We are given the equation:

$y^p = qe^{(x + 3)}$

Our goal is to manipulate it into the form $Y = mX + c$, which is the equation of a straight line.

Steps:

1. Take the natural logarithm (ln) on both sides:

   Since $e^{(x + 3)}$ has a base of $e$, taking the natural log will help simplify the expression. So, apply $\ln$ to both sides:

   $\ln(y^p) = \ln\left(qe^{(x + 3)}\right)$

2. Simplify both sides:

   On the left-hand side, using the logarithmic property $\ln(a^b) = b\ln(a)$, we get:

   $p \ln(y) = \ln(q) + (x + 3)$

3. Rewrite to isolate terms:

   Expand the right side:

   $p \ln(y) = \ln(q) + x + 3$

   Now, divide both sides by $p$ to isolate $\ln(y)$:

   $\ln(y) = \frac{\ln(q)}{p} + \frac{x}{p} + \frac{3}{p}$

4. Rearrange into a straight-line form $Y = mX + c$:

   Compare with the standard straight-line equation $Y = mX + c$, where $Y$ is the dependent variable, $X$ is the independent variable, $m$ is the slope, and $c$ is the y-intercept.

   Let:

   $Y = \ln(y), \quad X = x, \quad m = \frac{1}{p}, \quad \text{and} \quad c = \frac{\ln(q)}{p} + \frac{3}{p}$

Thus, the equation becomes:

$\ln(y) = \frac{1}{p}x + \left(\frac{\ln(q)}{p} + \frac{3}{p}\right)$

Which is in the form $Y = mX + c$ with the following:

• Slope $m = \frac{1}{p}$
• Intercept $c = \frac{\ln(q)}{p} + \frac{3}{p}$

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Would you like further clarification on any of the steps or have additional questions? Here are a few related questions to explore further:

1. How does the logarithmic transformation affect the slope of the line?
2. What happens if $p$ is a negative value?
3. Can this equation be used to solve for $y$ directly in terms of $x$?
4. How can we graph this equation?
5. What is the significance of $q$ in the equation after the transformation?

Tip: Whenever dealing with exponential or logarithmic equations, consider using logarithms to linearize them for easier interpretation and graphing.