Given that `log_2(3)=a,log_3(5)=b,log_7(2)=c`, express the logrithm of the number 63 to the base 140 in terms of a,b & c.

Answer

To solve the problem of expressing \(\log_{140}(63)\) in terms of \(a\), \(b\), and \(c\), we will follow a systematic approach using logarithmic properties.

### Step-by-Step Solution:

1. **Identify the given logarithmic values**:
\[ a = \log_2(3), \quad b = \log_3(5), \quad c = \log_7(2) \]

2. **Express \(\log_{140}(63)\)** using the change of base formula:
\[ \log_{140}(63) = \frac{\log(63)}{\log(140)} \]

3. **Factor the numbers**:
- \(63 = 7 \times 3^2\)
- \(140 = 7 \times 5 \times 2^2\)

4. **Use logarithmic properties** to expand \(\log(63)\) and \(\log(140)\):
\[ \log(63) = \log(7) + 2\log(3) \]
\[ \log(140) = \log(7) + \log(5) + 2\log(2) \]

5. **Substitute these into the expression for \(\log_{140}(63)\)**:
\[ \log_{140}(63) = \frac{\log(7) + 2\log(3)}{\log(7) + \log(5) + 2\log(2)} \]

6. **Divide the numerator and denominator by \(\log(7)\)**:
\[ \log_{140}(63) = \frac{1 + 2\frac{\log(3)}{\log(7)}}{1 + \frac{\log(5)}{\log(7)} + 2\frac{\log(2)}{\log(7)}} \]

7. **Express the fractions in terms of \(a\), \(b\), and \(c\)**:
- From the definitions:
- \(\frac{\log(3)}{\log(7)} = \log_7(3) = a \cdot c\)
- \(\frac{\log(5)}{\log(7)} = \log_7(5) = b \cdot c\)
- \(\frac{\log(2)}{\log(7)} = \log_7(2) = c\)

8. **Substituting these values**:
\[ \log_{140}(63) = \frac{1 + 2ac}{1 + bc + 2c} \]

### Final Answer:
Thus, we can express \(\log_{140}(63)\) in terms of \(a\), \(b\), and \(c\) as:
\[ \log_{140}(63) = \frac{1 + 2ac}{1 + bc + 2c} \]