Calculus - Which is bigger?

Put $f(x)=\log \frac{x(2x-1)}{2-x} = \log x + \log (2x-1) - \log (2-x)$ and $g(x)=2 \left( x-\frac{1}{x}\right)$ . Note that $f(1)=g(1)=0$ .

Differentiating, we have $f'(x)=\frac{1}{x}+\frac{2}{2x-1}+\frac{1}{2-x}$ and $g'(x)=2+\frac{2}{x^2}$ .

Subtracting, and rewriting as a single fraction, we find

$f'(x)-g'(x)=\frac{4(x-1)^2 (x^2-x+1)}{ (2-x)x^2 (2x-1)}$

Each term in this fraction is positive for $x \in (1,2)$ . So $f$ increases more rapidly than $g$ on this interval, and from the same starting point; therefore $\boxed{\log \frac{x(2x-1)}{2-x}>2 \left( x-\frac{1}{x}\right)}$ .