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The given equation is:

\[
e^{2x - 3} = 6
\]

We need to solve for \( x \) and approximate the solution to four decimal places.

### Steps to solve:

1. Take the natural logarithm (ln) of both sides to remove the exponential:
   \[
   \ln(e^{2x - 3}) = \ln(6)
   \]

2. Simplify using the property \( \ln(e^y) = y \):
   \[
   2x - 3 = \ln(6)
   \]

3. Solve for \( x \):
   \[
   2x = \ln(6) + 3
   \]
   \[
   x = \frac{\ln(6) + 3}{2}
   \]

Now, compute the value of \( x \):

\[
\ln(6) \approx 1.7918
\]
\[
x = \frac{1.7918 + 3}{2} = \frac{4.7918}{2} = 2.3959
\]

Thus, the solution to the equation is approximately:
\[
x \approx 2.3959
\]

Would you like more details or further clarification?

Here are some related questions to explore:

1. How do you apply natural logarithms to exponential equations in general?
2. What properties of logarithms are used to simplify such expressions?
3. How can we verify the solution by substituting it back into the original equation?
4. What happens if the base of the exponential function is different (e.g., not \( e \))?
5. How can the logarithmic approximation be calculated more accurately?

**Tip:** Always remember that the natural logarithm, \( \ln(x) \), is the inverse of the exponential function \( e^x \).

To four decimal places, approximate the solution to the equation e^(2x - 3) = 6.

Learn how to approximate the solution to the equation e^(2x - 3) = 6 using natural logarithms. This step-by-step guide demonstrates how to isolate x and compute its value to four decimal places. Suitable for students in Grades 10-12.

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