If a= `log_(24) 12, b = log_(36) 24, c = log_(48) 36 , ` then 1 + abc is equal to

Options

A. 2 ac
B. 2bc
C. 2ab
D. none of these

Answer

To solve the problem, we need to find the value of \(1 + abc\) where:

- \(a = \log_{24} 12\)
- \(b = \log_{36} 24\)
- \(c = \log_{48} 36\)

### Step 1: Express \(abc\) in terms of logarithms
Using the change of base formula, we can express each logarithm in terms of natural logarithms (or any common base):

\[ a = \frac{\log 12}{\log 24}, \quad b = \frac{\log 24}{\log 36}, \quad c = \frac{\log 36}{\log 48} \]

Now, we can multiply these together:

\[ abc = \left(\frac{\log 12}{\log 24}\right) \left(\frac{\log 24}{\log 36}\right) \left(\frac{\log 36}{\log 48}\right) \]

### Step 2: Simplify \(abc\)
Notice that the terms \(\log 24\) and \(\log 36\) cancel out:

\[ abc = \frac{\log 12}{\log 48} \]

### Step 3: Calculate \(1 + abc\)
Now we can substitute \(abc\) back into the expression for \(1 + abc\):

\[ 1 + abc = 1 + \frac{\log 12}{\log 48} \]

### Step 4: Rewrite \(1\) in terms of logarithms
We can express \(1\) as \(\frac{\log 48}{\log 48}\):

\[ 1 + abc = \frac{\log 48}{\log 48} + \frac{\log 12}{\log 48} = \frac{\log 48 + \log 12}{\log 48} \]

### Step 5: Use the property of logarithms
Using the property of logarithms that states \(\log a + \log b = \log(ab)\):

\[ 1 + abc = \frac{\log(48 \times 12)}{\log 48} \]

### Step 6: Calculate \(48 \times 12\)
Calculate \(48 \times 12\):

\[ 48 \times 12 = 576 \]

### Step 7: Final expression
Thus, we have:

\[ 1 + abc = \frac{\log 576}{\log 48} \]

### Step 8: Simplify \(\log 576\)
We can express \(576\) as \(24^2\):

\[ \log 576 = \log(24^2) = 2 \log 24 \]

### Step 9: Substitute back
Now substituting this back into our expression gives:

\[ 1 + abc = \frac{2 \log 24}{\log 48} \]

### Step 10: Express \(\log 48\) in terms of \(\log 24\)
We can express \(48\) as \(2 \times 24\):

\[ \log 48 = \log(2 \times 24) = \log 2 + \log 24 \]

### Step 11: Final expression for \(1 + abc\)
Now substituting this into our expression gives:

\[ 1 + abc = \frac{2 \log 24}{\log 2 + \log 24} \]

This is the final expression for \(1 + abc\).