The logarithm is an exponent

The statement $\log_b(c) = a$ and the statement $b^a = c$ are the same fact in two costumes. The base b is the number being raised; the argument c is the target; and the log's value a is the exponent that gets you there.

Read $\log_2 8$ aloud as a question: "2 to what power gives 8?" The answer is 3, so $\log_2 8 = 3$. Every conversion error in this topic comes from letting the three numbers wander out of their roles. When converting, anchor on the base: it is the base in both costumes, and the log's OUTPUT always lands in the exponent slot.

1. $\log_b(c) = a$: start here.
2. Keep the base: b stays on the bottom left.
3. The log's value becomes the exponent: $b^a$.
4. The argument is what you must hit: $b^a = c$. Check it numerically.

When the argument is a power of the base, the log is exact, no calculator involved. $\log_3 81 = 4$ because $3^4 = 81$. $\log_{10} 0.001 = -3$ because $10^{-3} = 0.001$. $\log_5\!\left(\tfrac{1}{25}\right) = -2$ because $5^{-2} = \tfrac{1}{25}$.

Two special values fall out of the exponent question instantly. $\log_b 1 = 0$ for every legal base, because $b^0 = 1$. And $\log_b b = 1$, because $b^1 = b$. Fractions and decimals between 0 and 1 have negative logs, because hitting a small target takes a negative exponent; the log itself is negative, but its argument is still positive.

When the argument is not a perfect power, trap the log between the powers on either side. For $\log_3 50$: since $3^3 = 27$ and $3^4 = 81$, and 50 sits between them, $\log_3 50$ is between 3 and 4. That estimate is often all a problem wants, and it is a built-in error detector: any claimed value outside the bracket is wrong before you check anything else.

Notice what the estimate is NOT: it is not $50 \div 3$. Division answers "how many threes add to 50," an additive question. The log asks "how many threes multiply to 50," and multiplying climbs much faster, which is why the answer is a small number like 3.5-ish rather than 16.7.

There is no property for the log of a sum. $\log_b(x + y)$ does not split, expand, or simplify in general; if you need its value, you add first and take the log of the total. The temptation comes from a half-remembered true rule: $\log_b(xy) = \log_b x + \log_b y$, which converts multiplication of arguments into addition of logs. Run it backward and you see the danger: adding two logs corresponds to MULTIPLYING their arguments, never adding them.

One numeric grenade to keep handy: $\log_2(8 + 8) = \log_2 16 = 4$, but $\log_2 8 + \log_2 8 = 3 + 3 = 6$. If the split were legal those would match. They never do, except by coincidence. When you catch yourself writing $\log(a+b) = \log a + \log b$, stop and add first.