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# (e^x+e^-x)/(e^x-e^-x)=2?

how do you solve for x?

### 3 Answers

- 1 decade agoFavorite Answer
e^x+e^-x/e^x-e^-x=2

or,(e^2x+1)/(e^2x-1)=2 (multiply up and down by e^x)

or,(e^2x-1)/(e^2x-1)+2/(e^2x-1)=2

or,1+{2/e^2x-1}=2

or,2/(e^2x-1)=1

or,e^2x-1=2

or,e^2x=3

or,log e^2x=log 3

or,2xlog e=log 3

or,2x=log 3 (;log e=1)

or,x=(log 3)/2

Solved.Thank you. Have a nice day.....

- 1 decade ago
One way

we know that (e^x+e^-x)/(e^x-e^-x)=coth(x) [hyperbolic cot function]

So, the equation reduces to

coth(x) = 2

=> x = (inverse coth)(2)

=> x = 0.1984 - 0.5673i

Other way

multiply and devide by e^x,you get

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

=>(e^2x+1)=2*(e^2x-1)

=>3=2*e^2x - e^2x

=>e^2x = 3

tanking log on both the sides

=>2x = log(3)

=>x = log(3) / 2

=>x = 0.5493

I hope that it is clear now.

All the best.

- 1 decade ago
write e^-x as 1/e^x and cross multiply

then u will get (e^x)^2 + 1 = 2{(e^x)^2 - 1}

then (e^x)^2 = 3

e^x = sqrt(3)

then x = ln (sqrt(3))

this is one method.

another one is go for hyperbolic functions

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

so coth(x) = 2

x = inverse of coth(2)

thats it