# Approximate a solution for a single variable exponential equation

Can anyone please help me fined (if it is possible) a closed-form solution or an approximation for the solution for the following equation (x is the only variable):

$$\frac{((a-1)b^{x+2}-(b-1)a^{x+2}+b-a)v+((a-1)b^{x+2}-(b-1)a^{x+2})u}{b-a}=0$$

What is known is:

1. a solution exists (the problem is finding a closed-form expiration).

2. $$x\ge0$$.

3. $$a,b,v,u$$ are parameters such that $$0, $$v>0$$, $$u>0$$.

Even an approximation for the solution will help. Since an expression is needed then numerical methods are not helpful here.

You can immediately solve for $$y=(a-1)b^{x+2}-(b-1)a^{x+2}$$ which occurs linearly twice in this expression, in the form:

$$y=-\frac{v(b-a)}{u+v}$$

However, you cannot get $$x$$ from $$y$$ in closed form unless for special values of $$a$$ and $$b$$, because the equation above has no solution in terms of elementary functions. Consider $$b^x=u$$ and $$a^x=u^{p}$$ where $$p=\ln a/\ln b>1$$. Then, you have a "polynomial" with a real exponent:

$$y=(a-1)b^2 u - (b-1)a^2 u^{p}$$

For numerical solution, I would start here (it's a reasonable expression where you can at least estimate the number and nature of solutions). Since you are asking for an analytical solution, this won't help much, because $$p$$ can be in principle anything from $$1$$ to $$\infty$$. Unless you have any other hints of the values -- if $$p$$ is very big, you can probably ignore the first term and get an analytical approximation. If $$p$$ is very close to $$1$$ ($$a$$ and $$b$$ very close together), you can do series expansion of all terms.

For the last case, you would take the expression for $$y$$ and compute the derivative with respect to $$b$$, and then take $$b=a$$, to get something like this:

$$y\approx y|_{b=a} + \frac{\partial y}{\partial b}|_{b=a} (b-a)$$

where of course you would have the painful problem of differentiating the $$p$$ term in the exponent. Unfortunately, I tried and even this equation contains combination $$u\ln u$$ and is therefore not solvable in terms of standard functions (you need Lambert's W function).

• Here is a link that tells about explicit solution of transcendental equations: "hindawi.com/journals/ijem/2015/523043" – Awe Kumar Jha Nov 27 '18 at 15:37
• How do you apply it? – Y.L Nov 27 '18 at 15:48
• Thanks @orion, regarding your last sentence, if $a$ and $b$ are very close together, can you please show how it can be done? – Y.L Nov 27 '18 at 15:58
• That link shows a numerical method that requires the function to be evaluated in a lot of points before "fitting" a polynomial to it. There is no miracles or free lunch. – orion Nov 27 '18 at 16:02
• Thank you @orion but I get stuck again at your last sentence lol. How do you apply this Lambert's W function method here? :) – Y.L Nov 27 '18 at 16:25
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