To solve the equation $2^{x^2 - 3x - 10} = 3^{x^2 - 3x - 10}$, let's analyze the situation step by step.

Step 1: Analyze the Equation

We have: $2^{x^2 - 3x - 10} = 3^{x^2 - 3x - 10}$ Since the bases (2 and 3) are different, the only way these two expressions can be equal is if the exponent $x^2 - 3x - 10$ equals zero.

Step 2: Set the Exponent Equal to Zero

Let's solve for $x$ by setting the exponent equal to zero: $x^2 - 3x - 10 = 0$

Step 3: Solve the Quadratic Equation

This is a quadratic equation, and we can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -3$, and $c = -10$. Plugging in these values: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 + 40}}{2}$ $x = \frac{3 \pm \sqrt{49}}{2}$ $x = \frac{3 \pm 7}{2}$ This gives us two possible solutions: $x = \frac{3 + 7}{2} = 5$ $x = \frac{3 - 7}{2} = -2$

Step 4: Verify the Solutions

To confirm, let's substitute $x = 5$ and $x = -2$ back into the original equation.

1. For $x = 5$: $2^{5^2 - 3(5) - 10} = 2^{25 - 15 - 10} = 2^0 = 1$ $3^{5^2 - 3(5) - 10} = 3^{25 - 15 - 10} = 3^0 = 1$ The equation holds true.

2. For $x = -2$: $2^{(-2)^2 - 3(-2) - 10} = 2^{4 + 6 - 10} = 2^0 = 1$ $3^{(-2)^2 - 3(-2) - 10} = 3^{4 + 6 - 10} = 3^0 = 1$ The equation holds true.

Final Answer:

The solutions to the equation are $x = 5$ and $x = -2$.

Would you like more details on any step or have any questions?

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Here are 5 related questions to further explore this topic:

1. How can you solve other types of exponential equations?
2. What happens if the bases on both sides of the equation are the same?
3. How does the quadratic formula work in different contexts?
4. Can you apply this method to equations with different powers of $x$?
5. What are some common mistakes to avoid when solving exponential equations?

Tip: Always check the solutions by substituting them back into the original equation to ensure they are correct.