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How to I solve the following exercise with a logarithm? I've forgotten the "trick" for doing so:

$x^{log_{10} x} =10^4$

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    $\begingroup$ Try applying log$_{10}$ on each side and use $\log(x^\alpha)=\alpha\log(x)$ $\endgroup$ – JJ Fleck May 25 '13 at 18:18
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To start, take $\log_{10}$ of both sides to get:

$$\log_{10}(x)\log_{10}(x) = \log_{10}(10)^{4} = 4$$

Then just solve $$(\log_{10}(x))^{2} = 4$$

$$\implies \log_{10}x = \pm 2$$

Exponentiating both sides, we get:

$$x = 10^{2} = 100$$ OR

$$x = 10^{-2} = \frac{1}{100}$$

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The trick: $a^b=e^{b\ln a}$. Applied to this problem: $$x^{\log_{10}x}=e^{\ln x \log_{10}x}=e^{\ln x \frac{\ln x}{\ln 10}}=e^{(\ln x)^2/(\ln 10)}$$

On the other hand $10^4=e^{4\ln 10}$, so we may take the log of both sides to get $$\frac{(\ln x)^2}{\ln 10}=4\ln 10$$

Cross-multiply, take the square root, exponentiate, and you're done.

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$x^{\log_{10} x} = 10^{\log_{10}x\log_{10}x }=10^4$

So $\log_{10}x=\pm 2$ and it follows that both $x=100$ or $x=0.01$ solve the equation.

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