I need help solving this equation
5 coshx + 4 sinh x =7
Any help would be greatly appreciated. Cheers!
Using the definition of cosh and sinh will help simpilfy the problem
$\displaystyle 5\left( \frac{e^{x}+e^{-x}}{2}\right)+4\left( \frac{e^{x}-e^{-x}}{2}\right)=7 \implies$
$\displaystyle 9e^{x}+e^{-x}=14$
Now this is quadratic in form if we multiply by $\displaystyle e^{x}$ we get
$\displaystyle 9(e^{x})^2+1=14e^{x} \iff 9(e^{x})^2-14e^{x}+1=0$
To simplify things let $\displaystyle u=e^{x}$
$\displaystyle 9u^2-14u+1=0$
You should be able to finish from here. Good luck