# Need help in solving for two unknowns involving a logarithmic function

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 !

• Hint. Divide the first equation by the second. Commented Mar 13, 2017 at 15:46
• Take the ratio between the two equations and derive an equation in the single unknown $\beta$.
– mlc
Commented Mar 13, 2017 at 15:47
• Get Wolfram alpha to draw the graph & get the answer for free! wolframalpha.com/input/… ... $\beta=-15.53 \cdots$ .. it is quite easy to set up an iteration to get this answer too. Commented Mar 13, 2017 at 16:22
• Alternative answer ... there is no solution in the range you specify. Commented Mar 13, 2017 at 16:23
• Thanks @DonaldSplutterwit , especially for the wolfram tool which I didn't know about Commented Mar 13, 2017 at 17:07

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.