Suppose we have an inequation as shown below:$$I_0:\space \ln (x) > \frac{x-2}{x}$$ Now we would like to find the largest set $S$ of real numbers such that any element $p\in S$ will satisfy $I_0$ if we put $x=p$.
I also came across an interesting problem in my textbook. Let us have an equation and an inequality:$$I_1:\space 4\space (\log_2 x)^2+1=2\space \log_2(y)$$ $$I_2:\space \log_2(x^2)≥\log_2(y)$$ How many ordered pairs $(x,y)$ will simultaneously solve $I_1$ and $I_2$ and what will they be?
The book wrote infinite number of ordered pairs will be the solution to $I_1$ and $I_2$ and all of them will be of the form $(\sqrt2,t)$ where $t\in R$. But I think there is only one solution, which is $(\sqrt2,2)$. Am I correct?
For $I_0$ we can consider $f(x):=x\ln(x)+2-x$, and note that $I_0$ holds for $x \in \mathbb{R}$ if and only if $f(x) > 0$. The domain of $f$ is clearly $\mathbb{R}_{>0}$, and its first and second derivatives are $f’(x) = \ln(x)$ and $f’’(x) = 1/x$. The only solution to $f’(x) = 0$ is clearly $x=1$, and since $f’’(1)>0$, $f$ reaches a global minimum here. Now noting that $f(1)=1$ (which is the minimum of $f$ on its domain) we see that $\forall x \in \mathbb{R}_{>0}:f(x)>0$. So $S = \mathbb{R}_{>0}$.
For your second question, assume that $(x,y)\in \mathbb{R}^2$ such that $I_1$ and $I_2$ hold. Then $4\log_2(x)^2+1\leq 2\log_2(x^2)$, which implies $(2\log_2(x)-1)^2\leq 0$. Since the square is non-negative this implies $\log_2(x)=1/2$, and thus $x = \sqrt{2}$. Using $I_1$ we have $1=\log_2(y)$, so $y=2$. There is indeed only one solution.