Proving a statement about a logarithmic function

Question

I'm trying to prove the function I've contracted is giving me a negative result in some equation but I'm having trouble doing it either with a direct way or by contradiction as I'm new to the log functions..

My function is $f:(0,\infty)->R$ such that $f\left(x\right)\ =\ \ln (x)-0.5$

I need to show that $$f\left(x\right)\cdot f\left(\frac{1}{x}\right)<0$$ $$\left(\ln\left(x\right)-0.5\right)\cdot\left(\ln\left(\frac{1}{x}\right)-0.5\right)<0$$

After I'm done with proving that (is it even correct?) I need to assume f(x) is continous at 1 and that f(1) does not equal zero..

Answer 1

$\ln \frac {1}{x} = -\ln x$

$(\ln x - \frac 12)(-\ln x - \frac 12)<0\\ -(\ln x)^2 + \frac 14<0\\ \frac 14< (\ln x)^2\\ \frac 12< |\ln x|$

Your proposition is only true when x is outside the interval $[e^{-\frac 12}, e^{\frac 12}]$

Answer 2

This is not true since we have

$$f(1) \cdot f\left(\frac11\right) = f(1)^2 \ge 0$$

To make things more fun, let's study when does $f(x) < 0$.

For $f(x) < 0$, we need $\ln (x) < 0.5$ and hence $x < e^\frac12$.

Hence $f\left(\frac1x\right) < 0$ we need $x > e^{-\frac12}$.

$$f(x) f\left(\frac1x \right) < 0 \iff x < e^{-\frac12} \lor x > e^{\frac12}$$