# Hyperbolic Tangent

• October 29th 2009, 07:28 PM
iheartthemusic29
Hyperbolic Tangent
Simplify the expression tanh(lnt).
I thought the answer was (t^2-1)/(t^2+1), but that's not correct, and I can't figure out why. Can somebody give me a hand? Thanks!
• October 29th 2009, 08:31 PM
tonio
Quote:

Originally Posted by iheartthemusic29
Simplify the expression tanh(lnt).
I thought the answer was (t^2-1)/(t^2+1), but that's not correct, and I can't figure out why. Can somebody give me a hand? Thanks!

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• October 29th 2009, 08:33 PM
tonio
Quote:

Originally Posted by iheartthemusic29
Simplify the expression tanh(lnt).
I thought the answer was (t^2-1)/(t^2+1), but that's not correct, and I can't figure out why. Can somebody give me a hand? Thanks!

Well, it is correct: $\tanh x = \frac{e^x-e^{-x}}{e^x+e^{-x}}$ , and aplying on $\ln x$ we get what you got.

Who says that's not the correct answer?

Tonio