1. ## Hyperbolics

Q: $10 \mathrm{cosh} x + 2 \mathrm{sinh} x = 11$

2. Originally Posted by Air
Q: $10 \mathrm{cosh} x + 2 \mathrm{sinh} x = 11$

$5 (e^x + e^{-x}) + (e^x - e^{-x}) = 11$

$\Rightarrow 6 e^x + 4 e^{-x} = 11$

$\Rightarrow 6 (e^x)^2 + 4 = 11 e^x$

$\Rightarrow 6 (e^x)^2 - 11 e^x + 4 = 0$

Substitute $u = e^x$:

$6 u^2 - 11u + 4 = 0$.

Solve for u and equate the solutions to $e^x$. Solve for x.