Solve for x with an expression of independent variable in the exponent

Question

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.

Answer 1

$$ \mathrm{e}^{kr(x-m)}\left[pk(l-r)\mathrm{e}^{-klx}-(1-p)k(r-h)\mathrm{e}^{-khx}\right] = 0 $$ 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.

Answer 2

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.

Answer 3

$$p(kl-kr)e^{(kr-kl)x-kmr}-(1-p)(kr-hk)e^{(kr-hk)x-kmr}=0$$ 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 $$\color{red}{x=\frac{1}{k(h-l)}\ln\left(\frac{(1-p)(r-h)}{p(l-r)}\right)}$$