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

Options

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

Answer

To solve the problem, we need to find the value of \(1 + abc\) where \(a = \log_{24} 12\), \(b = \log_{36} 24\), and \(c = \log_{48} 36\).

### Step-by-Step Solution:

1. **Convert the logarithms to a common base:**
We can express \(a\), \(b\), and \(c\) using the change of base formula:
\[ a = \frac{\log 12}{\log 24}, \quad b = \frac{\log 24}{\log 36}, \quad c = \frac{\log 36}{\log 48} \]

2. **Calculate \(abc\):**
Now, we multiply \(a\), \(b\), and \(c\):
\[ abc = \left(\frac{\log 12}{\log 24}\right) \left(\frac{\log 24}{\log 36}\right) \left(\frac{\log 36}{\log 48}\right) \]
Notice that the \(\log 24\) in the numerator of \(a\) cancels with the \(\log 24\) in the denominator of \(b\), and the \(\log 36\) in the numerator of \(b\) cancels with the \(\log 36\) in the denominator of \(c\):
\[ abc = \frac{\log 12}{\log 48} \]

3. **Find \(1 + abc\):**
Now we can find \(1 + abc\):
\[ 1 + abc = 1 + \frac{\log 12}{\log 48} \]
To combine these, we can express \(1\) as \(\frac{\log 48}{\log 48}\):
\[ 1 + abc = \frac{\log 48}{\log 48} + \frac{\log 12}{\log 48} = \frac{\log 48 + \log 12}{\log 48} \]

4. **Use properties of logarithms:**
Using the property of logarithms that states \(\log a + \log b = \log(ab)\):
\[ \log 48 + \log 12 = \log(48 \times 12) = \log 576 \]
Therefore:
\[ 1 + abc = \frac{\log 576}{\log 48} \]

5. **Simplify further:**
We can express \(576\) as \(24^2\):
\[ 1 + abc = \frac{\log(24^2)}{\log 48} = \frac{2 \log 24}{\log 48} \]

6. **Relate back to \(a\) and \(c\):**
We know:
\[ a = \frac{\log 12}{\log 24} \quad \text{and} \quad c = \frac{\log 36}{\log 48} \]
We can express \( \frac{\log 24}{\log 48} \) in terms of \(a\) and \(c\):
\[ \frac{\log 24}{\log 48} = 1 - c \]
Thus:
\[ 1 + abc = 2a \]

### Final Answer:
The value of \(1 + abc\) is \(2a\).