The exponent, found backward: aˣ = N ⟺ x = loga N — and products become sums.
Doubling reaches 8 in three steps, 16 in four, 1024 in ten — but how many doublings reach exactly 1000? That question is not asking for a count or a product; it is asking for the exponent, and exponents are exactly what a logarithm hands back. This lesson introduces loga N as the inverse of raising a to a power, translates fluently between the two languages, meets the everyday base-10 log (lg) and base-e log (ln), and earns the rules that turn multiplication into addition — the very trick that once powered slide rules and let astronomers multiply enormous numbers by simply adding.
In the previous lesson we held a base still and let the exponent run: y = ax. Reading it forward is easy — give me an exponent, I raise the base and report the result. 23 = 8; 210 = 1024. But now turn the question around. Suppose the result is known and the exponent is the mystery:
2x = 1000 — what power of 2 gives 1000?
We know 29 = 512 (too small) and 210 = 1024 (a hair too big), so the exponent we want sits just below 10. There is no whole-number answer, yet a perfectly good exponent does exist — somewhere near 9.97. Rather than leave it nameless, we give it a name. We call that exponent the logarithm of 1000 to the base 2 and write
x = log2 1000 (read "log base 2 of 1000")
In words: log2 1000 is the power you raise 2 to in order to get 1000. Nothing more mysterious than that. The logarithm is simply the exponent, retrieved by working backward from the result. In general:
For a base a > 0, a ≠ 1 and a number N > 0, the symbol loga N means the exponent x for which ax = N. So loga N answers one question: "to what power must I raise a to land on N?" The answer it returns is a number — an exponent.
Why the restrictions? The base must be a > 0 and a ≠ 1 for the same reasons the exponential needed them: a negative base has no smooth powers, and 1 raised to anything is just 1, so it could never reach any other N. And the input must be N > 0 because ax is always positive — no power of a positive base is ever zero or negative, so there is simply no exponent that produces N ≤ 0. We will return to that last point, because forgetting it is the most common slip with logs.
The widget above hints at the single most useful skill of this whole lesson: sliding back and forth between a power statement and a log statement. They are the same fact, seen from two angles.
ax = N ⟺ x = loga N
Left side: "a to the x equals N." Right side: "the exponent on a that gives N is x." Same a, same N, same x — just which one you solve for changes.
Let us build fluency with clean numbers. In each row the base stays put; only the framing changes:
| Exponential form | Logarithm form | Why |
|---|---|---|
| 23 = 8 | log2 8 = 3 | 3 is the power of 2 that gives 8 |
| 102 = 100 | lg 100 = 2 | 10² = 100, so lg 100 = 2 |
| 50 = 1 | log5 1 = 0 | anything to the 0 is 1 |
| 3−1 = ⅓ | log3(⅓) = −1 | 3−1 = ⅓, so the exponent is −1 |
| a1 = a | loga a = 1 | the power of a that gives a is 1 |
Two of those rows are worth memorizing, because they hold for every allowed base:
loga 1 = 0 (since a0 = 1) and loga a = 1 (since a1 = a)
loga N is not a · N, and it is not N ÷ a. It is the exponent that sits on a. When in doubt, rewrite it as a power: log2 16 = x means 2x = 16, so x = 4 — not 8, not 32.
Any allowed base gives a logarithm, but two bases come up so often they earn shorthand names.
The common log, base 10. Because we count in tens, base 10 fits the decimal world perfectly, and powers of ten are everywhere — 10, 100, 1000, and 0.1, 0.01. We write lg N = log10 N. Its values on the powers of ten are dead simple, because the logarithm just reads off the exponent:
lg 1000 = 3 (103 = 1000) · lg 100 = 2 · lg 10 = 1 · lg 1 = 0 · lg 0.1 = −1
The natural log, base e. A second base, the number e ≈ 2.71828, sits at the heart of continuous, natural growth (it is what compound interest approaches as you compound more and more often). Its logarithm gets its own name: ln N = loge N. So ln e = 1, ln 1 = 0, and ln 7 ≈ 1.946 — the exponent you put on e to reach 7. We will meet e again when we model growth in a later lesson; for now, treat ln as "just a log, with the favorite base e."
In most school work, "log" written with no base means lg (base 10). On many calculators the log button is base 10 and the ln button is base e — so the e-logarithm has its own key. When the base matters, write it: log2, log5, … leave nothing to guess.
Here is where logarithms become powerful instead of merely tidy. Recall the very first power rule: when you multiply two powers of the same base, the exponents add — am·an = am+n. Logarithms are exponents, so a logarithm of a product must turn into a sum of logarithms.
loga(M·N) = loga M + loga N
loga(M / N) = loga M − loga N
(valid for M > 0, N > 0). A logarithm trades the hard operation for the easy one underneath: multiply ↦ add, divide ↦ subtract.
Why is it true? Write M and N as powers of the base. Say M = ap and N = aq, so by definition loga M = p and loga N = q. Multiplying, M·N = ap·aq = ap+q, which says the exponent giving M·N is p + q — that is, loga(M·N) = p + q = loga M + loga N. The product rule for logs is just the product rule for exponents, wearing a different hat.
Check it on numbers you can see: 8·4 = 32, and
log2(8·4) = log2 32 = 5 = log2 8 + log2 4 = 3 + 2 = 5 ✓
The product rule splits log of a product, not log of a sum. So:
loga(M + N) ≠ loga M + loga N and loga(M·N) ≠ (loga M)(loga N)
A quick disproof of the first: log2(2 + 2) = log2 4 = 2, but log2 2 + log2 2 = 1 + 1 = 2… coincidence? Try 4 + 4: log2 8 = 3, while log2 4 + log2 4 = 2 + 2 = 4. Not equal — the "rule" fails. Adding inside the log is not the same as adding the logs.
