Prove that `(log_(a)n)/(log_(ab)n)=1+log_(a)b`

Answer

To prove that

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

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

### Step 1: Rewrite the logarithm in terms of natural logarithms
Using the change of base formula, we can express the logarithms in terms of natural logarithms (base \(e\)):

\[ \log_a n = \frac{\log_e n}{\log_e a} \]
and
\[ \log_{ab} n = \frac{\log_e n}{\log_e (ab)}. \]

### Step 2: Substitute into the left-hand side
Now, substituting these expressions into the left-hand side, we get:

\[ \frac{\log_a n}{\log_{ab} n} = \frac{\frac{\log_e n}{\log_e a}}{\frac{\log_e n}{\log_e (ab)}}. \]

### Step 3: Simplify the fraction
This simplifies to:

\[ \frac{\log_e n}{\log_e a} \cdot \frac{\log_e (ab)}{\log_e n} = \frac{\log_e (ab)}{\log_e a}. \]

### Step 4: Expand \(\log_e (ab)\)
Using the property of logarithms that states \(\log_e (xy) = \log_e x + \log_e y\), we can expand \(\log_e (ab)\):

\[ \log_e (ab) = \log_e a + \log_e b. \]

### Step 5: Substitute back into the expression
Now substituting this back into our expression gives:

\[ \frac{\log_e (ab)}{\log_e a} = \frac{\log_e a + \log_e b}{\log_e a}. \]

### Step 6: Split the fraction
This can be split into two separate fractions:

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

### Step 7: Rewrite \(\frac{\log_e b}{\log_e a}\)
Using the change of base formula again, we can rewrite \(\frac{\log_e b}{\log_e a}\) as \(\log_a b\):

\[ 1 + \log_a b. \]

### Conclusion
Thus, we have shown that:

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

This completes the proof.

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