Show that : `log_(a)m div log_(ab)m = 1 + log_(a)b`

Answer

To show that

\[ \frac{\log_a m}{\log_{ab} m} = 1 + \log_a b, \]

we will start with the left-hand side and manipulate it step by step.

### Step 1: Rewrite the logarithms using the change of base formula.

Using the change of base formula, we can express both logarithms in terms of a common base (let's use base 10):

\[ \log_a m = \frac{\log_{10} m}{\log_{10} a} \quad \text{and} \quad \log_{ab} m = \frac{\log_{10} m}{\log_{10} (ab)}. \]

### Step 2: Substitute the expressions into the left-hand side.

Now, we can substitute these expressions into the left-hand side:

\[ \frac{\log_a m}{\log_{ab} m} = \frac{\frac{\log_{10} m}{\log_{10} a}}{\frac{\log_{10} m}{\log_{10} (ab)}}. \]

### Step 3: Simplify the fraction.

This can be simplified by multiplying by the reciprocal of the denominator:

\[ = \frac{\log_{10} m}{\log_{10} a} \cdot \frac{\log_{10} (ab)}{\log_{10} m}. \]

The \(\log_{10} m\) in the numerator and denominator cancels out:

\[ = \frac{\log_{10} (ab)}{\log_{10} a}. \]

### Step 4: Expand \(\log_{10} (ab)\).

Using the property of logarithms that states \(\log_{10} (ab) = \log_{10} a + \log_{10} b\), we can rewrite the expression:

\[ = \frac{\log_{10} a + \log_{10} b}{\log_{10} a}. \]

### Step 5: Separate the terms.

Now, we can separate the terms in the fraction:

\[ = \frac{\log_{10} a}{\log_{10} a} + \frac{\log_{10} b}{\log_{10} a} = 1 + \frac{\log_{10} b}{\log_{10} a}. \]

### Step 6: Rewrite \(\frac{\log_{10} b}{\log_{10} a}\) as \(\log_a b\).

By the change of base formula, we know that:

\[ \frac{\log_{10} b}{\log_{10} a} = \log_a b. \]

### Final Result:

Thus, we have:

\[ \frac{\log_a m}{\log_{ab} m} = 1 + \log_a b. \]

This shows that the left-hand side equals the right-hand side, confirming the identity.