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I was wondering how one would multiply two logarithms together?

Say, for example, that I had:

$$\log x·\log 2x < 0$$

How would one solve this? And if it weren't possible, what would its domain be? Thank you!

(I've uselessly tried to sum the logs together but that obviously wouldn't work. Was just curious as to what it would give me)

EDIT: Thank you everyone for the answers!

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As $\log2x=\log2+\log x,$

$$0>\log x\cdot\log2x=\log x(\log x+\log 2)$$

Now if $(y-a)(y-b)<0$ with $a<b,$ we can prove $$a<y<b$$

So, here we have $$-\log 2<\log x<0$$

$$2^{-1}<x<1$$

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There is no particular rule for the product of logarithms, unlike for the sum.

Applying the latter, you can rewrite

$$\log(x)\log(2x)=\log(x)(\log(x)+\log(2))=t(t+\log(2))$$

and proceed as usual to find the domain of $t$. Then $x=e^t$.

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You want the product of two things to be negative. That means one of them is positive and the other negative. Use what you know about logarithms to interpret what that means in this case.

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See $\log(a)\log(b)\neq \log(a)+\log(b)$ now $\log$ is negative from $0-1$ but here we will get both terms negative giving us positive answer implying greater than $0$ . But now here's what we missed if $x>0.51$ then $2x$ is greater than $1$ so we such set of numbers ie $0.5<x<1$ thus domain of this function is $(0.5,1)$

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It works as for most products of two quantities: you have $\log x \log 2x < 0 $ if they ($ \log x$ and $ \log 2x$) are of opposite signs

This amounts to: $0<x<1$ and $2x>1$, or $x>1$ and $0<2x<1$, hence $x\in ]\frac{1}{2},1[$, or $\emptyset$ (respectively), hence the solution is $x\in ]\frac{1}{2},1[ \cup \emptyset = ]\frac{1}{2},1[ $.

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