As you mentioned it is easier to compare these numbers using logarithms rather than their true values. So instead of using $m^n$ we can look at $\log m^n = n\log m$.
If you are writing a program then you would start by setting MAX = $a\log b$ and MIN = $a\log b$. Then for every number after, $m^n$, divide MAX by $n\log m$. If the result is less than one then the new number is larger so replace MAX with the new number. If the result is greater than one then the new number is smaller and there is nothing to do. Do the similar thing with MIN. After you have gone through all the numbers MAX should hold the largest value and MIN should hold the smallest value.
This is of course assuming what you are using to calculate these values can compute $n\log m$ for the large values you need.
You have
$$(18\log_32\log_43-3\log_32\log_43)(2\log_23\log_32-2\log_32\log_83)\;,$$
which is correct. Clearly
$$18\log_32\log_43-3\log_32\log_43=15\log_32\log_43\;,$$
so we can immediately simplify it to
$$15\log_32\log_43(2\log_23\log_32-2\log_32\log_83)\;.\tag{1}$$
Now if $4^x=3$, then $2^{2x}=3$, so $\log_23=2\log_43$, or $\log_43=\frac12\log_23$. Similarly, you can verify that $\log_83=\frac13\log_23$. Thus, we can further simplify $(1)$ to
$$\frac{15}2\log_32\log_23\left(2\log_23\log_32-\frac23\log_32\log_23\right)\;.\tag{2}$$
This is extremely easy to evaluate if you know something about products of the form $\log_ab\log_ba$. If not, use the fact that in general $\log_bx=\frac{\log_ax}{\log_ab}$. I’ll complete the calculation below but leave it spoiler-protected; mouse-over to see it.
A useful general fact is that $\log_ab\log_ba=1$; this follows easily from the fact that in general $\log_bx=\frac{\log_ax}{\log_ab}$. Thus, $\log_23\log_32=1$, and $(2)$ is simply $\dfrac{15}2\left(2-\dfrac23\right)=\dfrac{15}2\cdot\dfrac43=10$.
Best Answer
Try the steps below:
$$\log_2x >\log_3x$$ $$\frac{\ln x}{\ln2}>\frac{\ln x}{\ln 3}$$ $$\ln x \left(\frac{1}{\ln2} - \frac{1}{\ln3} \right)>0$$
Since
$$\frac{1}{\ln2} - \frac{1}{\ln3} >0$$
the following must hold,
$$\ln x > 0$$
Thus,
$$x>1$$