How to solve the transcendental equation $a^h=bh+c$ with a parameter

Question

I've got a random Rayleigh variable $\xi$ with $p_\xi(x)=\frac{x}{\sigma^2}\exp\{-\frac{x}{2\sigma^2}\},x\geq0$

There are two hypotheses: $H_0:\sigma=\sigma_0$ and $H_1:\sigma=\sigma_1$

I have built the procedure for Wald sequential analysis and now I have to find $P\{\text{deny } H_0\Bigr|\sigma=a\}=\frac{1-B^h}{A^h-B^h}$

Now I have to find h from the following equation:

$$\int\limits_0^{+\infty}\left (\frac{p_1(x)}{p_0(x)}\right)^hp(x;a)=1$$

There are given $\sigma_0=2.2$ and $\sigma_1=2.7$ in the task.

After I had simplified the integral, I got the following equation:

$$\left (\frac{22}{27}\right)^h=1-\frac{6125}{8809}ha^2$$, that I have to solve for $h$.

I think I must get W-Lambert function somehow but I have no ideas so far. I would appreciate any hints.

Answer 1

In general, the solution for $a^x=bx+c$ is: $$x = -\frac{c}{b}-W\left(\frac{-a^{-c/b}\log a}{b}\right)\log^{-1}a$$

Answer 2

I was typing when appeared nbubis's answer; so I shall not repeat. You must be aware that, in R, W(x) does not have a value if x < -1/E (proof by Euler and Lambert). For the case of your last equation, the solution exists only if a < 0.542712. If this is the case, solve your equation using Newton method. It seems that the solution is almost exponential with "a" [h=575 for a=0.05, h=144 for a=0.10, h=64 for a=0.15, h=36 for a=0.20, h=23 for a=0.25, h=15 for a=0.30, h=10 for a=0.35, h=7 for a=0.40, h=4 for a=0.45, h=2 for a=0.50).

http://en.wikipedia.org/wiki/Lambert_W_function gives you two methods (second and third orders) for the computation of W(x) as well as approximations by series.