The task involves solving equations using the zero-product property, which states that if a product of factors equals zero, then at least one of the factors must be zero. I'll go through each part individually:

Problem 9: Solve each equation using the zero-product property

(a) $x \cdot e^x = 0$

Since this is a product of $x$ and $e^x$, set each factor equal to zero:

1. $x = 0$
2. $e^x = 0$: This is impossible because the exponential function $e^x$ is never zero.

Solution: $x = 0$

(b) $e^{3x} (x^2 - 9) = 0$

1. $e^{3x} = 0$: Again, this is impossible because $e^{3x}$ is never zero.

2. $x^2 - 9 = 0$: Solve this by factoring as $(x - 3)(x + 3) = 0$.

   Solutions: $x = 3$ and $x = -3$

Solution: $x = 3$ or $x = -3$

(c) $e^{-2x} (x^3 + 1) = 0$

1. $e^{-2x} = 0$: Impossible, as $e^{-2x}$ is never zero.

2. $x^3 + 1 = 0$: Factor as $(x + 1)(x^2 - x + 1) = 0$.

   Solution for $x + 1 = 0$: $x = -1$

   The quadratic $x^2 - x + 1 = 0$ has no real solutions.

Solution: $x = -1$

(d) $-2(e^{2x} - 6)x^3 = 0$

1. $-2 = 0$: This is irrelevant for solving.
2. $e^{2x} - 6 = 0$: Solve for $x$: $e^{2x} = 6 \Rightarrow 2x = \ln 6 \Rightarrow x = \frac{\ln 6}{2}$
3. $x^3 = 0$: Solution is $x = 0$.

Solution: $x = 0$ or $x = \frac{\ln 6}{2}$

(e) $x^2 (5 - e^{3x}) = 0$

1. $x^2 = 0$: Solution is $x = 0$.
2. $5 - e^{3x} = 0$: Solve for $x$: $e^{3x} = 5 \Rightarrow 3x = \ln 5 \Rightarrow x = \frac{\ln 5}{3}$

Solution: $x = 0$ or $x = \frac{\ln 5}{3}$

(f) $x e^x - e^x = 0$

Factor out $e^x$: $e^x (x - 1) = 0$

1. $e^x = 0$: Impossible.
2. $x - 1 = 0$: Solution is $x = 1$.

Solution: $x = 1$

(g) $x - x e^x = 0$

Factor out $x$: $x (1 - e^x) = 0$

1. $x = 0$: Solution is $x = 0$.
2. $1 - e^x = 0$: Solve for $x$: $e^x = 1 \Rightarrow x = 0$

Solution: $x = 0$

(h) $4e^{-3x} - x^2 e^{-3x} = 0$

Factor out $e^{-3x}$: $e^{-3x} (4 - x^2) = 0$

1. $e^{-3x} = 0$: Impossible.

2. $4 - x^2 = 0$: Factor as $(2 - x)(2 + x) = 0$.

   Solutions: $x = 2$ and $x = -2$

Solution: $x = 2$ or $x = -2$

Summary of Solutions

• (a) $x = 0$
• (b) $x = 3$ or $x = -3$
• (c) $x = -1$
• (d) $x = 0$ or $x = \frac{\ln 6}{2}$
• (e) $x = 0$ or $x = \frac{\ln 5}{3}$
• (f) $x = 1$
• (g) $x = 0$
• (h) $x = 2$ or $x = -2$

Would you like further details on any particular equation, or do you have any questions?

Additional Practice Questions

1. How can we solve exponential equations like $e^{kx} = a$ in general terms?
2. What are the implications of the zero-product property in more complex expressions?
3. How can factoring be used to simplify polynomial equations before solving?
4. Why are there no real solutions for some quadratic equations?
5. What steps should be followed when exponential terms are present in equations?

Tip

Remember, the exponential function $e^x$ never equals zero, which is a critical property when solving equations involving exponential terms.