We are tasked with solving the equation:

\[
e^x - 10e^{-x} = 3
\]

### Step 1: Multiply the entire equation by \( e^x \)
To eliminate the \( e^{-x} \), we can multiply both sides by \( e^x \), which will give us:

\[
e^x \cdot e^x - 10e^x \cdot e^{-x} = 3e^x
\]

This simplifies as follows:
\[
e^{2x} - 10 = 3e^x
\]

### Step 2: Rearrange the equation
Move all terms to one side to set the equation to 0:
\[
e^{2x} - 3e^x - 10 = 0
\]

### Step 3: Substitution
Let \( u = e^x \). This simplifies the equation into a quadratic form:
\[
u^2 - 3u - 10 = 0
\]

### Step 4: Solve the quadratic equation
Now solve the quadratic equation using the quadratic formula:

\[
u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}
\]

Simplifying:
\[
u = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2}
\]

This gives two possible solutions for \( u \):
\[
u = \frac{3 + 7}{2} = 5 \quad \text{or} \quad u = \frac{3 - 7}{2} = -2
\]

### Step 5: Back-substitute \( u = e^x \)
Now, we solve for \( x \) by substituting \( u = e^x \):

1. \( e^x = 5 \) → \( x = \ln(5) \)
2. \( e^x = -2 \) → No solution, because \( e^x \) is always positive.

### Final Answer:
The only solution is:
\[
x = \ln(5)
\]

Would you like more details or have any questions?

### Related Questions:
1. How does multiplying by \( e^x \) help solve equations with both \( e^x \) and \( e^{-x} \)?
2. What are the steps to solve a quadratic equation using the quadratic formula?
3. Why can't \( e^x \) be negative, like \( -2 \)?
4. How can we verify the solution \( x = \ln(5) \)?
5. What other methods can be used to solve exponential equations?

**Tip:** Always check for extraneous solutions, especially when solving exponential equations, since some steps may introduce invalid answers.

Learn how to solve the exponential equation e^x - 10e^-x = 3 with a detailed step-by-step solution. This problem covers key mathematical concepts such as solving exponential equations, converting to a quadratic form, and using logarithms to find the solution. Suitable for high school students in Grades 10-12.

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