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Math Help - Hyperbolic Functions (not difficult)

  1. #1
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    Hyperbolic Functions (not difficult)

     cosh(x) = \frac{5}{3} , solve for x.

    This is how I go about it (not sure if it's the right start).

     \frac{1}{2}(e^x + e^{-x}) = \frac{5}{3}

     e^x + e^{-x} = \frac{10}{3}

    Is this next step right?

     x + \frac{1}{x} = log_e10 - log_e3

    If so, how can I solve for x from here? Cheers,
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  2. #2
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    Quote Originally Posted by U-God View Post
     cosh(x) = \frac{5}{3} , solve for x.

    This is how I go about it (not sure if it's the right start).

     \frac{1}{2}(e^x + e^{-x}) = \frac{5}{3}

     e^x + e^{-x} = \frac{10}{3}

    Is this next step right? Mr F says: NO!

     x + \frac{1}{x} = log_e10 - log_e3

    If so, how can I solve for x from here? Cheers,
     e^x + e^{-x} = \frac{10}{3}

    Let u = e^x:

    u + \frac{1}{u} = \frac{10}{3}

    Multiply both sides by u:

    \Rightarrow u^2 + 1 = \frac{10}{3} u.

    Solve this quadratic equation for u and hence for e^x \Rightarrow x.
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