A calculator offers two bases and there are infinitely many. One identity closes the gap: log_b(x) = log_c(x) / log_c(b) for any base c you can actually compute. It follows in two lines from the definition and the power law, and the derivation is what keeps the argument and the base the right way up.
Core ideas
One theorem, proved from the definition, plus a clear view of the two expressions it is constantly confused with.
log_c(x) / log_c(b) is a quotient of two logarithms. “The Laws of Logarithms” has no law for such a quotient: it is not log_c(x/b), and it is not log_c(x) − log_c(b). Writing the division deliberately, with the base of the answer underneath, is most of the work of this lesson.
Take c = e and the identity reads log_b(x) = (1 / ln b) · ln x. So the graph of any logarithm is the graph of ln stretched vertically by a fixed factor — and reflected as well when 0 < b < 1, since ln b is then negative. The bases differ by a scale, never by a shape.
The identity holds for every legal c. Three different helper bases give three quotients that look nothing alike and must agree to every digit. When a numeric answer looks wrong, that is the cheapest test available.
The material
The statement, its two-line proof, the corollaries, and what it is not.
log_2(10) is a perfectly definite number: the exponent that turns 2 into 10. It is between 3 and 4, since 2^3 = 8 and 2^4 = 16. No calculator has a base-2 logarithm key, and no table was ever printed for base 2. The identity below is how the number is obtained anyway.
Let b and c be positive with b ≠ 1 and c ≠ 1, and let x > 0. Then log_b(x) = log_c(x) / log_c(b).
Let y = log_b(x). By the definition in “What a Logarithm Is”, that says exactly b^y = x.
Equal numbers have equal logarithms, so apply log_c to both sides: log_c(b^y) = log_c(x). The power law of “The Laws of Logarithms” turns the left side into y · log_c(b).
Divide both sides by log_c(b). That step is legal because log_c(b) = 0 happens only when b = 1, and the hypothesis excluded it. What is left is y = log_c(x) / log_c(b), which is the claim.
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Read the result once more for placement. The argument goes on top and the base of the answer goes underneath. log_2(10) is ln 10 / ln 2, not ln 2 / ln 10. Both quotients are real numbers and neither looks broken: the first is about 3.3219 and the second about 0.3010.
Put c = x. Then log_x(x) = 1, and the identity collapses to log_b(x) = 1 / log_x(b). The pair log_2(8) = 3 and log_8(2) = 1/3 is that statement with numbers in it, and it is a useful sanity check whenever a logarithm and its mirror image both appear in a problem.
Put c = e. Then log_b(x) = ln x / ln b = (1 / ln b) · ln x, so each logarithm function is a fixed multiple of the natural one. For base 2 that multiple is 1 / ln 2 ≈ 1.442695; for base 10 it is 1 / ln 10 ≈ 0.434294. “Common and Natural Logarithms”'s ln x ≈ 2.302585 · log x is the same statement with the two bases swapped.
log_8(32) is not a decimal problem. Both numbers are powers of 2, so change to base 2: log_8(32) = log_2(32) / log_2(8) = 5/3. Check it against the definition: 8^(5/3) = (2^3)^(5/3) = 2^5 = 32. Changing base does not force an approximation; choosing a helper base in which both numbers are recognisable powers avoids one.
Now the trap, on numbers. Take log_5(40). The identity gives log 40 / log 5 = 1.602060 / 0.698970 ≈ 2.292030. The two expressions it is confused with are log(40/5) = log 8 ≈ 0.903090 and log 40 − log 5, which is the same log 8. Neither is close to the answer, and neither is a law.
There is one more reason to know the identity even on a calculator that accepts any base. It is what proves that two people computing the same logarithm in different bases must agree, and in “Solving Exponential Equations” it is what turns x = log_5(17) from a name for the answer into a number.
Interactive
Derived, then checked through three different bases
Method
Four steps, and the first one often ends the problem with an exact fraction.
Write x and b as powers of one common base. log_8(32) becomes log_2(2^5) / log_2(2^3) = 5/3, with no decimal anywhere.
On a calculator that means 10 or e; either is fine and both give the same value. By hand it means whichever base makes the two numbers recognisable.
log_b(x) = log_c(x) / log_c(b). Say it in that order before writing it. The reversed quotient is a real number and the calculator will not object, so nothing downstream will look wrong.
Find k with b^k < x < b^(k+1); the logarithm lies between k and k + 1. This catches a reversed quotient immediately, since the reciprocal is almost never in the same interval.
40 is not a power of 5, so no exact answer exists and a decimal is the honest output. Three helper bases must still produce the same one.
The three quotients agree to every digit shown, which is what the theorem promises: the helper base cancels out of the answer. All six logarithms on those three lines are irrational, and every decimal here is rounded to six places.
The last two lines are not near-misses, they are answers to different questions. log_40(5) is the reciprocal — the exponent that turns 40 into 5 — and log 8 is the logarithm of a quotient rather than a quotient of logarithms. The bracket at the top rejects both without any arithmetic: neither lies between 2 and 3.
For a faded change of base, take log_9(50). The supplied first move is the bracket 9^1 < 50 < 9², so the result must lie between 1 and 2. Complete ln 50 / ___, round only the final quotient, and use the bracket to reject the reversed fraction. Then state why changing the helper from ln to log cannot change the answer.
The proof is the protection. If you can get there from b^y = x and the power law, you cannot put log b on top by accident, because the derivation shows where the denominator came from. Memorising the fraction alone leaves a fifty-fifty guess.
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