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The given equation is \( y = ae^{bx} \), and the task is to rewrite this equation in a linear form \( Y = A + BX \). We need to determine the variables \( X \), \( Y \), \( A \), and \( B \) accordingly.

### Step-by-step Solution:
1. Start with the equation:
   \[
   y = ae^{bx}
   \]
2. Take the natural logarithm of both sides to linearize the equation:
   \[
   \ln(y) = \ln(ae^{bx})
   \]
3. Use logarithmic properties:
   \[
   \ln(y) = \ln(a) + \ln(e^{bx})
   \]
   \[
   \ln(y) = \ln(a) + bx
   \]
   
Now, the equation is in the form of:
   \[
   \ln(y) = \ln(a) + bx
   \]
   
4. Comparing this with the linear form \( Y = A + BX \), we can identify:
   - \( Y = \ln(y) \)
   - \( A = \ln(a) \)
   - \( B = b \)
   - \( X = x \)

### Conclusion:
This matches the option **(A)**:
\[
Y = \log y, A = \log a, B = b, X = x
\]

Would you like any further clarification or details on this? Here are five related questions you might explore:

1. How does taking the logarithm help linearize exponential equations?
2. Can logarithms be used to linearize other types of functions, such as polynomials?
3. How would you interpret the slope and intercept in the context of this linearized equation?
4. What happens if you take the logarithm of both \( y \) and \( x \) in the equation \( y = ae^{bx} \)?
5. How would you plot \( y = ae^{bx} \) in its original and linearized forms?

**Tip:** Always check if an equation can be simplified by taking logarithms, especially when dealing with exponentials!

If the equation y = ae^(bx) can be written in linear form Y = A + BX, what are X, Y, A and B?

Learn how to linearize the exponential equation y = ae^(bx) by applying logarithms. Discover the transformation into linear form, Y = A + BX, where Y = ln(y), A = ln(a), B = b, and X = x. Perfect for students in Grades 10-12.

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