# Thread: Hyperbolic Functions (not difficult)

1. ## 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,

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