If `log_(7)12=a` `,log_(12)24=b`, then find value of `log_(54)168` in terms of a and b.

Answer

To find the value of \( \log_{54} 168 \) in terms of \( a \) and \( b \), where \( \log_{7} 12 = a \) and \( \log_{12} 24 = b \), we can follow these steps:

### Step 1: Express \( \log_{54} 168 \) using the change of base formula
Using the change of base formula, we can express \( \log_{54} 168 \) as:
\[ \log_{54} 168 = \frac{\log_{7} 168}{\log_{7} 54} \]

### Step 2: Simplify \( \log_{7} 168 \)
We can break down \( 168 \) as follows:
\[ 168 = 7 \times 24 \]
Thus,
\[ \log_{7} 168 = \log_{7} (7 \times 24) = \log_{7} 7 + \log_{7} 24 = 1 + \log_{7} 24 \]

### Step 3: Express \( \log_{7} 24 \) in terms of \( a \) and \( b \)
Using the property of logarithms, we can express \( \log_{7} 24 \) as follows:
\[ \log_{7} 24 = \frac{\log_{12} 24}{\log_{12} 7} \]
From the problem, we know that \( \log_{12} 24 = b \). Now we need to find \( \log_{12} 7 \).

Using the property of logarithms:
\[ \log_{12} 7 = \frac{1}{\log_{7} 12} = \frac{1}{a} \]
Thus,
\[ \log_{7} 24 = \frac{b}{\frac{1}{a}} = ab \]

### Step 4: Substitute back into \( \log_{7} 168 \)
Now substituting \( \log_{7} 24 \) back into the equation for \( \log_{7} 168 \):
\[ \log_{7} 168 = 1 + ab \]

### Step 5: Simplify \( \log_{7} 54 \)
Next, we need to express \( \log_{7} 54 \). We can break down \( 54 \) as follows:
\[ 54 = 2 \times 27 = 2 \times 3^3 \]
Thus,
\[ \log_{7} 54 = \log_{7} (2 \times 27) = \log_{7} 2 + \log_{7} 27 = \log_{7} 2 + 3 \log_{7} 3 \]

### Step 6: Express \( \log_{7} 2 \) and \( \log_{7} 3 \) in terms of \( a \) and \( b \)
To express \( \log_{7} 2 \) and \( \log_{7} 3 \), we can use the relationships we have:
- \( \log_{12} 2 = \log_{12} 12 - \log_{12} 6 = 1 - \log_{12} 6 \)
- \( \log_{12} 3 = \frac{1}{\log_{3} 12} = \frac{1}{\frac{1}{\log_{12} 3}} \)

However, we can also express \( \log_{7} 2 \) and \( \log_{7} 3 \) using the known values of \( a \) and \( b \).

### Step 7: Combine the results
Finally, we can combine the results to express \( \log_{54} 168 \):
\[ \log_{54} 168 = \frac{1 + ab}{\log_{7} 54} \]

### Final Result
Thus, the value of \( \log_{54} 168 \) in terms of \( a \) and \( b \) is:
\[ \log_{54} 168 = \frac{1 + ab}{\log_{7} 54} \]