The Laws of Logarithms

Why the laws exist

Every law of logarithms is a law of exponents wearing a different hat. Recall the product rule of exponents: b^m · b^n = b^(m+n) — to multiply powers of the same base you *add* the exponents. Since a logarithm *is* an exponent, this tells us that the log of a product should equal the *sum* of the logs. Multiplication on one side becomes addition on the other. That trade — turning a hard operation into an easy one — is the entire reason logarithms were invented.

Proof of the product law (it's just exponents):

Let  M = b^x   so  log_b(M) = x
Let  N = b^y   so  log_b(N) = y

Then  M·N = b^x · b^y = b^(x+y)   <- exponent product rule
So    log_b(M·N) = x + y = log_b(M) + log_b(N).

That's the whole story: multiply inside,
add the logs outside.

The three laws

1. [[product-law-of-logarithms|Product law]]: log_b(M·N) = log_b M + log_b N. A log of a product splits into a sum of logs. (Mirror of the exponent product rule.)
2. [[quotient-law-of-logarithms|Quotient law]]: log_b(M/N) = log_b M − log_b N. A log of a quotient splits into a difference of logs. (Mirror of the exponent quotient rule.)
3. [[power-law-of-logarithms|Power law]]: log_b(M^p) = p · log_b M. A power inside a log comes out front as a multiplier. (Mirror of the exponent power rule.) This one is the workhorse for solving equations.

Using the laws to expand and simplify:

Expand  log( x^3 · y / z )
  = log(x^3) + log(y) - log(z)     (product & quotient)
  = 3·log(x) + log(y) - log(z)     (power law on x^3)

Condense  2·ln(a) - ln(b)
  = ln(a^2) - ln(b)                (power law)
  = ln( a^2 / b )                  (quotient law)

Note what is NOT allowed:
  log(M + N) does NOT equal log M + log N.
  The laws only touch products, quotients, powers.

Change of base

Your calculator has only two log buttons: log (base 10) and ln (base e). So how do you find log₂ 50? The change-of-base formula rewrites a log in any base using logs in a base you do have: log_b(x) = log_c(x) / log_c(b), where c is any convenient base — usually 10 or e. Pick either; the answer is the same.

Evaluate  log_2(50)  using base-10 logs:

  log_2(50) = log(50) / log(2)
            = 1.69897 / 0.30103
            = 5.6439...

Check:  2^5.6439 ≈ 50.  Good.

With natural logs you'd get the same:
  log_2(50) = ln(50) / ln(2)
            = 3.91202 / 0.69315
            = 5.6439...