Question

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

Answer 1

No real solutions.

Explanation:

The equation is `sin(e^x)=2(e^(lambda x)+e^(-lambda x))/2=2cosh(lambda x)`

with `lambda = log_e5` but

`2cosh(lambda x) ge 2` and `-1 le sin(e^x) le 1` so no real solution is possible.

The equation `sin(e^x)=2cosh(lambdax)` surely has infinite complex solutions arranged in a capricious pattern.

You can obtain complex roots proceeding as follows.

(1) Define the complex function `f(z)=sin(e^(x+iy))-2cosh(lambda(x+iy))=0`
(2) Determine the real and imaginary components. In this case we have

`Re(f(z))=r(x,y)=-2 Cos(lambda x) Cosh(lambda y) + Cosh(e^x Siny)Sin(e^x Cosy)`

`Im(f(z))=s(x,y)=-2 Sin(lambda x) Sinh(lambda y) + Cos(e^x Cosy)Sinh(e^x Siny)`
(3) Using an iterative procedure such as Newton-Raphson, determine the solutions for

`{(r(x,y)=0),(s(x,y)=0):}`

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

The complex roots lay into the intersection of both curves.

and a zoomed region

A sample of solutions