If `a = log_(2)3, b = log_(2) 5` and `c = log_(7)2`, then `log_(140) 63` in terms of a, b, c is

Options

A. `(2ac + 1)/(2c +bc + 1)`
B. `(2ac + 1)/(2a + c + a)`
C. `(2ac + 1)/(2c + ab + a)`
D. None of these

Answer

To solve the problem of finding \( \log_{140} 63 \) in terms of \( a = \log_{2} 3 \), \( b = \log_{2} 5 \), and \( c = \log_{7} 2 \), we can follow these steps:

### Step 1: Use the Change of Base Formula
We start with the change of base formula for logarithms:
\[ \log_{140} 63 = \frac{\log_{2} 63}{\log_{2} 140} \]

### Step 2: Factor 63 and 140
Next, we factor 63 and 140:
- \( 63 = 3^2 \times 7 \)
- \( 140 = 2^2 \times 5 \times 7 \)

### Step 3: Apply Logarithm Properties
Using the properties of logarithms, we can express the logarithms of the products:
\[ \log_{2} 63 = \log_{2} (3^2 \times 7) = \log_{2} 3^2 + \log_{2} 7 = 2 \log_{2} 3 + \log_{2} 7 \]
\[ \log_{2} 140 = \log_{2} (2^2 \times 5 \times 7) = \log_{2} 2^2 + \log_{2} 5 + \log_{2} 7 = 2 + \log_{2} 5 + \log_{2} 7 \]

### Step 4: Substitute Known Values
Now, we substitute \( a \), \( b \), and \( c \):
- \( \log_{2} 3 = a \)
- \( \log_{2} 5 = b \)
- To express \( \log_{2} 7 \), we can use the change of base formula:
\[ \log_{2} 7 = \frac{1}{\log_{7} 2} = \frac{1}{c} \]

Substituting these into our expressions:
\[ \log_{2} 63 = 2a + \frac{1}{c} \]
\[ \log_{2} 140 = 2 + b + \frac{1}{c} \]

### Step 5: Substitute Back into the Change of Base Formula
Now we substitute these back into the change of base formula:
\[ \log_{140} 63 = \frac{2a + \frac{1}{c}}{2 + b + \frac{1}{c}} \]

### Step 6: Simplify the Expression
To simplify, we can multiply the numerator and denominator by \( c \):
\[ \log_{140} 63 = \frac{2ac + 1}{2c + bc + 1} \]

### Final Answer
Thus, we have:
\[ \log_{140} 63 = \frac{2ac + 1}{bc + 2c + 1} \]