log functions?

Question

how do you solve the equation for x...?

(e^x - e^-x) / (e^-x + e^x) = -t

Answer 1

multiply both sides by the denominator on the left:
e^x-e^-x=t(e^-x+e^x)
e^x-e^-x=te^-x+te^x
e^x-te^x=te^-x+e^-x
e^x(1-t)=e^-x(t+1)
e^x(1-t)=(t+1)/e^x
e^2x=(t+1)/(1-t)
2x=ln ((t+1)/(1-t))
x=ln ((t+1)/(1-t))/2

Hope that helps!

Answer 2

If you study the function x |----> F(x) = (e^{x}-e^{-x})/(e^{x}+e^{-x}), you can easily see that - 1 < F(x) < 1 for every x in R. Therefore, for t out of the interval (-1,1), there is no solution to your equation.
You can also see that F is a bijection of R onto (-1, 1), therefore for every t in (-1,1) there is a unique x in R such that F(x) = -t. Now let's solve your equation for t in (-1, 1).

F(x)=-t <==> e^{x}-e^{-x}+t(e^{x}+e^{-x})=0

<==> (1+t)e^{x}-(1-t)e^{-x}=0

<==> (1+t)e^{2x}-(1-t) =0

<==> e^{2x} = (1-t)/(1+t)

<==> 2x=ln{(1-t)/(1+t)} since (1-t)/(1+t) > 0 for t in (-1; 1)

<==> x =1/2ln{(1-t)(1=t)}.

ps: I need my points!!!

Answer 3

Substitute a=e^x, then
(a-1/a) / (1/a + a) = -t
LHS = (a^2-1)/a divided by (1+a^2)/a = (a^2-1)/(a^2+1) = -t
a^2-1 = -ta^2 - t
a^2 = (1-t)/(1+t)
e^x = a = sqrt((1-t)/(1+t))
x = ln[sqrt((1-t)/(1+t))] = ln[(1-t)/(1+t)] / 2. = ln(1-t) / 2 - ln (1+t) / 2... pick any form you like...

D (below me) solved it for RHS = t instead of (-t), I believe that's why our answers are different...

Answer 4

The first thing you do is multiply the top and the bottom by e^x. Note that e^x * e^(-x) yields e^(x - x) = e^0 = 1.
Also note that e^x * e^x = e^(2x). So now we have

(e^(2x) - 1) / (1 + e^(2x)) = -t

Multiply both sides by the denominator to get

e^(2x - 1) = -t (1 + e^(2x))
e^(2x - 1) = -t + te^(2x)

e^(2x) * e^(-1) = -t + te^(2x)
e^(2x) * e^(-1) - te^(2x) = -t
e^(2x) [e^(-1) - t] = -t

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

Therefore,

ln [-t / [e^(-1) - t] = 2x

and

x = (ln [-t / [e^(-1) - t])/2