To solve the given equation:
[tex]\[ 125^{3x - 8} = 5^{x^2 + 13x - 45} \][/tex]
we start by expressing both sides of the equation with the same base. Notice that [tex]\( 125 \)[/tex] can be written as a power of [tex]\( 5 \)[/tex], specifically:
[tex]\[ 125 = 5^3 \][/tex]
Thus, we can rewrite the left side of the equation as:
[tex]\[ 125^{3x - 8} = (5^3)^{3x - 8} \][/tex]
Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex], we get:
[tex]\[ (5^3)^{3x - 8} = 5^{3(3x - 8)} = 5^{9x - 24} \][/tex]
So, the equation becomes:
[tex]\[ 5^{9x - 24} = 5^{x^2 + 13x - 45} \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 9x - 24 = x^2 + 13x - 45 \][/tex]
To solve this quadratic equation, first bring all terms to one side of the equation:
[tex]\[ 0 = x^2 + 13x - 45 - 9x + 24 \][/tex]
[tex]\[ 0 = x^2 + 4x - 21 \][/tex]
We now solve the quadratic equation [tex]\(x^2 + 4x - 21 = 0\)[/tex]. We can factor the quadratic equation:
[tex]\[ x^2 + 4x - 21 = (x + 7)(x - 3) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ x + 7 = 0 \quad \text{or} \quad x - 3 = 0 \][/tex]
[tex]\[ x = -7 \quad \text{or} \quad x = 3 \][/tex]
Therefore, the solutions to the equation [tex]\( 125^{3x - 8} = 5^{x^2 + 13x - 45} \)[/tex] are:
[tex]\[ x = -7, 3 \][/tex]
