Three laws turn a product into a sum, a quotient into a difference, and a power into a multiple. Each one is a single exponent law from Foundations of Algebra written in the other notation, and each is proved in two lines from the definition. Two more rules that look like laws are not.
Core ideas
Nothing here is new mathematics. It is the exponent laws you already have, said in the notation of “What a Logarithm Is”.
Multiplying powers of the same base adds the exponents. A logarithm is an exponent. So the logarithm of a product is the sum of the logarithms — not a coincidence and not a separate fact, but the same statement written twice.
log_b(M + N) does not simplify. Not to log_b M + log_b N, not to anything. Addition inside the argument has no counterpart outside it, and inventing one is the most expensive habit available in this course.
log_b(M·N) = log_b M + log_b N requires M > 0 and N > 0, or the right-hand side names nothing. The usual casualty is log_b(x²) = 2·log_b(x), which fails for negative x: the correct identity is log_b(x²) = 2·log_b|x|.
The material
Where each law comes from, what it is allowed to touch, and the two rules people invent in their place.
Fix a base b and take M > 0, N > 0. Write m = log_b M and n = log_b N, which by the definition means M = b^m and N = b^n. Every proof below is that substitution followed by one exponent law.
log_b(M·N) = log_b(M) + log_b(N) for all M > 0 and N > 0.
M·N = b^m·b^n = b^(m+n) by the exponent law for a product. So m + n is the exponent that produces M·N, and by the definition of a logarithm that is exactly what log_b(M·N) means. Hence log_b(M·N) = m + n = log_b M + log_b N.
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log_b(M/N) = log_b(M) − log_b(N) for all M > 0 and N > 0.
M/N = b^m/b^n = b^(m−n). So m − n is the exponent that produces M/N, which gives log_b(M/N) = m − n = log_b M − log_b N. The subtraction on the right is where the sign is lost in practice, and the order matters: log_b(N/M) is the negative of this.
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log_b(M^p) = p·log_b(M) for all M > 0 and every real p.
M^p = (b^m)^p = b^(mp) by the exponent law for a power of a power. So mp is the exponent that produces M^p, and log_b(M^p) = mp = p·log_b M. Note that p is unrestricted: it may be negative, or a fraction, which is what lets a root come out as a coefficient.
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A number check is not a proof, but it is a fast way to catch a misremembered law. log_2(8·4) = log_2(32) = 5, and log_2(8) + log_2(4) = 3 + 2 = 5. If a rule you are about to use fails on numbers you can evaluate by hand, it is not a rule.
The first is the sum error and there is nothing to rescue in it. The second is more interesting, because the quotient log_b(M)/log_b(N) is not meaningless — it equals log_N(M). Here log_2(8)/log_2(4) = 3/2, and log_4(8) is indeed 3/2. A ratio of two logarithms is a change of base, which “Change of Base” makes into a rule; it is not the logarithm of a ratio.
Expanding means using the laws left to right, until every logarithm has a single letter inside it. Condensing means the reverse: coefficients go back up as exponents, sums become products, differences become quotients, and the whole expression ends as one logarithm. Both directions are needed — expanding to differentiate later, condensing to solve equations in “Solving Logarithmic Equations”.
log_b(x²) = 2·log_b(x) is true for x > 0 and false for x < 0. At x = −5 the left side is log_b(25), a perfectly good number, and the right side is 2·log_b(−5), which is not defined at all. The identity that holds everywhere except x = 0 is log_b(x²) = 2·log_b|x|. Expanding a logarithm can shrink its domain, and the answer should say so.
Interactive
Three that hold, two that people invent
Method
Work from the outside in. Doing it the other way re-creates brackets you then have to remove.
If the argument is a quotient, use the quotient law before anything else; then split the products; only then bring exponents down. The outermost operation is the one the logarithm is applied to.
√y is y^(1/2) and ∛y is y^(1/3). The power law does not recognise a root sign, so convert first — the coefficient 1/2 comes out of that conversion, not out of nowhere.
log_b(x³) becomes 3·log_b(x). The exponent must sit on the whole argument of that logarithm: in log_b((x + 1)³) the cube is on the bracket and comes down, while in log_b(x³ + 1) it is on x alone and nothing comes down.
The expansion is valid where every logarithm in it is defined. For log_b(x³·√y / z²) that means x > 0, y > 0 and z > 0, and the answer is incomplete without it.
The expansion is four lines. The check is four more, and it is worth every one of them.
The check catches the two failures worth catching: a coefficient that never came down, and a sign lost on the quotient. If the right-hand column had come to 15 you would know the −2·log_2(4) had been added instead of subtracted, before the mistake travelled any further.
It confirms the expansion at one point and nowhere else. The proof that it holds for every positive x, y and z is the two-line derivation of each law above — a numeric check can refute a false rule but cannot establish a true one.
For a faded expansion, take log_b(√(x^5·y) / z³) with x, y, z > 0. The supplied first line is the quotient-law split. Rewrite the root as a power, finish the expansion, then condense your answer back to the original expression. Check that every exponent lands on the whole factor it came from and that the minus sign stays with the denominator.
Given log_b(x²·√y / z³) you should produce 2·log_b(x) + (1/2)·log_b(y) − 3·log_b(z) with x, y, z > 0, and turn it back again. If the order of the laws is ever in doubt, work outermost first and the doubt does not arise.
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