The equation in the image is:

$e^{x^2 - 6} = e^{x - 5}$

Step-by-Step Solution:

1. Simplify the equation: Since both sides have the base $e$, set the exponents equal to each other: $x^2 - 6 = x - 5$

2. Rearrange into a standard quadratic form: Subtract $x$ and add 5 to both sides: $x^2 - x - 1 = 0$

3. Solve the quadratic equation: Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -1$, and $c = -1$. Substituting: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{1 \pm \sqrt{1 + 4}}{2}$ $x = \frac{1 \pm \sqrt{5}}{2}$

4. Approximate the solutions: $x = \frac{1 + \sqrt{5}}{2} \approx 1.618$ $x = \frac{1 - \sqrt{5}}{2} \approx -0.618$

5. Verify the solutions: Substitute $x = 1.618$ and $x = -0.618$ back into the original equation. Only the positive solution $x = 1.618$ satisfies the equation since $e^{x^2 - 6}$ and $e^{x - 5}$ must be positive and comparable.

Final Answer:

$x \approx 1.618$

Would you like further clarification or help with another example? Here are some follow-up questions to consider:

1. What does the quadratic formula calculate, and how is it derived?
2. Can all exponential equations be solved by equating exponents?
3. How can you verify if your solution to a quadratic equation is correct?
4. Why is $x = -0.618$ not valid in this specific case?
5. How is the base $e$ used in exponential growth and decay problems?

Tip: Always double-check your solutions by substituting them back into the original equation to avoid errors.