# Solving equation that involves nepper number

I have the following equation.

$$\frac{1}{5} = \large e^{\frac{10^{-4}}{x}}$$

I assume that x will be $0.0621$ but I can't remember how does one solves such equation. I recall that it involves using logarithm or LN... can you please help me remember?

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"nepper"=Napier? –  Hans Lundmark Jan 16 '13 at 18:26

You can do the logarithm with base $e$ on both sides, and use the property $\log_a a^x=x$, then:

$$\ln 1/5 = 10^{-4}/x$$ $$x=\frac{10^{-4}}{\ln\frac{1}{5}}$$

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You can take the logarithm on both side and get

$$\ln\frac{1}{5}=\frac{10^{-4}}{x}$$

so that

$$x = \frac{10^{-4}}{-\ln 5}=-6.21\times 10^{-5}$$

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\begin{align} \ln\left(\frac{1}{5}\right)=-\ln(5)&=\ln\left(e^{10^{-4}}/x\right)\\ &\Rightarrow -\ln(5)=10^{-4}/x\\ &\Rightarrow x=\frac{10^{-4}}{-\ln(5)} \end{align}

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