Why is arcsin converted into logs in this equation?

[cathy]

Homework Statement

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.

 

[tiny-tim]

hi caty!

(hey, what's an h ? )

2sinh[ln(e^x + √(e^2x + 1))]

= exp[ln(e^x + √(e^2x + 1))] - exp[-ln(e^x - √(e^2x + 1))]

= e^x + √(e^2x + 1)) - 1/[e^x + √(e^2x + 1))]

= [e^2x + e^2x + 1 + 2e^x√(e^2x + 1)) - 1]/[e^x + √(e^2x + 1))]

= 2e^x

alternatively, put e^x = sinhy

then sinh[ln(e^x + √(e^2x + 1))]

= sinh[ln(sinhy + coshy)]

= sinh[ln(e^y)]

= sinh[y] = e^x​

 

[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.

 

[cathy]

Ahh got it! Thank you very much!