Change of Base Formula
We might have noticed that a scientific calculator has only "log" and "ln" buttons. Also, we know that "log" stands for a logarithm of base 10 and "ln" stands for a logarithm of base e. But there is no option to calculate the logarithm of a number with any other bases other than 10 and e. The change of base formula solves this issue. Also, it is used in solving several logarithms problems. Let us learn the change of base formula along with its proof and a few solved examples.
What Is Change of Base Formula?
The change of base formula is used to write a logarithm of a number with a given base as the ratio of two logarithms each with the same base that is different from the base of the original logarithm. Here,
 The argument of the logarithm in the numerator is the same as the argument of the original logarithm.
 The argument of the logarithm in the denominator is the same as the base of the original logarithm.
 The bases of both logarithms of numerator and denominator should be the same and this base can be any positive number other than 1.
The change of base formula is,
\(\log _{b} a=\dfrac{\log _{c} a}{\log _{c} b}\)
Another form of this formula is, \(\log _{b} a \cdot \log _{c} b=\log _{c} a\), which is also widely used in solving the problems.
Proof of Change of Base Formula
Let us assume that
\(\log_b a = p\), \(\log_c a= q\), and \(\log_c b =r\).
By converting each of these into exponential form, we get,
\(a = b^p\), \(a = c^q\), \(b=c^r\).
From the first two equations,
\(b^p=c^q\)
Substituting \(b=c^r\) (which is from third equation) here,
\((c^r)^p=c^q\)
\(c^{pr} = c^q\)
\(pr = q\)
\(p = \dfrac q r\)
Substituting the values of \(p, q\), and r here,
\(\log _{b} a=\dfrac{\log _{c} a}{\log _{c} b}\)
Solved Examples Using Change of Base Formula

Example 1
Evaluate the value of \(\log_{64} 8\) using the change of base formula.
Solution:
We will apply the change of base formula (by changing the base to 10),
\( \begin{align} \log_{64} 8 &= \dfrac{\log 8}{\log 64}\\[0.2cm] &= \dfrac{\log 8}{\log 8^2}\\\\[0.2cm] &= \dfrac{\log 8}{2 \log 8}\,\,\, [\because \log a^m = m \log a]\\[0.2cm] &= \dfrac{1}{2} \end{align}\)
Answer: \(\log_{64} 8= \dfrac 1 2\)

Example 2
Evaluate the value of \(\log_3 2 \cdot \log_4 3 \cdot \log_5 4\).
Solution:
By alternate form of the change of base formula, \(\log _{b} a \cdot \log _{c} b=\log _{c} a\). We apply this twice to evaluate the given expression.
\(\begin{align} \log_3 2 \cdot \log_4 3 \cdot \log_5 4 & = (\log_3 2 \cdot \log_4 3) \, \log_5 4\\[0.2cm]
&= \log_4 2 \cdot \log_5 4\\[0.2cm]
&= \log_5 2 \end{align}\)Answer: \(\log_3 2 \cdot \log_4 3 \cdot \log_5 4 = \log_5 2\)