Solve logarithmic inequality I need help solving $-\log_2(x-1)*\log_2(3x-4)&gt;0$ 

The logarithms have a base number of $2$.
Set $u=\log_2(x-1)$ and $v=\log_2(3x-4)$. Then you need to find when $-uv&gt;0$, that is, $uv&lt;0$. This becomes


  either $u&gt;0$ and $v&lt;0$ or $u&lt;0$ and $v&gt;0$.


Now replace $u$ and $v$ by their definition and solve the inequalities.

For instance, the first case becomes
$$
\begin{cases}
\log_2(x-1)&gt;0 \\
\log_2(3x-4)&lt;0
\end{cases}
$$
so you get $x-1&gt;1$ from the first and $0&lt;3x-4&lt;1$ from the second. Can you go on?
Not a mathematician, but I can solve this using grade 11 math. All you need is to find when you get a positive number as a total. It does not matter what the base of the logarithm is. 

$\log(x)$ is negative when $0 &lt; x &lt; 1$, $\log (x)$ when $x &gt; 1$ is positive no matter what the base of the log is.

A negative number times a negative number is positive
A negative number times a positive number is negative
A positive number times a positive number is positive

$0 &lt; x - 1 &lt; 1$ when $1 &lt; x &lt; 2$
$0 &lt; 3x - 4 &lt; 1$ when $3/4 &lt; x &lt; 1$

your inequality is therefore true when $1 &lt; x &lt; 2$