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We are given the equation \( 7^y = 3^x \) and need to show that \( y = kx \) for some constant \( k \).

### Steps to solve:

1. **Take the natural logarithm of both sides**:
   \[
   \ln(7^y) = \ln(3^x)
   \]
   
2. **Apply the logarithmic property** \( \ln(a^b) = b \ln(a) \):
   \[
   y \ln(7) = x \ln(3)
   \]

3. **Solve for \( y \)**:
   \[
   y = \frac{\ln(3)}{\ln(7)} x
   \]
   
4. **Identify the constant \( k \)**:
   The constant \( k \) is given by the ratio \( \frac{\ln(3)}{\ln(7)} \). Hence,
   \[
   y = kx \quad \text{where} \quad k = \frac{\ln(3)}{\ln(7)}
   \]

This shows that \( y = kx \), where \( k \) is the constant ratio of the natural logarithms of 3 and 7.

Would you like to see more details or have further questions?

### Related Questions:
1. What are the properties of logarithms used in solving exponential equations?
2. How do you simplify logarithmic expressions involving different bases?
3. Can \( k \) be expressed in terms of common logarithms instead of natural logarithms?
4. What other methods could we use to solve equations involving powers of different bases?
5. How do you graph the equation \( 7^y = 3^x \)?

### Tip:
When dealing with exponential equations involving different bases, logarithms are often the key to simplifying and solving the problem efficiently.

If 7^y = 3^x, show that y = kx for some constant k which is to be determined.

Learn how to solve the exponential equation 7^y = 3^x and prove that y = kx. The solution covers key logarithmic properties and methods for solving exponential equations. Suitable for students in Grades 10-12.

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