$$\boxed{\log_b a = \dfrac{\log_k a}{\log_k b}}$$

The change of base formula for logarithms states that any $\log_b a$ can be expressed as $\dfrac{\log_k a}{\log_k b}$ where $k$ is any positive number.

This formula is often used when calculating logarithms with the base that is inconvenient.

**What is a logarithm?**

A logarithm $\log_b a$ asks the question, ”To what power must $b$ be raised to obtain $a$?” The number $a$ and the base $b$ both could be some arbitrary numbers. The change of base formula allows us to convert such logarithms into logarithms with some commonly used base, often base 10 (common logarithm), base \( e \) (natural logarithm), or whatever base is needed for the problem.

**The Change of Base Formula:**

\[\begin{align*} \log_b a = \frac{\log_k a}{\log_k b} \end{align*}\]

where \( k \) can be any positive number.

Examples of the Change of Base Formula

**Example 1:**

Find the value of \( \log_2 34 \cdot \log_{34} 4 \) using the change of base formula.

**Solution:**

We can use the change of base formula for both logarithms. Rewrite each logarithm in terms of base \(10\):

\[\begin{align*} \log_2 34 \cdot \log_{34} 4 = \frac{\log_{10} 34}{\log_{10} 2} \cdot \frac{\log_{10} 4}{\log_{10} 34} \end{align*}\]

Now simplify each term individually. Notice that we can cancel out \(\log_{10} 34\) in the numerator and denominator:

\[\begin{align*} = \frac{\cancel{\log_{10} 34}}{\log_{10} 2} \cdot \frac{\log_{10} 4}{\cancel{\log_{10} 34}} \end{align*}\]

This leaves us with:

\[\begin{align*} = \frac{\log_{10} 4}{\log_{10} 2} \end{align*}\]

Since \(2^2 = 4\), we conclude:

\[\begin{align*} \log_2 34 \cdot \log_{32} 4 =\frac{\log_{10} 2^2}{\log_{10} 2} =\frac{2\log_{10} 2}{\log_{10} 2} = \boxed{2} \end{align*}\]

**Example 2:**

Given that \( \ln 5 \approx 1.6 \) and \( \ln 7 \approx 1.9 \), express \( \log_7 25 \) in terms of natural logarithms and approximate its value.

**Solution:**

We start by using the change of base formula to express \( \log_7 25 \) in terms of natural logarithms. According to the change of base formula:

\[\begin{align*} \log_7 25 = \frac{\ln 25}{\ln 7} \end{align*}\]

Next, we express \( \ln 25 \) in terms of \( \ln 5 \), since \( 25 = 5^2 \):

\[\begin{align*} \ln 25 = \ln(5^2) = 2 \ln 5 \end{align*}\]

Substitute \( \ln 25 = 2 \ln 5 \) into the expression for \( \log_7 25 \):

\[\begin{align*} \log_7 25 = \frac{2 \ln 5}{\ln 7} \end{align*}\]

Now we use the given values \( \ln 5 \approx 1.6 \) and \( \ln 7 \approx 1.9 \):

\[\begin{align*} \log_7 25 \approx \frac{2 \times 1.6}{1.9} = \frac{3.2}{1.9} \approx 1.684 \end{align*}\]

Thus, \( \log_7 25 \approx \boxed{1.684} \).