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To solve the equation \( 3e^{2x} + 5 = 16e^x \) for \( x \), we can proceed with the following steps:

### Step 1: Rewrite the equation
We start by rewriting the equation in a standard quadratic form:
\[
3e^{2x} - 16e^x + 5 = 0
\]

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

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

Substituting these values into the formula:
\[
u = \frac{16 \pm \sqrt{(-16)^2 - 4(3)(5)}}{2(3)}
\]
\[
u = \frac{16 \pm \sqrt{256 - 60}}{6}
\]
\[
u = \frac{16 \pm \sqrt{196}}{6}
\]
\[
u = \frac{16 \pm 14}{6}
\]

### Step 4: Simplify the roots
This gives us two potential solutions for \( u \):
\[
u_1 = \frac{16 + 14}{6} = \frac{30}{6} = 5
\]
\[
u_2 = \frac{16 - 14}{6} = \frac{2}{6} = \frac{1}{3}
\]

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

Taking the natural logarithm on both sides:
\[
x_1 = \ln(5)
\]
\[
x_2 = \ln\left(\frac{1}{3}\right) = -\ln(3)
\]

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

Would you like a more detailed explanation or do you have any questions?

### Relative Questions:
1. How would you solve the equation if the constant on the right side was different, such as \( 3e^{2x} + 5 = 18e^x \)?
2. What is the significance of using the natural logarithm to solve exponential equations?
3. How would the solution change if the equation was \( 3e^{2x} + 5 = 16e^{3x} \)?
4. Can you explain the relationship between exponential functions and logarithms?
5. What would happen if you substituted \( u = e^{-x} \) instead of \( u = e^x \)?

### Tip:
When solving exponential equations, substituting \( u = e^x \) can transform the problem into a more manageable quadratic form, which can then be solved using familiar methods.

Learn how to solve the equation 3e^(2x) + 5 = 16e^x with a detailed step-by-step solution. This problem involves exponential functions, quadratic equations, and the use of natural logarithms. Suitable for advanced high school students and beyond.

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