To solve the exponential equation [tex]\(5^{x-3} = 5^{x^2-9}\)[/tex], we can use the fact that if [tex]\(a^b = a^c\)[/tex] and [tex]\(a\)[/tex] is a positive number different from 1, then [tex]\(b = c\)[/tex]. Here, both bases are 5, which is a positive number different from 1, so we can set the exponents equal to each other:
[tex]\[
x - 3 = x^2 - 9
\][/tex]
Now, we have a quadratic equation to solve. Let's move all terms to one side of the equation:
[tex]\[
x - 3 - (x^2 - 9) = 0
\][/tex]
Simplify the equation:
[tex]\[
x - 3 - x^2 + 9 = 0
\][/tex]
Combine like terms:
[tex]\[
-x^2 + x + 6 = 0
\][/tex]
Rearrange the terms to write the quadratic equation in standard form:
[tex]\[
-x^2 + x + 6 = 0 \quad \implies \quad x^2 - x - 6 = 0
\][/tex]
Now we solve the quadratic equation [tex]\(x^2 - x - 6 = 0\)[/tex]. We can solve this using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the equation [tex]\(x^2 - x - 6 = 0\)[/tex], the coefficients are [tex]\(a = 1\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -6\)[/tex]. Plugging these values into the quadratic formula gives:
[tex]\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{1 \pm \sqrt{1 + 24}}{2}
\][/tex]
[tex]\[
x = \frac{1 \pm \sqrt{25}}{2}
\][/tex]
[tex]\[
x = \frac{1 \pm 5}{2}
\][/tex]
This gives us two potential solutions:
[tex]\[
x = \frac{1 + 5}{2} = \frac{6}{2} = 3
\][/tex]
[tex]\[
x = \frac{1 - 5}{2} = \frac{-4}{2} = -2
\][/tex]
Now we need to check these solutions in the original equation to verify they are valid.
For [tex]\(x = 3\)[/tex]:
[tex]\[
5^{3-3} = 5^{3^2-9} \quad \implies \quad 5^0 = 5^0 \quad \implies \quad 1 = 1
\][/tex]
For [tex]\(x = -2\)[/tex]:
[tex]\[
5^{-2-3} = 5^{(-2)^2-9} \quad \implies \quad 5^{-5} = 5^{4-9} \quad \implies \quad 5^{-5} = 5^{-5}
\][/tex]
Both solutions satisfy the original equation. So, the solutions to the equation [tex]\(5^{x-3} = 5^{x^2-9}\)[/tex] are:
[tex]\[
x = 3 \text{ and } x = -2
\][/tex]
