Solve for x with an expression of independent variable in the exponent I am helping a friend in research. I wish to solve for x, in this equation. 

$p, k, m, l, r\text{ and }h$ are all constants. They may vary, depending on the user's input, but they would all be determined to be constants, before $x$ is sought. None of them would be zero. If the range of likely values will be of some relevance : 
$p\text{ and }k$ will be between 0 and 1. 
$r, h\text{ and }l$ will be between 0 and 2. 
Online 'solve for x' tools such as Wolfram Alpha, Cymath and QuickMath don't seem to be able to solve it. Cymath said : Unable to solve. QuickMatch said : Takes too much CPU time. I would appreciate if someone could solve this. 
 A: Too long for a hint.
If I read your equation correctly you can rearrange it so that it looks like this:
$$
A e^{Bx-C} = De^{Ex-C}
$$
Then take the (natural) log of both sides and continue.
I note that others here have done the algebra for you.
A: $$
\mathrm{e}^{kr(x-m)}\left[pk(l-r)\mathrm{e}^{-klx}-(1-p)k(r-h)\mathrm{e}^{-khx}\right] = 0
$$
if we can ignore for one second $x=-\infty$ we require 
$$
pk(l-r)\mathrm{e}^{-klx}-(1-p)k(r-h)\mathrm{e}^{-khx} = 0
$$
or
$$
pk(l-r)\mathrm{e}^{-khx}\left[\mathrm{e}^{-k(l-h)x}-\frac{(1-p)(r-h)}{p(l-r)}\right] = 0
$$
once again lets focus on 
$$
\mathrm{e}^{-k(l-h)x}-\frac{(1-p)(r-h)}{p(l-r)} = 0 \implies -k(l-h)x = \ln\left[\frac{(1-p)(r-h)}{p(l-r)}\right]
$$
so we can find 
$$
x = -\frac{1}{k(l-h)}\ln\left[\frac{(1-p)(r-h)}{p(l-r)}\right]
$$
assuming that you have a valid value for the log.
A: $$p(kl-kr)e^{(kr-kl)x-kmr}-(1-p)(kr-hk)e^{(kr-hk)x-kmr}=0$$
Multiplying the both sides by $e^{kmr}$, we have
$$pk(l-r)e^{k(r-l)x}-(1-p)k(r-h)e^{k(r-h)x}=0$$
Here, let $X=e^{kx}$. Then, we have
$$pk(l-r)X^{r-l}-(1-p)k(r-h)X^{r-h}=0$$
Multiplying the both sides by $X^{h-r}$ to get$$pk(l-r)X^{h-l}-(1-p)k(r-h)=0,$$
i.e.
$$X^{h-l}=\frac{(1-p)(r-h)}{p(l-r)}$$
Hence, we have
$$X=e^{kx}=\left(\frac{(1-p)(r-h)}{p(l-r)}\right)^{\frac{1}{h-l}}$$
Hence, we have
$$\color{red}{x=\frac{1}{k(h-l)}\ln\left(\frac{(1-p)(r-h)}{p(l-r)}\right)}$$
