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I understand the inverse of e^{x} is the natural logarithm. However I don't understand how the following expression is true:

$e^{-\ln x} = e^{\ln(1/x)}$

Any assistance is appreciated.

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    $\begingroup$ Hint: $a\log(x) = \log(x^a)$ $\endgroup$
    – CBBAM
    Jun 3, 2021 at 3:02
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    $\begingroup$ $\ln(1/x)=-\ln x$ $\endgroup$ Jun 3, 2021 at 3:12
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    $\begingroup$ $e^a = e^b \iff a = b.$ By definition, $\ln(1/x) = -\ln(x)$, because (in general) $r^{(-s)} = \frac{1}{r^s},$ by definition. $\endgroup$ Jun 3, 2021 at 3:20

3 Answers 3

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One of the properties of logarithms is the following:

$$\log({x^k}) = k\log{x}$$

Therefore when you have $-\ln x$, you essentially go backwards:

$$-\ln x = -1 \times \ln x = \ln(x^{-1}) = \ln \left( \frac{1}{x} \right) $$

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$$ e^{-\ln(x)} = e^{\ln(x^{-1})} = e^{\ln(1/x)} $$

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Because of the fact that $\ln(x)$ and $e^x$ are inverses:

$$\frac{1}{e^{\ln(x)}}=\frac{1}{x}=e^{\ln\left(\frac{1}{x}\right)}$$

Altering the first expression with the identity that $\frac{1}{e^{x}}=e^{-x}$ yields:

$$e^{-\ln x}=\frac{1}{x}=e^{\ln\left(\frac{1}{x}\right)}$$

Which is the expression that you are looking for.

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