Change of Base Formula

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Change of Base Formula: meaning, method and solved example (Algebra 2) – Maths PDF


In short: The change of base formula rewrites a log in any base as a quotient of two logs in a base your calculator knows: log base b of a equals ln a divided by ln b.

Algebra 2Functions and GraphsF-LE.A.4

What is Change of Base Formula?

The change of base formula exists because calculators only have two log keys, base 10 and base e. When a problem asks for something like the log of 50 in base 7, you convert it into a fraction of two keys you can press.

For \(a > 0\), \(b > 0\) and \(b \neq 1\): \(\log_b a = \frac{\ln a}{\ln b} = \frac{\log a}{\log b}\). Any new base \(c\) works, as long as the same one is used on top and bottom. The numerator is the number’s log, and the denominator is the old base’s log.

How to spot it

  • A calculator has no key for the base in the problem, so the change of base formula is needed, as with log base 7 or base 12.
  • A logarithm of a non-power must be approximated, like log_6 100, rather than written exactly.
  • Two logs with different bases must be compared or combined in one expression.
  • An exponential equation like 7^x = 50 asks for x, and the answer is a log in an unfamiliar base.

The method

  1. Write the log you need as \(\log_b a\) and identify the number \(a\) and the base \(b\).
  2. Apply the change of base formula: \(\ln a\) on the top and \(\ln b\) on the bottom (common logs work equally well).
  3. Compute both logs and divide, keeping at least four decimal places.
  4. Check by raising the base \(b\) to your answer; it should return about \(a\).
log base 7 of 50ln 50ln 73.912 / 1.946 = 2.010 (about)
Because 7^2 = 49 is just below 50, the log must be a little over 2.

Change of Base Formula: solved example

Problem

Evaluate \(\log_7 50\) to three decimal places and check the result.

Solution

Apply the formula with natural logs: \(\log_7 50 = \frac{\ln 50}{\ln 7} \approx \frac{3.91202}{1.94591} \approx 2.0104\).

Check: \(7^2 = 49\), which is just below 50, so an exponent slightly above 2 makes sense. The value is about 2.010.

Try it

Question 1

Approximate \(\log_5 200\).

Show the answer

\(\frac{\ln 200}{\ln 5} \approx \frac{5.2983}{1.6094} \approx 3.292\).

Question 2

Use base 2 to evaluate \(\log_8 32\) exactly.

Show the answer

\(\frac{\log_2 32}{\log_2 8} = \frac{5}{3}\).

Change of Base Formula vs. Logarithm Rules

The rules for products, quotients and powers keep the base the same and rearrange what is inside. This formula does the opposite: it keeps the inside number and switches the base. A mix-up shows up as writing \(\frac{\ln a}{\ln b}\) as \(\ln\frac{a}{b}\), which is a different quantity.

Logarithm Rules

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