I have been solving a microeconomics problem related to optimization and ended up with the following equation:
$$\left(100\alpha\right)^{\alpha}\left(25\beta\right)^{\beta}-\left(65\alpha+5\beta\right)^{\alpha}\left(\frac{130}{3}\beta\right)^{\beta}>0$$
Being unable to express $\beta$ in terms of $\alpha$ or vice versa, I plugged it into desmos and got this graph, which is somehow linear. My question is: how is it possible for this monstrosity to somehow simplify to a linear function? Furthermore, how do I simplify this equation to a linear one, or is it only possible computationally?
Some context: the microeconomics problem was to find the optimum of two Cobb-Douglas utility functions and corresponding budget limitations for some arbitary $\alpha, \beta$ within the restrictions below:
$$U_1(x,y) = x^{\alpha} y^{\beta},\quad\alpha, \beta \in (0;1), \quad 100x+400y-10000\leqslant0$$
and
$$U_2(x,y)=(x+5)^{\alpha} y^{\beta},\quad\alpha, \beta \in (0;1),\quad 100x+150y-6000\leqslant0$$
The linearity of the graph means that there is a directly proportional correlation between the preference for either utility function and exponent value, and I really want to figure out why this is the case, but I am unable to find any information about it online.
If you let $\alpha = \gamma\beta$ then the equation becomes $$ \begin{aligned} 0 &< (100\gamma\beta)^{\gamma\beta}(25\beta)^\beta - (65\gamma\beta + 5\beta)^{\gamma\beta}\left(\frac{130}{3}\beta\right)^\beta \\ &= \left((100\gamma\beta)^\gamma\cdot25\beta\right)^\beta - \left((65\gamma+5)^\gamma \beta^\gamma\cdot\frac{130}{3}\beta\right)^\beta \\ &=\left(25(100\gamma)^\gamma\beta^{\gamma+1}\right)^\beta - \left(\frac{130}{3}(65\gamma+5)^\gamma\beta^{\gamma+1}\right)^\beta. \end{aligned} $$
This inequality is true iff the ratio of the first term to the second term is $>1$, which gives us $$ \begin{aligned} 1&< \frac{\left(25(100\gamma)^\gamma\beta^{\gamma+1}\right)^\beta}{\left(\frac{130}{3}(65\gamma+5)^\gamma\beta^{\gamma+1}\right)^\beta} \\ &= \frac{\left(25(100\gamma)^\gamma\right)^\beta}{\left(\frac{130}{3}(65\gamma+5)^\gamma\right)^\beta} \\ \implies 1&< \frac{\left(25(100\gamma)^\gamma\right)}{\left(\frac{130}{3}(65\gamma+5)^\gamma\right)}. \end{aligned} $$ This results in a transcendental equation for $\gamma$, but if a solution exists, it implies that the original inequality is satisfied with $\alpha$ and $\beta$ in a linear relation $\alpha = \gamma\beta$.