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

how do you solve for x?

3 Answers

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  • 1 decade ago
    Favorite 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

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