Showing that $\ln \frac{1+e^x}{1+e^{-x}} = x$

Question

Is the below a valid approach?

$$\frac{1+e^x}{1+e^{-x}} = \frac{1+e^x}{1+e^{-x}} \times \frac{e^{-x}}{e^{-x}}$$

We know that $\frac{e^{-x}}{1+e^{-x}} = \frac{1}{e^x+1}$, so

$$\frac{1+e^x}{1+e^{-x}} \times \frac{e^{-x}}{e^{-x}} = \frac{1+e^x}{1+e^x} \times \frac{1}{e^{-x}}.$$

So, $\ln \frac{1+e^x}{1+e^{-x}} = \ln \frac{1}{e^{-x}} = - \ln e^{-x} = x$.

Answer 1

That looks correct. It becomes a tiny bit simpler if you expand the fraction with $e^x$ instead of expanding with $e^{-x}$: $$ \frac{1+e^x}{1+e^{-x}} = \frac{1+e^x}{1+e^{-x}} \frac{e^x}{e^x} = \frac{1+e^x} {e^x+1} e^x = e^x $$

Answer 2

This looks correct to me but remember that "$\log$" is typically an abbreviation for "$\log_{10}$" in certain contexts such as physics and engineering. It's pretty clear that you are referring to the natural logarithm here, but if you would like there to be zero ambiguity as what you are referring to, then I suggest using the more widespread notation for the natural log given by "$\ln$."

Answer 3

Yes it is correct, indeed

$$\log \frac{1+e^x}{1+e^{-x}} = \log (1+e^x)-\log(1+e^{-x})=\\=\log (1+e^x)-\log(1+e^{-x})-\log e^x+\log e^x=\log (1+e^x)-\log(e^x+1)+x=x$$