Domain for real logarithmic square root

Question

If $x$ is a real number, then $\sqrt{\log_e \left(\frac{4x-x^2}{3}\right)}$ is a real number if and only if

Accepted Answer

C. For the square root of a logarithm to be real, the quantity inside the square root must be non-negative: $$\log_e \left(\frac{4x-x^2}{3}\right) \ge 0$$ $$\frac{4x-x^2}{3} \ge e^0 = 1$$ $$4x - x^2 \ge 3$$ $$x^2 - 4x + 3 \le 0$$ $$(x - 1)(x - 3) \le 0 \implies 1 \le x \le 3$$

Answer

$-3 \le x \le 3$

Answer

$1 \le x \le 2$

Answer

$1 \le x \le 3$

Answer

$-1 \le x \le 3$

Domain for real logarithmic square root

Explanation

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