The given relationship is $y = a \cdot b^x$, which is an exponential model. To fit this into a linear regression model, we apply a logarithmic transformation to both sides. Let's transform the equation step by step.

1. Start with the original equation: $y = a \cdot b^x$

2. Apply the natural logarithm ($\ln$) to both sides: $\ln(y) = \ln(a \cdot b^x)$

3. Using logarithmic properties, split the terms: $\ln(y) = \ln(a) + \ln(b^x)$

4. Further simplify using the property $\ln(b^x) = x \cdot \ln(b)$: $\ln(y) = \ln(a) + x \cdot \ln(b)$

5. Now, let $y^* = \ln(y)$ and $b_0 = \ln(a)$, and rewrite the equation: $y^* = b_0 + b_1 \cdot x$

Here, $b_1$ is the slope of the linearized model, and we see from the transformed equation that $b_1 = \ln(b)$.

Thus, the correct answer is:

$\ln(b)$

Do you want more details or have any other questions?

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Here are five related questions:

1. What is the role of $b_0$ in this transformed linear model?
2. How can we determine $a$ after performing the logarithmic transformation?
3. What does the slope $b_1 = \ln(b)$ represent in the context of exponential growth?
4. How would the transformation change if the base of the exponent were different from $e$?
5. How does this transformation affect the interpretation of residuals in linear regression?

Tip: Always remember that when fitting models, transformations like logarithms help linearize non-linear relationships, making it easier to apply linear regression techniques.