Let's take a closer look at your calculations:
$$x^6\geq x^8 \Leftrightarrow \ln(x^6)\geq \ln(x^8)$$
Here we must have $x\neq 0$, because $\ln(0)$ is not well-defined. So you have to check if the inequality holds for $x=0$ separately. (It does hold).
Now you want to use $$\ln(x^6)=6\cdot\ln(x)\geq 8\cdot\ln(x)=\ln(x^8).$$ This only holds for $x>0$, as $\ln(x)$ is only well-defined for $x>0$. Be sure to always check the requirements when using such rules. So for now we can only continue with the assumption $x>0$.
Your next step would be to divide both sides by $\ln(x)$. We can only do this, if $\ln(x)\neq 0$ which is equivalent to $x\neq 1$. So you also have to check if the inequality holds for $x=1$ separately (it does hold).
Assuming that we have $x>0$ and $x\neq 1$ we can now divide by $\ln(x)$. But because we are dealing with an inequality, we have to be careful and take a look at the sign of $\ln(x)$:
$$\ln(x)>0 \Leftrightarrow x>1,\quad \ln(x)<0 \Leftrightarrow 0<x<1.$$
So we have to distinguish two cases:
If $x>1$ we have:
$$x^6\geq x^8 \Leftrightarrow \ln(x^6)\geq \ln(x^8) \Leftrightarrow 6\ln(x)\geq 8\ln(x) \Leftrightarrow 6\geq 8.$$
This is not true, thus the inequality doesn't hold for $x>1$.
For $0<x<1$ we get:
$$x^6\geq x^8 \Leftrightarrow \ln(x^6)\geq \ln(x^8) \Leftrightarrow 6\ln(x)\geq 8\ln(x) \Leftrightarrow 6\leq 8.$$
This is true, thus the inequality holds for $0<x<1$.
We now have to do the same steps again for $x<0$, using that in this case we have $$\ln(x^6)=6\cdot\ln(-x),\ln(x^8)=8\cdot\ln(-x).$$ (I don't feel like writing the same steps again...)
The problem that arose with your calculations is, that you just applied some rule you heard of without carefully checking if you are allowed to apply this rule. Also to "just cancel out" often leads to problems, when one doesn't check that there is no $0$ involved that might get canceled.
Another (personal) remark: why would one use logarithms to solve this inequality? Just because it is possible to solve this question, doesn't mean that one should do it (you can see that a correct use of logarithms leads to a very long solution with lots of cases one has to think about).