To solve the equation \( 6 + 3e^{x+1} = 21 \), follow these steps:

1. Subtract 6 from both sides of the equation:
   \[
   3e^{x+1} = 21 - 6 = 15
   \]

2. Divide both sides by 3:
   \[
   e^{x+1} = \frac{15}{3} = 5
   \]

3. Take the natural logarithm (ln) of both sides:
   \[
   \ln(e^{x+1}) = \ln(5)
   \]

   Since \( \ln(e^{y}) = y \), this simplifies to:
   \[
   x + 1 = \ln(5)
   \]

4. Solve for \( x \):
   \[
   x = \ln(5) - 1
   \]

5. Using a calculator, we get:
   \[
   x = \ln(5) - 1 \approx 1.6094 - 1 = 0.6094
   \]

So, the solution to the equation, rounded to four decimal places, is \( x \approx 0.6094 \).

Would you like more details or have any questions?

Here are 5 related questions for further practice:
1. Solve \( 4 + 2e^{x-2} = 10 \).
2. Find \( x \) if \( e^{2x+1} = 12 \).
3. Solve \( 3e^{x+1} - 7 = 8 \).
4. What is the solution to \( e^{x} + 5 = 20 \)?
5. Solve \( 2 + e^{3x} = 18 \).

**Tip:** When dealing with exponential equations, taking the natural logarithm (ln) is often the key to isolating \( x \).

6 + 3e^(x+1) = 21. round to 4 decimal place

Learn how to solve the exponential equation 6 + 3e^(x+1) = 21, with a step-by-step solution using natural logarithms. This problem covers key mathematical concepts like logarithms and exponential equations, suitable for students in Grades 9-12. The final answer is rounded to 4 decimal places.

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