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Change of Base Formula

Change of Base Formula – definition and example, Algebra 2 math glossary – Mathovore

Key takeaway: The change of base formula says log_b x = log_c x ÷ log_c b, which lets a calculator with only base 10 and base e keys find a logarithm for any other base.

Grade 11 – Algebra 2F-LE.A.4Functions and Graphs

Definition of Change of Base Formula

Definition

The change of base formula exists because calculators usually offer only two logarithm keys: base 10 and base \(e\). It trades a logarithm in an awkward base for a fraction of two logarithms in a friendly base, and the value does not change.

The idea comes from exponents. If the answer to \(\log_b x\) is \(y\), then \(b^y=x\). Taking a logarithm in any new base on both sides and solving for \(y\) produces the fraction. Any new base works, as long as it is positive and not 1; base 10 and base \(e\) are simply the convenient ones.

Formula and notation

For positive numbers \(x\), \(b\), \(c\) with \(b\neq 1\) and \(c\neq 1\):

\[\log_b x=\frac{\log_c x}{\log_c b}=\frac{\ln x}{\ln b}.\]

The number being evaluated goes on top and the old base goes on the bottom.

exponent 25² = 25exponent 35³ = 12580 at 2.723log₅ 80 = ln 80 ÷ ln 5 = 4.382 ÷ 1.609 ≈ 2.723
80 lies between 5² and 5³, so its logarithm base 5 lies between 2 and 3.

Key facts

  • The old base lands in the denominator; the argument stays in the numerator.
  • Using base 10 or base e gives the same result to every decimal place.
  • Setting x = c gives the handy fact that log_b c equals 1 divided by log_c b.
  • Graphing calculators need this rewrite to plot a logarithm with an arbitrary base.

Change of Base Formula example

Problem

Estimate \(\log_5 80\) with a calculator that has only a natural-log key, rounding to thousandths.

  1. Choose base \(e\): \(\log_5 80=\frac{\ln 80}{\ln 5}\).

  2. The calculator reports about 4.38203 for the top and 1.60944 for the bottom.

  3. Dividing the two values gives roughly 2.723.

  4. Sense check: \(5^2=25\) and \(5^3=125\), and 80 sits between them.

\(\log_5 80\approx\)2.723.

Common mistake

Watch out

Callum computes \(\log_5 80\) as \(\frac{\ln 5}{\ln 80}\approx 0.367\).

Why it is wrong

He flipped the fraction. The change of base formula keeps the base on the bottom: \(\frac{\ln 80}{\ln 5}\approx 2.723\). An answer below 1 is a warning sign here, because 80 is larger than 5.

Related terms and lessons

Browse the glossary: A–Z list · Algebra 2 vocabulary · Functions and Graphs

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