1. **State the problem:** Verify if the simplification of $\log_2 \left( \frac{x}{3} + 5 \right) - \log_2 (16)$ to $\log_2 \left( \frac{x}{48} + \frac{5}{16} \right)$ is correct. 2. **Recall the logarithm subtraction rule:** For positive $a,b$ and base $c>0$, $c\neq1$, we have $$\log_c(a) - \log_c(b) = \log_c \left( \frac{a}{b} \right).$$ 3. **Apply the rule:** $$\log_2 \left( \frac{x}{3} + 5 \right) - \log_2 (16) = \log_2 \left( \frac{\frac{x}{3} + 5}{16} \right).$$ 4. **Simplify the fraction inside the logarithm:** $$\frac{\frac{x}{3} + 5}{16} = \frac{x}{3 \times 16} + \frac{5}{16} = \frac{x}{48} + \frac{5}{16}.$$ 5. **Addressing the "keep, switch, flip" rule:** This rule applies when dividing two fractions: keep the first fraction, switch the division to multiplication, and flip the second fraction. Here, the expression inside the logarithm is a sum, not a division of two fractions. The subtraction of logarithms converts to division of the arguments, but the numerator is $\frac{x}{3} + 5$, a sum, so we do not flip any terms inside the sum. 6. **Conclusion:** The simplification is correct as is. The "keep, switch, flip" rule does not apply inside the sum; it only applies to division of fractions. Therefore, $$\log_2 \left( \frac{x}{3} + 5 \right) - \log_2 (16) = \log_2 \left( \frac{x}{48} + \frac{5}{16} \right).$$