# Nature of zeros of $4\cdot2^{2p}\cdot3^p-4\cdot3^{2p}-7\cdot2^{3p}+8\cdot2^{2p}+8\cdot3^p-4=0$

Recently while coming up with an example for a paper I'm writing I find myself wanting something to say about how 'awful' the first positive root of the equation $$4\cdot2^{2p}\cdot3^p-4\cdot3^{2p}-7\cdot2^{3p}+8\cdot2^{2p}+8\cdot3^p-4=0$$ is. Numerically I know it's about 1.576, but I suspect that it's no only irrational but cannot be solved for using elementary functions, e.g. $\log_2(3+\sqrt{5})$.

I've not much / any background in number theory of things like this and am kind of stumped. Are there any simple arguments in this direction?

Note: This is the simplest of the awful equations I could come up with. I'm also aware that it has a root at $p=2$

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Let $x=2^p$, $y=3^p$, then $4x^2y-4y^2-7x^3+8x^2+8y-4=0$ and $y=x^{\log_23}$. The first equation is a quadratic in $y$ with coefficients involving $x$. Solve it for $y$, and equate the two expressions for $x$. You may get something that can be handled by standard theorems on irrationality/transcendence (or maybe not - I haven't tried it). – Gerry Myerson Jun 28 '11 at 7:22
@Gerry Myerson. Thanks for spotting that. I'll have a further bash at it. – Stephen Sanchez Jul 1 '11 at 5:54