Logarithm on inequality: Find the domain of $u$ where $u$ is the 1-1 transformation $ u = -2\log{y}$, where $0<y<1$.

Question

I have $0<y<1$. I need to find the domain of $u$ where $u$ is the $1$-$1$ transformation $ u = -2\log{y}$.

Can I perform the following steps? Take Logarithm: $\log{0}<\log{y}<\log{1}$, Multiply by $-2$: $-2\log{0}>-2\log{y}>-2\log{1}$ and obtain $0<u<\infty$?

Otherwise how can I find it? Cause logarithm looks like an increasing function, so if something has an order before, it should also have it after, or is it wrong? In general how do I deal with logarithms or with taking functions to all sides of inequalities?

Answer 1

Since $$ \lim_{y\to0}\log y=-\infty $$ and $\log 1=0$, due to the intermediate value theorem and the fact that the logarithm (in base $e$) is increasing, $\log y$ takes on any negative value and the same is true for $2\log y$.

Thus $u=-2\log y$ takes on any positive value:

$-2\log y>0$ if and only if $0<y<1$.

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Note. Here's my definition for $\lim_{x\to c}f(x)=-\infty$, for $f$ defined on $D(f)$, which $c$ is an accumulation point of:

for every $M$, there exists $\delta>0$ such that, for every $x$, if $0<|x-c|<\delta$ and $x\in D(f)$, then $f(x)<M$.

Hence, no $x\to 0^+$ is necessary.