Is this exponential equation solvable? natural logarithms, exponential

Question

$$\displaystyle{a=\frac{e^{-cos(\frac{b}{x})}-e^{-\frac{1}{x}}}{(1-e^{-\frac{1}{x}})}}$$

I'm trying to solve for $x$. $a$ and $b$ are constants.

Any help is really appreciated.

Thanks Ghassan

Answer 1

$\displaystyle{a=\frac{e^{-cos(\frac{b}{x})}-e^{-\frac{1}{x}}}{(1-e^{-\frac{1}{x}})}}$

We can make things a bit more manageable by letting $t = -\dfrac{1}{x}$ giving us

$\displaystyle{a=\frac{e^{-cos(bt)}-e^{t}}{(1-e^{t})}}$

which we can write as

$\displaystyle{cos(bt)=-\ln{(e^t + a(1-e^t))}}$

As you can see in this animated graph there is always at least one solution, and possibly infinitely many more, depending on the value of $a$.

Answer 2

As said, there is no solution which can be expressed using elementary function and root finding methods should be used. For illustration purposes, I shall consider $$f(x)=\displaystyle{\pi+\frac{e^{-cos(\frac{e}{x})}-e^{-\frac{1}{x}}}{(1-e^{-\frac{1}{x}})}}$$ and use Newton iterative scheme $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Starting far away from the solution (say, at $x_0=2$), the successive iterates are $4.35491$, $6.30492$,$6.41182$, $6.41195$ which is the solution for six significnt figures.