We are given the equation:

\[
Be^{8x} + Cxe^{8x} = 0
\]

To solve this equation, follow these steps:

1. Factor out the common term \( e^{8x} \):

\[
e^{8x}(B + Cx) = 0
\]

2. Since \( e^{8x} \neq 0 \) for any value of \(x\) (because the exponential function is never zero), we focus on the factor \( B + Cx \):

\[
B + Cx = 0
\]

3. Solve for \(x\):

\[
x = -\frac{B}{C}
\]

Thus, the solution to the equation is:

\[
x = -\frac{B}{C}
\]

Let me know if you'd like further details or have any questions.

Here are 5 related questions to explore further:
1. How would the solution change if the equation involved \( e^{kx} \) instead of \( e^{8x} \)?
2. What happens if both \( B \) and \( C \) are zero?
3. Can you find the solution for specific values of \( B \) and \( C \)?
4. How would you solve a similar equation involving logarithms instead of exponentials?
5. What if the equation had a higher-order term like \( x^2e^{8x} \)?

**Tip:** Exponential equations often allow factoring, which simplifies solving. Always check if a common term can be factored out.

Discover how to solve the exponential equation (Be^(8x)) + (Cxe^(8x)) = 0 using factoring techniques. This solution involves understanding key algebraic concepts, such as factoring exponential terms and solving linear equations, suitable for students in Grades 10-12.

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