I have the following equations:
$$ 0.59 = \alpha\log_2(1 + 0.063\,\beta\,) $$ $$ 0.23 = \alpha\log_2(1 + 0.05\,\beta\,) $$
such that, $\alpha\in (0,\infty)$ and $\beta\in (0,\infty)$.
I tried to solve them by expressing $\beta$ as an exponential function and then using a Taylor series expansion as below:
$$ 0.59 = \alpha\,\frac{\ln(1 + 0.063\,\beta)}{\ln2}$$ $$=> 1 + 0.063\,\beta = e^{0.41/\alpha} = 1 + \frac{0.41}{\alpha} + \frac{0.41^2}{\alpha^22!} + \frac{0.41^3}{\alpha^33!} + \, \ldots$$
Similarily for the second term. I finally end up with a quadratic equation (after equating the two equations for $\beta\,)$ in $\alpha$ which results in complex roots.
Where am I going wrong ? Is there any other way to solve this without involving approximations ?
Any help is appreciated. Thanks !
After elimination of $\alpha$,
$$(1+0.063\beta)^{0.23}=(1+0.05\beta)^{0.59}.$$
This is a transcendental equation, which must be solved numerically. Besides the trivial $\beta=0$, there is another solution in the negatives, which you reject.