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    I need to work out the value of x when sinhx = 1

    I got to this point:

    e^(x) - e^(-x) = 2

    Now I know I am meant to multiply everything by e^(x) but everything seems to go wrong from there so can someone finish off my working out please. cheers
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    (Original post by badatmaths321)
    I need to work out the value of x when sinhx = 1

    I got to this point:

    e^(x) - e^(-x) = 2

    Now I know I am meant to multiply everything by e^(x) but everything seems to go wrong from there so can someone finish off my working out please. cheers
    Yes, multiply through by e^x

    You get e^{2x}-1=2e^x \Rightarrow (e^x)^2-2e^x-1=0 which is just a quadratic in e^x so you can solve for x once you solve for the exponential.
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    (Original post by badatmaths321)
    I need to work out the value of x when sinhx = 1

    I got to this point:

    e^(x) - e^(-x) = 2

    Now I know I am meant to multiply everything by e^(x) but everything seems to go wrong from there so can someone finish off my working out please. cheers
    If  \sinh x=a then  x=\arsinh a=\ln (a+\sqrt{a^2+1}) , this definition for the inverse of sinh is well worth remembering.
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    (Original post by B_9710)
    If  \sinh x=1 then  x=\arsinh x=\ln (x+\sqrt{x^2+1}) , this definition for the inverse of sinh is well worth remembering.
    I think that's in the formula booklet as far as the exam is concerned; though two very simple definitions which can be useful to remember for the exam are:

    \cosh[\sinh^{-1}(x)]=\sqrt{x^2+1}

    \sinh[\cosh^{-1}(x)]=\sqrt{x^2-1}

    (Original post by badatmaths321)
    ...
 
 
 
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