Solve logarithmic inequality

Question

I need help solving $-\log_2(x-1)*\log_2(3x-4)>0$

The logarithms have a base number of $2$.

Answer 1

Set $u=\log_2(x-1)$ and $v=\log_2(3x-4)$. Then you need to find when $-uv>0$, that is, $uv<0$. This becomes

either $u>0$ and $v<0$ or $u<0$ and $v>0$.

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

For instance, the first case becomes $$ \begin{cases} \log_2(x-1)>0 \\ \log_2(3x-4)<0 \end{cases} $$ so you get $x-1>1$ from the first and $0<3x-4<1$ from the second. Can you go on?

Answer 2

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 $0 < x < 1$, $\log (x)$ when $x > 1$ 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

$0 < x - 1 < 1$ when $1 < x < 2$
$0 < 3x - 4 < 1$ when $3/4 < x < 1$

your inequality is therefore true when $1 < x < 2$