The Laws of Logarithms
Use matching logarithm bases and positive arguments to turn products into sums, quotients into differences and powers into multipliers. Expand or combine logarithms, and use change of base to evaluate them.
What a learner can do afterwards
- Write log x + 2 log y as a single logarithm for positive x and y
- Expand log(a²b / c) into separate terms for positive a, b and c
- Use a change of base to evaluate a logarithm the calculator has no key for
1 · Read
You have used logs to name exponents. Here log means base 10; ln means base e. A log base must be positive and not 1, and its argument must be positive. The product and quotient laws combine logs with the same base. Expanding log(x^2) into 2 log x requires x > 0: a negative x has a positive square, but log x has no real value.
For positive M and N, log(MN) = log M + log N and log(M/N) = log M - log N. Also log(M^r) = r log M for any real r. Read these laws backwards to condense. Move coefficients into powers first, then multiply or divide the arguments. Each subtracted log gives a denominator factor. For positive s, t and u, 3 log s + log t - 2 log u = log(s^3 t/u^2).
To expand, separate products and quotients, then bring exponents to the front. For positive r, s and t, log(r^3 s/t^2) = log(r^3) + log s - log(t^2) = 3 log r + log s - 2 log t. A power of a product affects both factors: log((rs)^3) = 3(log r + log s). If a denominator contains a product, subtract its whole log expansion.
The product law applies to multiplication inside a log. It does not split terms joined by + inside the argument. For example, log(r+s) must stay together. Check a proposed identity with positive numbers: log(1+1) = log 2 > 0, while log 1 + log 1 = 0. One counterexample is enough to show that a claim does not hold for every allowed value.
Change of base divides the log of the argument by the log of the original base: log base b of M = log(M)/log(b) = ln(M)/ln(b). Here M > 0, b > 0 and b != 1. Both logs in a fraction use the same new base. For example, log base 5 of 125 = log(125)/log(5). Since 125 = 5^3, the power law gives 3 log(5)/log(5) = 3.
If the argument and base are powers of one number, apply the power law to both logs. For example, log base 8 of 16 = log(2^4)/log(2^3) = 4 log(2)/(3 log(2)) = 4/3. You can also divide given log values. If log base 3 of A = 7 and log base 3 of B = 4, then A = 3^7 and B = 3^4. Changing to base 3 gives log base B of A = 7/4.
Use matching bases and positive arguments. Products give sums, quotients give differences, and powers give multipliers. Change base by dividing by the log of the original base.
Lesson, questions and flashcards written for this dataset.
The readings and six extra questions were written for this dataset, and the flashcards were reviewed and revised for it. Video references credit Khan Academy's Using the logarithmic product rule and Using the logarithmic power rule. The video license terms are not established here.
More to read
Why adding logs multiplies their arguments
Take a base b with b > 0 and b != 1. Suppose log base b of M = p and log base b of N = q. This says M = b^p and N = b^q, so both M and N are positive.
Multiplying gives MN = b^(p+q). The exponent needed to make MN is therefore p+q. This gives the product law: adding the two logs gives the log of MN.
Division gives M/N = b^(p-q), so the quotient law subtracts the logs. Raising M to a real power r gives M^r = b^(pr), so the power law multiplies the log by r.
All these calculations use one shared base. They do not split a sum inside a logarithm. For example, log(1+1) is positive, but log 1 + log 1 is zero.
Check new arguments when expanding. A negative x gives a positive x^2, so log(x^2) exists. But log x has no real value there. The expansion log(x^2) = 2 log x therefore needs x > 0.
Finding the denominator in change of base
Let t be log base b of M. In power form, b^t = M. Here M > 0, b > 0 and b != 1.
Take base 10 logs of both sides. The power law gives t log(b) = log(M). Since b != 1, log(b) is not zero. Dividing by it gives t = log(M)/log(b).
The argument M stays in the numerator. The original base b goes in the denominator. Both logs use the same new base. Natural logs give the same answer: t = ln(M)/ln(b).
For example, log base 16 of 8 = log(8)/log(16). Write 8 = 2^3 and 16 = 2^4. The fraction becomes 3 log(2)/(4 log(2)) = 3/4.
Reversing that fraction answers a different question: log base 8 of 16. Identify the argument and original base before entering the fraction into a calculator.
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