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

For $\sqrt{\log_e \frac{4x-x^2}{3}}$ to be a real number, we need:

The expression inside the square root must be non-negative

$\log_e \frac{4x-x^2}{3} \geq 0$

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$\log_e a \geq 0$ when $a \geq 1$

This is because $\log_e 1 = 0$ (since $e^0 = 1$) and $\log_e a > 0$ when $a > 1$ (since $e^{\text{positive}} > 1$)

Therefore, we need: $\frac{4x-x^2}{3} \geq 1$

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$\frac{4x-x^2}{3} \geq 1$

$4x - x^2 \geq 3$

$4x - x^2 - 3 \geq 0$

$-x^2 + 4x - 3 \geq 0$

$x^2 - 4x + 3 \leq 0$

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We need to factor $x^2 - 4x + 3$:

We need two numbers that multiply to 3 and add to -4

These numbers are -3 and -1

$x^2 - 4x + 3 = (x-3)(x-1)$

Our inequality becomes: $(x-3)(x-1) \leq 0$

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For $(x-3)(x-1) \leq 0$, we need the product to be negative or zero.

The zeros are at $x = 1$ and $x = 3$.

When $x < 1$: both factors are negative, so product is positive

When $1 < x < 3$: $(x-1) > 0$ and $(x-3) < 0$, so product is negative

When $x > 3$: both factors are positive, so product is positive

Including the boundary points where the product equals zero:

At $x = 1$: $(1-3)(1-1) = (-2)(0) = 0$

At $x = 3$: $(3-3)(3-1) = (0)(2) = 0$

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Therefore, $\sqrt{\log_e \frac{4x-x^2}{3}}$ is real when $1 \leq x \leq 3$.

This problem combines three important concepts:

1. Domain requirements for square roots (non-negative argument)
2. Properties of logarithms (when they're non-negative)
3. Solving quadratic inequalities