<s>[INST] Find all $a$ such that $y=\log_\frac{1}{\sqrt3} (x-2a) = \log_3(x-2a^3-3a^2) $ 
Find all values of parameter $a\in \mathbb{Z}$ such that $$y= \log_\frac{1}{\sqrt3} (x-2a)$$ $$and$$ $$y = \log_3(x-2a^3-3a^2)$$ intersect at points with whole coordinates.

This is what I did:
$$\log_\frac{1}{\sqrt3}(x-2a) = \frac{log_3(x-2a)}{\log_33^{-1/2}}$$
$$\frac{\log_3(x-2a)}{\log_33^{-1/2}} = \log_3(x-2a^3-3a^2)$$
$$-2\log_3(x-2a) = \log_3(x-2a^3-3a^2)$$
$$\log_3\frac{1}{(x-2a)^2} = \log_3(x-2a^3-3a^2)$$
$$\frac{1}{(x-2a)^2} = (x-2a^3-3a^2)$$
I got up to this point, not sure how to proceed.
 [/INST] After getting a common base, $3$, and exponentiating, we have
$$
\frac{1}{(x-2 a)^2} = -3 a^2 - 2 a^3 + x
$$Since we are requiring $a,x$ to be whole numbers, the RHS is an integer which means the LHS must be as well: then $x-2a=\pm 1$. However, since these came from logarithms, only the case $x-2a=1$ is allowed. Then the equation becomes $1=-3a^2-2a^3+1+2a$, whose solutions are $a=\{-2,0,1/2\}$.
So in summary, the point $(a,x)=(0,1)$ is the only whole number solution , and there is also the solution $(a,x)=(-2,-3)$ if one admits negative integers.
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