- #1
ognik
- 643
- 2
This is one of those 'huh?' moments where I can follow what is said, but don't understand it at all.
From $ coth(y) = x = \frac{{e}^{2y+1}}{{e}^{2y-1}}, $ we extract: $
y=\frac{1}{2} ln \frac{x+1}{x-1} $.
I've looked at graphs and definitions online, I follow what is done (kind of swapping x and y) - but would like to understand the details instead of just parroting it.
So I tried: $\ln\left({x}\right) = \ln\left({{e}^{2y}+1}\right) - \ln\left({{e}^{2y}-1}\right) $ - and am stuck here...
From $ coth(y) = x = \frac{{e}^{2y+1}}{{e}^{2y-1}}, $ we extract: $
y=\frac{1}{2} ln \frac{x+1}{x-1} $.
I've looked at graphs and definitions online, I follow what is done (kind of swapping x and y) - but would like to understand the details instead of just parroting it.
So I tried: $\ln\left({x}\right) = \ln\left({{e}^{2y}+1}\right) - \ln\left({{e}^{2y}-1}\right) $ - and am stuck here...