Logarithm problem 2 by Dhaval Furia

Let $f(x) = \sqrt{\ln\left(\dfrac {4x-x^2}3\right)}$ . For $f(x)$ to be real,

$\begin{aligned} \ln\left(\frac {4x-x^2}3\right) & \ge 0 \\ \frac {4x-x^2}3 & \ge 1 \\ x^2 - 4x + 3 & \le 0 \\ (x-1)(x-3) & \le 0 \end{aligned}$

For $f(x)$ to be real, $\boxed{1 \le x \le 3}$ .