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To solve the equation \( 3e^{2x} + 5 = 16e^x \) using natural logarithms, follow these steps:

### Step 1: Rewrite the equation
Start by moving all terms to one side of the equation:
\[
3e^{2x} - 16e^x + 5 = 0
\]

### Step 2: Substitute \( u = e^x \)
Let \( u = e^x \). Then \( e^{2x} = u^2 \). Substituting these into the equation gives:
\[
3u^2 - 16u + 5 = 0
\]

### Step 3: Solve the quadratic equation
We solve the quadratic equation \( 3u^2 - 16u + 5 = 0 \) using the quadratic formula:
\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3 \), \( b = -16 \), and \( c = 5 \).

Substituting these values:
\[
u = \frac{16 \pm \sqrt{256 - 60}}{6}
\]
\[
u = \frac{16 \pm \sqrt{196}}{6}
\]
\[
u = \frac{16 \pm 14}{6}
\]

This gives us two solutions:
\[
u_1 = \frac{16 + 14}{6} = \frac{30}{6} = 5
\]
\[
u_2 = \frac{16 - 14}{6} = \frac{2}{6} = \frac{1}{3}
\]

### Step 4: Solve for \( x \) using natural logarithms
Recall that \( u = e^x \), so:
\[
e^x = 5 \quad \text{or} \quad e^x = \frac{1}{3}
\]

Now, take the natural logarithm (ln) on both sides to solve for \( x \):
\[
x_1 = \ln(5)
\]
\[
x_2 = \ln\left(\frac{1}{3}\right) = -\ln(3)
\]

### Final Answer:
The solutions for \( x \) are:
\[
x_1 = \ln(5)
\]
\[
x_2 = -\ln(3)
\]

Would you like further explanation on how logarithms were used here or have any other questions?

### Relative Questions:
1. How would the solution change if we used a different base for the logarithm, like base 10?
2. Can you explain why taking the natural logarithm is effective in solving exponential equations?
3. How would the approach differ if the equation was \( 3e^{3x} + 5 = 16e^x \)?
4. What are some real-world applications where solving equations involving \( e^x \) and natural logarithms is necessary?
5. Could you solve this equation graphically, and how would the graph look?

### Tip:
When dealing with equations involving exponentials, remember that logarithms, especially natural logarithms, are powerful tools to solve for the variable in the exponent.

Learn how to solve the equation 3e^(2x) + 5 = 16e^x using natural logarithms. This problem involves exponential equations and demonstrates the application of natural logarithms to solve for the variable in the exponent. Suitable for high school students in Grades 10-12.

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