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Thread: Hyperbolic Functions (not difficult)

  1. #1
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    Hyperbolic Functions (not difficult)

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

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

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

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

    Is this next step right?

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

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

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

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

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

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

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

    Let $\displaystyle u = e^x$:

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

    Multiply both sides by u:

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

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