Chapter 11 asked forward questions: start at 3, multiply by 3 four times, where do you land? The answer was 81. A logarithm asks that same question backwards: start at 3, land on 81, how many times did you multiply? The answer is 4. Same three numbers, same relationship — only the unknown has moved.
So log3 81 = 4. Read the notation piece by piece and it stops being strange: the small number tucked underneath is the base you are multiplying by, the number after it is the result you are trying to land on, and the value of the whole expression is the exponent that gets you there.
Notice what does not move: the base stays the base. It sits under the log on one side and underneath the exponent on the other, and it never becomes anything else. What swaps is the other two: the number inside the log is the result of the power, and the value of the log is the exponent. If you can place those three numbers correctly, you can flip any statement in either direction, and the figure above shows exactly where each one lands.
Neither form is more correct or more advanced than the other. They are translations of one sentence, the way “12 ÷ 4 = 3” and “4 · 3 = 12” are translations of one sentence. When a problem is hard to see in log form, flip it; when it is hard to see in exponential form, flip it back.
This restriction is not decoration — it is forced by Chapter 11. Every power of a positive base is positive, so when you ask by = x, the right side simply has no way to be zero or negative:
A fraction with 1 on top is never 0, no matter how large the bottom grows. So log2 0 and log2(−8) do not have hard answers — they have no answers at all. They are undefined, the way 5 ÷ 0 is undefined. The domain of logb x is x > 0, and that is the first thing to write down in any problem where a variable sits inside a log.
Say it carefully, because this is the most common confusion in the chapter: the number inside the log must be positive, but the value of the log may be negative. log10 0.001 = −3. The input 0.001 is a perfectly respectable positive number; it is the answer that came out negative.
b ≠ 1. One raised to any power at all is still one, so the flip breaks in both directions at once:
b > 0. A negative base cannot support the fractional exponents a log has to produce. (−4)1/2 asks for √−4, which is not a real number, and the same failure repeats at every exponent whose fraction has an even denominator. Meanwhile the whole-number powers of −4 alternate in sign — −4, 16, −64, 256 — hopping across the axis instead of tracing anything. There is no continuous curve to invert, so negative bases are ruled out. Base 0 fails too: 0−3 means dividing by 0.
Each of these is the flip applied to something you already know. None of them is worth memorizing separately — derive them in your head in two seconds instead.
The fourth one is the same idea run the other way, and it is worth doing slowly because it looks like magic and is not. Let y = logb x. The flip says that means by = x. Now put the first statement into the second in place of y:
For any base bigger than 1, the sign of the answer is decided entirely by whether the input is above or below 1 — because the powers of b pass through 1 exactly at the exponent 0.
| the number inside | example, base 10 | the log | because |
|---|---|---|---|
| bigger than 1 | log 100 | 2 | positive |
| exactly 1 | log 1 | 0 | always zero |
| between 0 and 1 | log 0.01 | −2 | negative |
| 0 or less | log 0, log(−5) | — | undefined |
Use it as a free error check. If a problem asks for log5 0.04 and your answer comes out positive, you already know it is wrong before checking anything else, since 0.04 sits below 1. (It is −2: 5−2 = 1/25 = 0.04.)
Two bases come up so often that they are written without a base at all, which trips people up the first time they meet it.
Neither is a different kind of object. The flip is identical: ln x = y means ey = x, so ln e = 1, ln 1 = 0, and ln(e5) = 5 for exactly the reasons above. Be careful with a calculator, though — the button marked log is base 10 and the button marked ln is base e, and they give different numbers for the same input: log 20 ≈ 1.301 while ln 20 ≈ 2.996.
Because the flip swaps which number is the input and which is the output, the point (3, 8) on y = 2x becomes the point (8, 3) on y = log2 x. Swapping the coordinates of every point is precisely what reflecting across the line y = x does — switch the figure above to Same curve and watch the two dots trade places across that diagonal.
| function | domain | range | asymptote |
|---|---|---|---|
| y = bx | all real x | y > 0 | y = 0, horizontal |
| y = logb x | x > 0 | all real y | x = 0, vertical |
The two rows are the same row with the columns swapped, and that is not a coincidence — it is the flip, drawn. The exponential can never output 0, which is exactly why the log can never accept 0. Its graph dives down the y-axis without ever touching it, so logb x has no maximum and no minimum: it takes every real value eventually, just very slowly. Going from log10 100 = 2 up to log10 1000 = 3 costs you nine hundred units of input to buy one unit of output.
Here is a piece of good news that is rarely said out loud. When a problem has one log, alone on one side, flipping it can never produce an illegal answer — because the flip sets the inside of the log equal to by, and by is automatically positive. The domain takes care of itself:
Trouble arrives later in the chapter, the moment two logs are combined into one. Combining changes the domain, so the new equation can have solutions the original never allowed. Those are the extraneous ones, and they must be thrown out:
The habit that protects you costs one line: before solving, write the restriction that every log's inside imposes, and at the end substitute back into the original equation, not into any rearranged version of it. For log2 x + log2(x − 6) that restriction is x > 6, which alone would have caught −2 immediately.
One more place a false answer sneaks in, and it appears in this very section: solving for an unknown base. logb 81 = 4 flips to b4 = 81, and that equation has two real solutions, 3 and −3, since (−3)4 = 81 as well. Only b = 3 survives, because a base must be positive. Work through the Flip to solve case above to see it happen.