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

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.

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

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:

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

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

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.