log 2 + log 3 is 0.778. log 5 is 0.699. The difference is not rounding: 0.778 is log 6, because adding two logarithms multiplies the numbers inside them.
a is fixed at 2. Move b and watch the second bar track log(2b), never log(2 + b).
The laws you already have hold for fractional powers, and a fraction in the exponent means a root and a power:
am/n = (n√a)m = n√(am)
so for 163/4 you may take the fourth root first and cube it, or cube first and take the fourth root:
Both give 8, and the first is the one to use by hand because the numbers stay small. Take the root first, every time.
A negative exponent is a reciprocal, so 16−3/4 = 1/8, and the two ideas combine without any new rule.
163/4 is not 16 × 3/4. That gives 12, and it treats the exponent as a multiplier. An exponent says how many times to multiply the base by itself, and a fractional one extends that to roots. Check the size: 161/2 = 4 and 161 = 16, so 163/4 must sit between 4 and 16. Both 8 and 12 pass that test, which is why this one needs the method and not just a glance.
| Law | In words | Why |
|---|---|---|
| logaxy = logax + logay | Adding logs multiplies the numbers. | Because adding exponents multiplies powers. |
| loga(x/y) = logax − logay | Subtracting logs divides them. | Because subtracting exponents divides powers. |
| logaxm = m logax | A power comes out to the front. | Because xm is x multiplied by itself m times. |
Two values worth knowing before any law. Since a1 = a and a0 = 1:
logaa = 1 and loga1 = 0
in every base. So log 10 = 1, ln e = 1, log22 = 1, and log of 1 is zero whatever the base. They come up constantly as the last step of a rearrangement, and a student who recomputes them each time loses time they do not have on Paper 1. Note also that an even root means the positive one, so √(a²) = |a| rather than a.
Each law is an exponent law wearing different clothes. A logarithm is an exponent: logay is the power you raise a to in order to get y. If you can state that sentence you can rebuild all three laws, and you will not invent a fourth.
There is no law for log of a sum. Try it with a = 2, b = 3:
log 2 + log 3 = 0.301 + 0.477 = 0.778
and that is not log 5. It is log 6, because the product law says so. The false step does not produce an approximation; it answers a different question:
The gap is 0.079 here and it grows with the numbers. The two agree only when a + b happens to equal ab, and the one case with both numbers equal is a = b = 2, because 2 + 2 and 2 × 2 are both 4. There are others with a and b different, such as 3 and 1.5, where both sides come to 4.5. They are accidents of the numbers, not a law, and you cannot tell from looking at a question whether you have landed on one.
log(x + y) cannot be simplified. If a question gives you a sum inside a logarithm, the sum stays there, or you factorise it first into a product and then the product law applies.
Your calculator has log base 10 and ln. For anything else:
logax = logbx / logba
and b is whichever base you actually have. So
log47 = ln 7 / ln 4 = 1.404
and the same rule turns an awkward pair into an exact answer:
log25125 = log5125 / log525 = 3/2 = 1.5
Choosing base 5 there, rather than 10, is what makes it exact. Look for a base that both numbers are powers of before reaching for the calculator.
When the unknown is in the exponent, take logs of both sides. For 2x−1 = 10:
x − 1 = log210 = 3.32, so x = 4.32
When both sides can be written to the same base, no logarithm is needed. For (1/3)x = 9x+1, write everything as a power of 3:
3−x = 32x+2, so −x = 2x + 2, giving x = −2/3
Try the same-base route first. It is exact, it is quicker, and on Paper 1 it is the only route available.
Both machines have a base entry on the log key, so change of base is a convenience rather than a necessity. Knowing the formula still matters, because Paper 1 has no calculator at all.
When you may use it. Analysis Paper 1 is non-calculator, so every logarithm there must be exact: log25125 = 3/2, not 1.5 off a screen. Paper 2 and Paper 3 allow the calculator, and that is where 1.404 is an acceptable answer.
The mark people lose. Writing log a + log b as log(a + b). It is log(ab), and the two differ by a real amount: 0.778 against 0.699 on the smallest pair of distinct integers above 1. The habit that prevents it: say the law out loud as "adding logs MULTIPLIES", every time you use it. And note the argument order trap above, where the Casio wants the base first and the Nspire wants it second, which silently returns log74 = 0.712 instead.
1. Evaluate 163/4.
2. Find log 2 + log 3, to 3 decimal places.
3. Now find log(2 + 3), to 3 decimal places.
4. Solve 2x−1 = 10, to 2 decimal places.
5. Why is log a + log b not log(a + b)?
1 markThe correct law, named or used.
1 markThe algebra, with the logarithm removed correctly.
1 markThe answer, exact on Paper 1 and to 3 significant figures on Paper 2.
An invented law loses every mark after the step it appears in, because nothing that follows it can be right. Examiners mark the method, and a law that does not exist is not a method.
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