# Math Help - Hyperbolic Functions (not difficult)

1. ## 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,

2. Originally Posted by U-God
$cosh(x) = \frac{5}{3}$, 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 $e^x \Rightarrow x$.