We need to solve the equation:

$e^x + 4e^{-x} - 5 = 0$

Step 1: Substitution

Let $y = e^x$. Then $e^{-x} = \frac{1}{y}$. Substituting these into the equation gives:

$y + 4\frac{1}{y} - 5 = 0$

Multiply through by $y$ (noting $y > 0$ since $e^x > 0$):

$y^2 + 4 - 5y = 0$

Step 2: Rearrange into standard quadratic form

$y^2 - 5y + 4 = 0$

Step 3: Solve the quadratic equation

The quadratic formula is given by:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -5$, and $c = 4$. Substituting these values:

$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$

$y = \frac{5 \pm \sqrt{25 - 16}}{2}$

$y = \frac{5 \pm \sqrt{9}}{2}$

$y = \frac{5 \pm 3}{2}$

This gives two solutions:

$y = \frac{5 + 3}{2} = 4 \quad \text{and} \quad y = \frac{5 - 3}{2} = 1$

Step 4: Back-substitute for $y = e^x$

We solve $e^x = 4$ and $e^x = 1$:

1. $e^x = 4$ gives $x = \ln(4)$.
2. $e^x = 1$ gives $x = \ln(1) = 0$.

Final Solution:

The real solutions are:

$x = \ln(4) \quad \text{and} \quad x = 0$

Would you like a detailed graph or further clarifications?

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1. What is the relationship between $e^x$ and $e^{-x}$?
2. How do you simplify equations involving exponents with opposite signs?
3. Why is $y > 0$ in this substitution?
4. What are some techniques for solving quadratic equations?
5. How can we check if our solutions are correct?

Tip: Always verify your solutions by substituting them back into the original equation to ensure they satisfy it.