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Arcosh in terms of logs

Right. In my textbook, it says:

arsinh(x) = u
...
=> eu = x +/- √(x2 + 1)

But we ignore the negative root since:
√(x2 + 1) > x
=> eu < 0 , which is not possible

giving arsinh x = ln[x + &#8730;(x2 + 1)]

However, this is not the case for arcosh

arcosh(x) = u
...
=> eu = x +/- &#8730;(x2 - 1)

but &#8730;(x2 - 1) is not necessarily more than x

but the identity given for arcosh is:

arcosh = ln[x + &#8730;(x2 - 1)]

so, if we exclude the -ve root, will we not lose some answers?
Reply 1
You lose the negative answer.

Taking the positive sign would give say arccosh(2) whilst taking the negative just gives -arccosh(2).

This is because

x+ rt(x^2-1)

and

x - rt(x^2-1)

are reciprocals.

If you want arccosh to be a function on x >= 1 then you need to choose one of the roots.
Reply 2
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