# Turning arcsin into logs

1. Feb 24, 2014

### cathy

1. The problem statement, all variables and given/known data

Hello. Will someone please explain to me why this is true?
arcsinh(e^x) = ln(e^x + √(e^(2x) + 1))

2. The attempt at a solution

I cannot figure out why arcsin is able to be put into terms of ln. Thank you in advance.

2. Feb 24, 2014

### tiny-tim

hi caty!

(hey, what's an h ? )

2sinh[ln(ex + √(e2x + 1))]

= exp[ln(ex + √(e2x + 1))] - exp[-ln(ex - √(e2x + 1))]

= ex + √(e2x + 1)) - 1/[ex + √(e2x + 1))]

= [e2x + e2x + 1 + 2ex√(e2x + 1)) - 1]/[ex + √(e2x + 1))]

= 2ex

alternatively, put ex = sinhy

then sinh[ln(ex + √(e2x + 1))]

= sinh[ln(sinhy + coshy)]

= sinh[ln(ey)]

= sinh[y] = ex

3. Feb 24, 2014

### LCKurtz

In addition to Tiny's explanation you could also note $y =\sinh^{-1}(e^x)$ is the same as $e^x =\sinh(y)= \frac{e^y - e^{-y}} 2$ or $2e^x = e^y - \frac 1 {e^y}$. Solve that for $y$ in terms of $x$ and you will get your expression and you will see where the logarithms come from.

Last edited: Feb 24, 2014
4. Feb 25, 2014

### cathy

Ahh got it! Thank you very much!