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sorry, how to solve it???

Posted 10 years ago

cosh^-1 (3x)= ln [3x + (?9x2_1)]

POSTED BY: salina azhar
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And to confirm Daniel's result you can compair the plots of the two functions.

For example, in WolframAlpha you can ask:

plot ArcCosh[3 x] and Log[3 x + Sqrt[9 x^2 - 1]]

And you will see where the two plots coincide.

POSTED BY: David Reiss

Taking some guesses as to the actual input, maybe something like this?

Reduce[ArcCosh[3 x] == Log[3 x + Sqrt[9 x^2 - 1]], x, Reals]

(* Out[517]= x == 1/3 || x > 1/3 *)

So for real values this is apparently an identity for x>=1/3.

POSTED BY: Daniel Lichtblau

It is unclear from your posting what the actual equation is since it is not formatted properly. In particular it is not clear what ?9x2_1 means. Try using the formatting tools at the top of the posting area. Also could you give us an example of what you have tried to do to solve the equation in Mathematica or Wolfram Alpha?

POSTED BY: David Reiss
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