1. State the problem. Problem: Show that $\log_5 25 = 2$. 2. Formula and rules. Use the definition of logarithm: $$\log_b a = c \iff b^c = a$$ Important rule: The logarithm $\log_b$ is the inverse of the exponential function $b^x$ for $b>0$, $b\neq 1$, so to evaluate a logarithm we find the exponent that returns the argument when using the base. 3. Apply the definition. Set $a=25$ and $b=5$ and use the equivalence $\log_5 25 = c \iff 5^c = 25$. 4. Compute the exponent. Evaluate $5^2 = 25$ so the exponent $c$ that satisfies $5^c=25$ is $c=2$. 5. Alternative using inequalities. Observe $5^1 = 5 < 25$ and $5^2 = 25$, and since $5^x$ is strictly increasing for $5>1$, we have $1 < \log_5 25 \le 2$, and because $5^2=25$ the logarithm equals $2$. 6. Final answer. Therefore $$\log_5 25 = 2$$.