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To solve the given equation \( e^{5x - 3} \cdot e^{-x} = 2e \), let’s simplify and solve for \( x \).

### Step 1: Simplify the Exponential Equation

The equation is:
\[
e^{5x - 3} \cdot e^{-x} = 2e
\]

Using the property of exponents \( e^a \cdot e^b = e^{a + b} \), we can combine the exponents on the left-hand side:
\[
e^{(5x - 3) + (-x)} = 2e
\]

Simplify the exponent:
\[
e^{4x - 3} = 2e
\]

### Step 2: Divide Both Sides by \( e \)

To make it easier to isolate \( x \), divide both sides by \( e \):
\[
e^{4x - 3} = 2
\]

### Step 3: Take the Natural Logarithm of Both Sides

To solve for \( x \), take the natural logarithm of both sides:
\[
4x - 3 = \ln(2)
\]

### Step 4: Solve for \( x \)

Now, add 3 to both sides:
\[
4x = \ln(2) + 3
\]

Then, divide by 4 to isolate \( x \):
\[
x = \frac{\ln(2) + 3}{4}
\]

### Step 5: Calculate the Value

Using approximate values, with \( \ln(2) \approx 0.693 \):
\[
x \approx \frac{0.693 + 3}{4} = \frac{3.693}{4} \approx 0.923
\]

### Final Answer

Rounded to the nearest thousandth, the solution is:
\[
x \approx 0.923
\]

So, the solution set is:
- **Option A**: The solution set is \(\{ 0.923 \}\).

Would you like more details on any step or have any questions?

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**Relative Questions:**

1. How do you combine exponents with the same base?
2. Why is taking the natural logarithm useful in solving exponential equations?
3. What are common properties of logarithms that simplify solving equations?
4. How do you determine if an exponential equation has no solution?
5. Why is rounding necessary in expressing irrational solutions?

**Tip:** Always check the solution by substituting it back into the original equation to ensure accuracy.

Solve the exponential equation. Express irrational solutions as decimals correct to the nearest thousandth.

e^(5x-3) * e^(-x) = 2e

Learn how to solve the exponential equation e^(5x-3) * e^(-x) = 2e by combining exponents, taking natural logarithms, and simplifying. This solution is expressed to the nearest thousandth and demonstrates the properties of exponents and logarithms. Suitable for students in Grades 11-12.

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