# How many real solutions do the equation sin(e^x)-5^x-5^(-x)=0 have? Thank you!

##### 1 Answer
Nov 24, 2016

No real solutions.

#### Explanation:

The equation is $\sin \left({e}^{x}\right) = 2 \frac{{e}^{\lambda x} + {e}^{- \lambda x}}{2} = 2 \cosh \left(\lambda x\right)$

with $\lambda = {\log}_{e} 5$ but

$2 \cosh \left(\lambda x\right) \ge 2$ and $- 1 \le \sin \left({e}^{x}\right) \le 1$ so no real solution is possible.

The equation $\sin \left({e}^{x}\right) = 2 \cosh \left(\lambda x\right)$ surely has infinite complex solutions arranged in a capricious pattern.

You can obtain complex roots proceeding as follows.

(1) Define the complex function $f \left(z\right) = \sin \left({e}^{x + i y}\right) - 2 \cosh \left(\lambda \left(x + i y\right)\right) = 0$
(2) Determine the real and imaginary components. In this case we have

$R e \left(f \left(z\right)\right) = r \left(x , y\right) = - 2 C o s \left(\lambda x\right) C o s h \left(\lambda y\right) + C o s h \left({e}^{x} S \in y\right) S \in \left({e}^{x} C o s y\right)$

$I m \left(f \left(z\right)\right) = s \left(x , y\right) = - 2 S \in \left(\lambda x\right) S \in h \left(\lambda y\right) + C o s \left({e}^{x} C o s y\right) S \in h \left({e}^{x} S \in y\right)$
(3) Using an iterative procedure such as Newton-Raphson, determine the solutions for

$\left\{\begin{matrix}r \left(x y\right) = 0 \\ s \left(x y\right) = 0\end{matrix}\right.$

To have an idea about the roots placement there is an attached plot showing in blue the contour for $r \left(x , y\right) = 0$ and in red the contour for $s \left(x , y\right) = 0$

The complex roots lay into the intersection of both curves. and a zoomed region A sample of solutions 