Analysis only. It builds on the introduction to logarithms at SL 1.5 and feeds the logarithmic and exponential graphs at SL 2.9.
Set b to 3 and ask the class which bar is log 5. Most will point at the taller one, because 0.778 is the number they have just calculated. It is log 6. The label underneath each bar says which number it is the logarithm of, and that is the whole figure.
Then sweep b and watch the gap grow. The false law is not a bad approximation, it is an exact answer to a different question, and the figure is built to make that distinction visible rather than stated.
Stop at b = 2, where the two bars coincide. Ask why. 2 + 2 and 2 × 2 are both 4, and it is the only pair of equal positive numbers for which that happens. A class that works that out has understood the product law properly.
| Question | Answer |
|---|---|
| 1. 163/4 | (4√16)³ = 2³ = 8. |
| 2. log 2 + log 3 | 0.301 + 0.477 = 0.778, which is log 6. |
| 3. log(2 + 3) | log 5 = 0.699. |
| 4. 2x−1 = 10 | x − 1 = log210 = 3.32, so x = 4.32. |
| 5. Why the false law fails | B. log a + log b is log(ab), so it answers a different question. |
Questions 2 and 3 are deliberately adjacent and deliberately the smallest numbers available. If the gap is visible at 2 and 3, nobody can argue it away as rounding. Have them write both answers down before you say anything about either.
1 markThe correct law, named or used.
1 markThe algebra, with the logarithm correctly removed.
1 markThe answer, exact on Paper 1 and to three significant figures on Paper 2.
An invented law costs every mark after the line it appears in, because nothing downstream of it can be credited. Worth saying in those words: the cost is not one mark, it is the rest of the question.
| They wrote | What happened |
|---|---|
| 12 | On question 1, 16 × 3/4. The exponent has been read as a multiplier. Note that the size check does not catch it: 8 and 12 both sit between 161/2 = 4 and 161 = 16. |
| 64 | Used the SQUARE root: 4³. The denominator of the exponent names the root, so 3/4 means fourth. |
| 2 | Took the root and forgot to cube. |
| 4096 | Cubed and forgot the root. Both halves of the work are there, in the wrong quantity. |
| 0.699 on question 2 | Computed log 5. They have applied the false law in reverse, which is rarer and worth a word. |
| 0.778 on question 3 | The law that does not exist, used. They added the logs on a question that asked for the log of the sum. |
| 0.144 | Multiplied the two logs rather than adding them. |
| 0.631 | Divided them, which is log32 by change of base. A nice accident to point out. |
| 3.32 | On question 4, stopped at log210. One step from correct. |
| 2 | On question 4, used a base ten log, so log 10 = 1. Watch for it: the working looks immaculate. |
"Why is there no law for log of a sum?" Because a logarithm is an exponent, and the three laws are the three exponent laws in disguise. Adding exponents multiplies; subtracting divides; a power multiplies out. There is no exponent law for adding the bases, so there is no logarithm law for adding the arguments. Giving the reason stops them inventing a fourth law next term.
"Can I take the log of a negative number?" Not on this course. logay is defined for y > 0, because no real power of a positive base is negative. It matters in practice: solving log(x) + log(x − 3) = 1 gives x = 5 and x = −2, and the −2 has to be rejected because log(−2) does not exist. That rejection is usually a mark.
"Which base does log mean?" On the button, ten. In writing, ten unless a base is given. ln is base e. If a question writes log2 it means it, and change of base is how you get there.
"Do I have to show the change of base?" On Paper 2, no: type it in. On Paper 1 it is the only route, and the questions there are chosen so the answer is exact. log25125 = 3/2 is the shape to expect, and the trick is choosing base 5 rather than base 10.
| Stage | What to do |
|---|---|
| Demonstrate | Put three lines on one screen: log(2)+log(3), log(5), log(6). 0.778, 0.699, 0.778. The first and third match and the second does not, and that is the entire lesson in three key presses. Do it before any algebra. |
| Where they stick | The argument order on a log with a base. The Casio's logab( wants the BASE first and the Nspire's log( wants it second, so the same two numbers typed the same way give log47 = 1.404 on one machine and log74 = 0.712 on the other. Neither errors. Have them check against a case they know, such as log28 = 3. |
| The check | Substitute back. If x = 4.32 solves 2x−1 = 10, then 23.32 should be 10. It is one entry and it catches the dropped +1, the wrong base and the division slip all three. |
Analysis Paper 1 has no calculator, so the fluency that matters here is recognising that log25125 is 3/2 by eye. Set a few of those with the machines face down.
| Step | What |
|---|---|
| 1 | Rational exponents from the laws, not as a new rule. 163/4 two ways, same answer. |
| 2 | Ask for log 2 + log 3 and log(2 + 3) in that order, on paper, before the figure. |
| 3 | Figure at b = 3. Which bar is log 5? |
| 4 | Build the three laws from the three exponent laws, out loud. |
| 5 | Sweep b, then stop at b = 2 and ask why the bars agree. |
| 6 | Change of base, with log25125 done exactly. |
| 7 | Exponential equations: the same-base route first, logs second. |
| 8 | The domain point, with log(x) + log(x − 3) = 1 and its rejected root. |
Do not say "log of a sum is roughly the sum of the logs". It is not roughly anything. It is exactly the log of the product, and calling it an approximation gives students permission to use it when the numbers are big, which is where the gap is largest.
Do not say "logs are the opposite of powers" and leave it there. It is true and too vague to use. The usable sentence is that logay is the power you raise a to in order to get y, because that sentence can be substituted into a problem and the vague one cannot.