Why don't logarithms work here? I have this problem:
For what values of $x$ is $x^6 \ge x^8$?
I solved this by the following.
\begin{align}
x^6 & \ge x^8 \\
\Longrightarrow \quad 1 & \ge x^2 \\
\end{align}
Therefore $$ -1 \le x \le 1 $$
When using the power identity for logarithms however, and dividing by $\ln x$, the answer is $6 \ge 8$. Why?
$$x^6 \ge x^8$$
$$6\ln x \ge 8\ln x$$
$$6 \ge 8$$
 A: 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).
A: Canceling the logs is not a valid step because it's possible that $\ln x = 0$.  This happens when $x=1$, which is a valid solution value for the original inequality.  So canceling like that can cause solutions to be lost.
Also, $\ln x$ can be negative.  In this case you'll need to reverse the inequality sign.  This happens when $0 < x < 1$.
If you're very careful, you can get part of the right answer by taking logs.  But this method should still be avoided because it loses solutions.  And the biggest reason of all, when you take logs and then use the log power rule as you did, you completely remove $x \le 0$ from the domain, and in general this can lead (and does in this case) to more lost solutions.
Note that dividing both sides by $x^8$ is also not valid because this is not allowed if $x=0$.  It still leads to the right answer but that's a happy accident.  If you do that method on an assignment, test, etc., you're likely to lose points.
The proper way to do this problem is as follows:
\begin{align}
x^6 &\ge x^8\\
x^6 - x^8 &\ge 0\\
x^6(1-x^2) &\ge 0
\end{align}
Equality is obtained when $x=0$ or $x=\pm 1$.  For the strict inequality $x^6(1-x^2) > 0$, this is true if $x^6$ and $1-x^2$ are both positive or if they're both negative.  Note that $x^6$ is never negative.  So the strict inequality is only satisfied when both factors are positive.  $x^6 > 0$ when $x \ne 0$ and $1-x^2 >0$ when $-1 < x < 1$.
Putting everything together tells us the solution is $-1\le x \le 1$, i.e., $[-1,1]$.
A: You can avoid division issues in the inequality by following this subtraction track. If $x=0$, the inequality is true. If $x\neq 0$, you can take the $\log$ because quantities $x^6$ and $x^8$ are strictly positive. 
$x^6 \ge x^8  \implies 6 \log |x| \ge 8 \log |x| \implies (6-8)\log |x| \ge 0$
which implies that $\log |x| \le 0$, or $x\in [-1\; 0[\cup]0\;1]$. Do not forget the absolute value when putting an even power out of a logarithm. Now your set of solutions is included in $\{0\}\cup[-1\; 0[\cup]0\;1] = [-1\;1]$.
If $x\in [-1\;1]$, $x^2\le 1$, hence $x^6 x^2 \le x^6$, so the set of solutions is the whole segment $[-1\;1]$.
A: From $$x^6\ge x^8,$$ taking the logarithms
$$\log x^6\ge\log x^8$$
follows, provided that both arguments are positive. So we need to handle $x=0$ separately, and this is trivial.
Next, as the exponents are even, we can rewrite (with a mandatory absolute value)
$$6\log|x|\ge8\log|x|,$$ which is just
$$\log|x|\le 0,$$ or
$$0<|x|\le1.$$

If you insist on using a division, you need to discuss the sign before simplifying


*

*$\log|x|>0\implies6\ge8$ which is impossible,

*$\log|x|=0\implies0\ge0$, so that $|x|=1$ are solutions,

*$\log|x|<0\implies6\le8$ so that $0<|x|\le1$ are solutions.
A: $\ln x$ is negative when $0<x<1$, so when you cancel that you may need to invert the inequality depending on what $x$ is.
(And, of course, taking logarithms does not work well for negative $x$ at all).
A: You can't divide both side to $x^8$ hence $0$ holds inequation.Furthermore,the solution of inequation is $${ x }^{ 6 }\left( x-1 \right) \left( x+1 \right) \le 0$$

$$x\in [-1 ,1]$$

Next step is clear from the answer above
A: I would recommend to you never to “cancel”. Rather, you want to do the same thing to both sides of an equation; or, under suitable circumstances, to both sides of an inequality; or to multiply top and bottom of a fraction by the same thing.
So you took the inequality $x^6\ge x^8$. It’s not at all clear what you meant by the “logarithm identity”, but probably you took the log of both sides, to get $6\ln x\ge8\ln x$. Now what? If $\ln x$ is positive, you divide both sides by it and get the false inequality $6\ge8$. but if $\ln x$ is zero, you get a true equality $0=0$, and this is fine. Also, if $\ln x$ is negative, when you divide both sides of $6\ln x\ge8\ln x$ by $\ln x$, the inequality reverses, giving you the valid inequality $6\le  8$.
Conclusion? We get $\ln x\le0$, in other words $0<x\le1$.
A: A contrived argument:
By inspection, $x=0$ is a solution.
Then without loss of generality, $x>0$, because if $x$ is a solution, so is $-x$. This allows us to write
$$6\log x\ge8\log x.$$
By inspection, $\log x=0$ is a solution.
Then without loss of generality, $\log x> 0$ because if $\log x$ is a solution, $-\log x$ is a non-solution, and conversely. This allows us to write
$$6\ge8$$ and there are no solutions for $\log x$, so that all $-\log x>0$, i.e. $x<1$, are solutions.
Hence
$$x=0\lor(x>0\land x<1\land-x<1).$$
