Hyperbolic Tangent

**iheartthemusic29** · October 29th 2009, 06:28 PM

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!

**tonio** · October 29th 2009, 07:31 PM

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!

.

**tonio** · October 29th 2009, 07:33 PM

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

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