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

The logarithms have a base number of $2$.

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    Have you tried anything yet? We won't do homework for you – 2017-01-24
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    yes, I have... many times, but I still can't find the solution :/ – 2017-01-24
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    Mind sharing some of your approaches? – 2017-01-24
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    What I did was this: -log(2)(x-1)*log(2)(3x-4)>log(2)1 <=> -log(2)(x-1)>log(2)(1/(3x-4))... – 2017-01-24
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    then I subtracted the two logarithms in the right .. and then I switched one logarithm to the left member and I divided the arguments – 2017-01-24
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    We know that $log(u)$ is negative when $0 < u < 1$, and positive when $u >1$. Can you use the arguments of the two logarithms and this fact to help? – 2017-01-24
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    then I equaled the arguments.. but I didn't get the right solution :/ – 2017-01-24
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    I don't know ... – 2017-01-24
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    well we know the sign has to be < – 2017-01-24

2 Answers 2

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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?

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    thank you so much, I understood.. but now I have two different options, which one do I choose? because I know the solution is ]5/3;2[ – 2017-01-24
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    @SarahJones You have to put together the solutions of the two options. But the first one has no solution… – 2017-01-24
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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 $1 < x < 2$