Change of Base Formula for Logarithms

Every scientific or graphing calculator can evaluate \(\log_{10} x\) (usually just labeled "log") and \(\ln x\) (natural log, base \(e\)). But what happens when you need to evaluate something like \(\log_2 20\) or \(\log_5 7\)? There is no button for those bases. This is exactly the problem the change of base formula solves.

The change of base formula lets you rewrite a logarithm in one base as a ratio of two logarithms in a different, more convenient base. If you already know what a logarithm is, you can think of this formula as a bridge between the base you're given and the base your calculator understands.

For any positive numbers \(x\) and \(b\) (with \(b \ne 1\)), and any valid new base \(c\) (with \(c > 0\) and \(c \ne 1\)):

\(\log_b x = \dfrac{\log_c x}{\log_c b}\)

In words: to evaluate a log in base \(b\), pick a new base \(c\) you can actually compute, take the log of \(x\) in that new base, and divide by the log of \(b\) in that same new base.

In practice, \(c\) is almost always chosen to be 10 or \(e\), since those are the two bases every calculator supports:

\(\log_b x = \dfrac{\log x}{\log b} = \dfrac{\ln x}{\ln b}\)

The proof follows directly from the definition of a logarithm. Start by letting \(y = \log_b x\). By the definition of a logarithm, this means:

\(b^y = x\)

If you want to review how logarithms and exponents relate, see converting a logarithm to exponential form. Now take \(\log_c\) of both sides of \(b^y = x\):

\(\log_c(b^y) = \log_c x\)

The power rule of logarithms lets you bring the exponent \(y\) down in front:

\(y \cdot \log_c b = \log_c x\)

Solving for \(y\) by dividing both sides by \(\log_c b\):

\(y = \dfrac{\log_c x}{\log_c b}\)

Since \(y = \log_b x\), substituting back gives exactly the change of base formula:

\(\log_b x = \dfrac{\log_c x}{\log_c b}\)

Evaluate \(\log_2 20\).

Since a calculator does not have a base-2 log button, rewrite the expression using base 10:

\(\log_2 20 = \dfrac{\log 20}{\log 2}\)

Using a calculator, \(\log 20 \approx 1.301\) and \(\log 2 \approx 0.301\), so:

\(\log_2 20 \approx \dfrac{1.301}{0.301} \approx 4.32\)

You can check this is reasonable because \(2^4 = 16\) and \(2^5 = 32\), and 20 sits between them, so an answer close to 4.32 makes sense.

Evaluate \(\log_5 7\) using natural log this time, to show the base you choose does not change the final answer.

\(\log_5 7 = \dfrac{\ln 7}{\ln 5}\)

\(\ln 7 \approx 1.9459\) and \(\ln 5 \approx 1.6094\), so:

\(\log_5 7 \approx \dfrac{1.9459}{1.6094} \approx 1.21\)

Notice this method only involves a calculator, unlike simpler cases you may have tackled when evaluating logarithms without a calculator, where the base and argument are exact powers of each other.

The change of base formula is also how graphing calculators and software actually plot logarithmic functions with unusual bases. For example, \(y = \log_2 x\) is graphed by computing \(\dfrac{\log x}{\log 2}\) at every point:

Notice the graph passes through \((1, 0)\), since \(\log_2 1 = 0\), and through \((8, 3)\), since \(2^3 = 8\).

A few errors show up often when students first use this formula:

Do not divide \(x\) and \(b\) themselves. The formula uses \(\log_c x\) divided by \(\log_c b\), not \(x \div b\).

Do not mix up numerator and denominator. The argument \(x\) always goes on top, and the original base \(b\) always goes on the bottom.

Any valid new base \(c\) works, but 10 and \(e\) are used almost exclusively because that's what calculators provide.

Once this skill feels solid, it becomes a key tool for solving logarithmic equations that involve unusual bases.

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