(1+\log_2{x}) = 3(1+\log_2{3}). $$ The left-hand side is an increasing function of $x$, so it only has one solution, which is obviously $x=3$.

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We may rewrite the equation as,

$$216=(2x)^x$$

In the general case $x^x=c$ requires the lambert $W$ function to solve. So it's reasonable to be more hopeful and try to factor our number into its prime factorization,

$$216=2^3 3^3$$

$$216=(2(3))^3$$

Then it becomes clear that $x=3$. This can be the only solution because $\log_{2x} 216=\frac{\ln 216}{\ln 2x}$ is strictly decreasing and $x$ is strictly increasing.

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    $\begingroup$ By the definition of logarithm, $\log_{2x}216=x$ leads directly to $(2x)^x=216$. $\endgroup$ – Tom Zych Apr 23 '17 at 9:19
  • $\begingroup$ I've edited that, thanks. $\endgroup$ – Ahmed S. Attaalla Apr 23 '17 at 15:57
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In terms of using inspection, I think this makes the inspection easier. $$\ln_{2x}216=x \Rightarrow \ln 6^3 = x \ln 2x \Rightarrow 3 \ln 6 = x \ln2x $$

If you guess that the linear factors are equal ($x=3$) you immediately see the logarithmic factors are equal.

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