We are given the equation:

\[
5e^{0.2x} = 3
\]

### Step 1: Isolate the exponential term
To isolate \( e^{0.2x} \), divide both sides of the equation by 5:

\[
e^{0.2x} = \frac{3}{5}
\]

### Step 2: Take the natural logarithm
To eliminate the exponential, take the natural logarithm (ln) of both sides:

\[
\ln(e^{0.2x}) = \ln\left(\frac{3}{5}\right)
\]

Using the logarithmic property \( \ln(e^a) = a \), we simplify:

\[
0.2x = \ln\left(\frac{3}{5}\right)
\]

### Step 3: Solve for \( x \)
Now, divide by 0.2 to isolate \( x \):

\[
x = \frac{\ln\left(\frac{3}{5}\right)}{0.2}
\]

### Step 4: Simplify
Using a calculator, we find:

\[
\ln\left(\frac{3}{5}\right) \approx -0.5108
\]

Thus:

\[
x \approx \frac{-0.5108}{0.2} = -2.554
\]

### Final Answer:
\[
x \approx -2.554
\]

Would you like more details or have any further questions?

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Here are some related questions for practice:

1. How do you solve \( 4e^{0.3x} = 7 \)?
2. What is the value of \( \ln\left(\frac{4}{7}\right) \)?
3. How can logarithmic properties simplify solving \( e^{kx} = a \)?
4. What happens if the base of the exponential changes?
5. How does solving exponential equations differ from solving quadratic equations?

**Tip**: Always isolate the exponential term before applying the logarithm to simplify the solving process.

Learn how to solve the exponential equation 5e^0.2x = 3 with a step-by-step solution. This problem involves using logarithms and isolating exponential terms, ideal for students in Grades 10-12.

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