I can't understand my own solutions to $\log_5(3x-1)<1$ and $\log(6/x)>\log(x+5)$ Here I have two examples of logarithmic inequalities. Despite being able to solve, I simply couldn't fully understand my own process.

$\boxed{\text{Example 1: }\log_5(3x-1)<1}$
$\log_5(3x-1)<1 \Longleftrightarrow 3x-1<5 \Longleftrightarrow x<2$
But the solution is not $x\in(-\infty, 2)$
Now considering the values for $x$ where $\log_5(3x-1)$ is defined:  $ 3x-1>0 \implies x>\frac{1}{3}$
The solution is the intersection.
$$(-\infty, 2)\cap \left(\frac{1}{3}, \infty \right) \implies x\in \left(\frac{1}{3}, 2\right)$$

$\boxed{\text{Example 2: }\log \left(\frac{6}{x}\right)>\log(x+5)}$
Again, I solved
$\frac{6}{x}> x+5$ and $x+5>0$, as $x>-5$ being the range of defined values for the logarithms.
$$\frac{6}{x}> x+5 \Longleftrightarrow \frac{6}{x}-x-5 > 0  \Longleftrightarrow  \frac{x^2+5x-6}{x}<0 \Longleftrightarrow \frac{x^2+5x-6}{x}<0 \Longleftrightarrow \frac{(x+6)(x-1)}{x} < 0$$
Then, I just did the table and got $(-\infty, -6)\cup (0, 1) $
The solution for this problem is $ ((-\infty, -6)\cup (0, 1))\cap (-5, \infty) \implies x\in(0, 1) $


The aims of this question are:

*

*Understand how to solve inequalities better, understand it more intuitively;

*Understand how do inequalities work, understand it more intuitively, as well;

*Why the answer is the "solution" intersection with the defined values;


I'm sorry if the question is too elementary, but any hint would be welcome.
 A: You seem to be solving these inequalities just fine. Maybe it would be better, as suggested in the comments, to state restrictions first and then work your way from there.
In the first question, for example, you get a solution first ($x<2$) then apply restrictions from there. I think this is what may be leaving you confused with your process.
When you're given the logarithm $\log_5(3x-1)$, you should first find the values of $x$ satisfying $3x-1>0$, to ensure that you don't accidentally cause a negative number to be present in your logarithm. Once you get $x>\frac{1}{3}$, then you can start looking for a solution to the inequality. Once you get $x<2$, it'll be easy for you to apply the restriction in without having to think about it.
Same thing goes for the second one, but you didn't consider the logarithm on the left as well when determining restrictions (i.e. you got $x>-5$ but you didn't get $x>0$, which  gets you closer to the answer). I think this would have saved you some time.
Hopefully this helps you.
A: You seem to have a couple of the ideas down.
This is our basic definition $\log_b x = y \implies x = b^y$
If $y = 1$
$\log_b x = 1 \iff x = b$
There are a few basic characteristics of the function.
The function is "monotonically increasing."  That is $\log x > \log y \iff x > y$
The function is "injective":  $\log x = \log y \iff x = y$
And, the domain of $\log x = (0,\infty).$  If $x<0$ the function is not defined.
You don't need to know these vocabulary words.  You do need to understand the implications as it relates to the logarithm function.
To the problems at hand.
$\log_5 (3x-1) < 1 \implies 3x-1 < 5$ from the first two rules.  And $3x-1 > 0$ from the last rule
I think it is a good idea to list all of these constraints up front.
We might write it like: $0< 3x - 1 < 5$
$\frac 13 < x < 2$
For the second problem:
$\log \frac 6x > \log (x+5)\\
\frac 6x > x + 5 \text { and }\frac{6}x > 0 \text { and } x+5 > 0$
Fortunately, $\frac{6}x > 0 \implies x > 0 \implies x+5 > 0$ so we can drop the last constraint.
The contraint $x>0$ does us a service, in that, we can multiply through by $x$ without worrying about flipping the sign on the inequality.  If there were a possibility that x was negative, we could not do that.
$0 > x^2 + 5x - 6$ and $x>0$
$0>(x+6)(x-1)$ and $x>0$
The first inequality has a solution $(-6,1)$ and the second $(0,\infty)$
$(0,1)$ would be the interval where both hold.
