Finding the value of p while solving a recurrence equation using Akra-Bazzi method? I am learning to solve recurrence equations using Akra-Bazzi method. I am stuck in figuring out how to find out the value of p by solving equations like: 
$$\left(3\over4\right)^p + \left(1\over6\right)^p=1$$
$$\left(1\over2\right)^p + \left(1\over4\right)^p + \left(1\over8\right)^p=1.$$
I want to find it out with using a scientific calculator and not wolfram alpha as online tools are not allowed during my exam.
 A: First off, in general $p$ will be some transcendental number without a clean closed form in terms of familiar constants.  So I assume you just want to find $p$ to some reasonable accuracy.
I would attack this with  the following method, illustrated for the first of your equations. I would convert (by multiplying the whole equations by the least common multiple of the fractions on the left) to 
$$
9^p + 2^p = 12^p
$$
Then starting with a guess $p_{old} = 1$ I would iterate using $p_{old}$ on the left and the next guess $p_{new}$ on the 
$$
p_{temp} = \log_{12} (9^{p_{old}} + 2^{p_{old}}) \\
p_{new} = 2 p_{temp} - p_{old}
$$
This algorithm converges geometrically, that is, each new guess has half the error of the previous guess.  FOr this example, going through this step (which can easily be set up to need just a couple of button pushes) converges to better that $.001$ accuracy after seven steps.  ($p_7 = 0.852537$; the correct value is $.851838$.)
However, you have to be careful.  Sometimes (for example if all of the fractions are greater than $1$), $p$ needs to be negative, and then this method will not converge.  You can generally manipulate it into some related equation where it does converge.
For your second example, this method for some reason converges really quickly. The fourth iteration already gives 6-digit accuracy.
