0
6.9kviews
Solve the equation 7coshx+8sinhx = 1 for real values of x.
1
2.1kviews

7coshx+8sinhx = 1

$7\bigg(\dfrac{e^x+e^{-x}}{2}\bigg)+8\bigg(\dfrac{e^x-e^{-x}}{2}\bigg)=1$

$7e^x+7e^{-x}+8e^{x}-8e^{-x}=2$

$15e^x-e^{-x}=2$

multiplying the above equation by $e^{x}$ we get

$15e^{2x}-1=2e^{x} $

$15e^{2x}-2e^{x}=1$

$e^x=\dfrac{1}{3},-\dfrac{1}{5}$

$x=\log(\dfrac{1}{3})=-\log(3)$

Please log in to add an answer.