One rule remains, and it is the workhorse. When you raise a power to a power, exponents multiply: (am)n = amn. Translated into log language, an exponent inside a logarithm comes out front as a multiplier.
loga(Mn) = n · loga M
Quick check: log2 8 = log2(23) = 3 · log2 2 = 3 · 1 = 3 ✓. The exponent 3 simply steps down in front.
Changing the base. Calculators carry only lg and ln, yet we constantly want logs in other bases — log2 5, say. The bridge is the change-of-base formula: to find loga N, take logs in any convenient base b and divide.
loga N = logb Nlogb a = lg Nlg a = ln Nln a
Same answer whichever base b you choose — so use the one your calculator has.
Use it to evaluate something with no whole-number answer:
log2 5 = lg 5lg 2 ≈ 0.6990.301 ≈ 2.32
Sensible? 22 = 4 and 23 = 8, so the exponent giving 5 lies between 2 and 3, nearer 2 — and 2.32 lands right there. ✓
The three rules — product ↦ sum, quotient ↦ difference, power ↦ multiple — chain together to crack expressions that look forbidding. The strategy is always the same: break the complicated log into a sum or difference of simple logs, then evaluate the easy pieces.
(a) log2(32 · 48) = log2 32 + log2 4 − log2 8 = 5 + 2 − 3 = 4. (check: 32·4/8 = 16, and 24 = 16 ✓)
(b) lg 2 + lg 5 = lg(2·5) = lg 10 = 1. (the product rule run in reverse — collect, then evaluate)
(c) log3 18 − log3 2 = log3(182) = log3 9 = 2. (the quotient rule in reverse: 32 = 9 ✓)
Notice the rules run both ways. Splitting a hard log apart and collecting scattered logs into one are the same skill in opposite directions, and a good solver uses whichever direction makes the arithmetic land on something easy — a power of the base, ideally, whose log is just an integer.
Turn products / quotients / powers into sums / differences / multiples of logs; then evaluate the simple pieces (using loga a = 1, loga 1 = 0, and powers of the base). That is the entire art of working with logarithms — and it is why slide rules could multiply by sliding two log scales past each other and adding lengths.
A logarithm is an exponent retrieved backward. For a > 0, a ≠ 1 and N > 0, the number loga N is the power on a that yields N, captured by the conversion law ax = N ⟺ x = loga N. Always loga 1 = 0 and loga a = 1. Two bases earn names: lg = log10 (the common log) and ln = loge (the natural log, e ≈ 2.718). Because logs are exponents, they inherit the power rules backward: log(MN) = log M + log N, log(M/N) = log M − log N, log(Mn) = n log M, and any base converts to any other by loga N = ln N / ln a. Chain these to simplify hard expressions into easy integers. Remember the trap: log(M + N) is not log M + log N, and N must stay positive. Next we treat the log as a full function, y = loga x, and discover it is the mirror image of y = ax across the line y = x.
Write the exponential fact 34 = 81 in logarithm form.
log3 81 = 4. The base stays 3; the exponent 4 is what the log returns; 81 is the result. (ax = N ⟺ x = loga N.)
Evaluate log2 16.
Ask: 2 to what power is 16? Since 24 = 16, log2 16 = 4.
Evaluate lg 1000.
lg means log10. Since 103 = 1000, lg 1000 = 3.
Evaluate log5 1, and explain in one sentence why the base does not matter.
log5 1 = 0. For any allowed base a, a0 = 1, so loga 1 = 0 always — raising a base to the zero power is the only way to land on 1.
Simplify log2 6 + log2 4 − log2 3 to a single number.
Collect with the product/quotient rules: log2(6·4/3) = log2 8 = 3 (since 23 = 8).
Use the change-of-base formula to express log2 7 with the common log, and give its value to three decimals.
log2 7 = lg 7 / lg 2 ≈ 0.845 / 0.301 ≈ 2.807. (Sensible: 22 = 4, 23 = 8, so the exponent giving 7 is just under 3.)
Six questions to lock it in. Tap the answer you think is right.
The single big idea of this lesson is that a logarithm is an exponent found backward: loga N answers "to what power must I raise a to get N?" Everything else — the named bases lg and ln, the product/quotient/power rules, change of base — flows from that one definition plus the exponent rules students already own. The most reliable way to fix it is to insist on the conversion law ax = N ⟺ x = loga N, and to have students rewrite any unfamiliar log as a power before touching it.
Misconceptions to watch. The headline error is the linearity illusion: log(M + N) = log M + log N (false) confused with the true product rule log(MN) = log M + log N. A quick numerical disproof (log2(4+4) = 3, but log2 4 + log2 4 = 4) lands better than any warning. Two more: treating loga N as a · N or N ÷ a rather than an exponent; and forgetting that the input must be positive (N > 0), since no power of a positive base is ever ≤ 0. Encourage a sanity estimate every time — bracket the answer between the two nearest integer powers.
Common Core alignment. This lesson supports F-LE.A.4 (understand a logarithm as the solution of an exponential equation ax = N, and evaluate it), F-BF.B.5 (relate logarithms and exponents as inverse operations), and the structure standards HSA-SSE (use the properties of logarithms to rewrite and evaluate expressions). It builds directly on the rational-exponent and exponential-function work earlier in Stage 24 and sets up the logarithmic function and its mirror symmetry in the next lesson.