1. The problem states: Given $\log_5 25 = 2$, find the values of $f(4)$, $f(1)$, $f(\frac{1}{4})$, $f(16)$, and $f(2)$ where $f(x) = \log_4 x$. 2. Recall the definition of logarithm: $\log_a b = c$ means $a^c = b$. 3. Since $\log_5 25 = 2$, this is because $5^2 = 25$. 4. Now, for $f(x) = \log_4 x$, we find each value: - $f(4) = \log_4 4 = 1$ because $4^1 = 4$. - $f(1) = \log_4 1 = 0$ because any number to the power 0 is 1. - $f(\frac{1}{4}) = \log_4 \frac{1}{4} = -1$ because $4^{-1} = \frac{1}{4}$. - $f(16) = \log_4 16 = 2$ because $4^2 = 16$. - $f(2) = \log_4 2$ is not an integer, but can be expressed using change of base formula: $$f(2) = \log_4 2 = \frac{\log 2}{\log 4} = \frac{\log 2}{\log 2^2} = \frac{\log 2}{2 \log 2} = \frac{1}{2}.$$ 5. The first line with squares "Log_5 25 = 2, so □ □ = □" can be completed as "Log_5 25 = 2, so 5^2 = 25". Final answers: $f(4) = 1$, $f(1) = 0$, $f(\frac{1}{4}) = -1$, $f(16) = 2$, $f(2) = \frac{1}{2}$.