# Find the roots of e^(xcoshx)=3?

May 13, 2018

x=α in [-1;1] where e^(αcoshα)=0

#### Explanation:

${e}^{x \cosh x} = 3$
$x \cosh x = \ln 3$
$x \left({e}^{x} + {e}^{-} x\right) = \ln 9$
Let: $X = {e}^{x}$
$\ln X \left(X + \frac{1}{X}\right) = \ln 9$
$\frac{X \ln X + 1}{X} = \ln 9$
$X \ln X + X \ln \left({e}^{\frac{1}{X}}\right) - X \ln 9 = 0$
$X \ln \left(\frac{X {e}^{\frac{1}{X}}}{9}\right) = 0$
So:
$X = 0$
or: $\ln \left(\frac{X {e}^{\frac{1}{X}}}{9}\right) = 0$
$\cancel{{e}^{x} = 0}$, ${e}^{x}$ is never equal to 0.
${e}^{x} \cdot {e}^{{e}^{-} x} / 9 = 1$
${e}^{x + {e}^{x}} = \ln 9$
$x + {e}^{x} = \ln \left(\ln \left(9\right)\right)$
We cannot find the exact root, but because that $x + {e}^{x} - \ln \left(\ln \left(9\right)\right) = 0$ is a continuous and monotonic function and between [-1;1] this function is negative and then positive, using the intermediate values theorem, we know that there's an unique α in [-1;1] where e^(αcoshα)=0