To solve the equation [tex]\(\log \left(x^2 + 3x - 44\right) = 1\)[/tex], we can follow these steps:
1. Convert the logarithmic equation to its exponential form.
Given the equation [tex]\(\log \left(x^2 + 3x - 44\right) = 1\)[/tex], we understand that this is a common logarithm (base 10). The logarithmic form [tex]\(\log_b (y) = z\)[/tex] can be converted to the exponential form [tex]\(b^z = y\)[/tex].
So, we rewrite the equation as:
[tex]\[
x^2 + 3x - 44 = 10^1
\][/tex]
2. Simplify the equation.
Since [tex]\(10^1 = 10\)[/tex], we have:
[tex]\[
x^2 + 3x - 44 = 10
\][/tex]
3. Set the equation to zero to form a quadratic equation.
Subtract 10 from both sides to get:
[tex]\[
x^2 + 3x - 54 = 0
\][/tex]
4. Solve the quadratic equation.
The quadratic equation [tex]\(x^2 + 3x - 54 = 0\)[/tex] can be solved using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -54\)[/tex].
Compute the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 3^2 - 4(1)(-54) = 9 + 216 = 225
\][/tex]
Since the discriminant is positive, we will have two real solutions. Compute the square root of the discriminant:
[tex]\[
\sqrt{225} = 15
\][/tex]
Substitute the values into the quadratic formula:
[tex]\[
x = \frac{-3 \pm 15}{2}
\][/tex]
This gives us the two solutions:
[tex]\[
x = \frac{-3 + 15}{2} = \frac{12}{2} = 6
\][/tex]
[tex]\[
x = \frac{-3 - 15}{2} = \frac{-18}{2} = -9
\][/tex]
5. Verify the solutions in the context of the original equation.
Substitute [tex]\(x = 6\)[/tex]:
[tex]\[
x^2 + 3x - 44 = 6^2 + 3(6) - 44 = 36 + 18 - 44 = 10
\][/tex]
Check:
[tex]\[
\log (10) = 1 \quad \text{(True)}
\][/tex]
Substitute [tex]\(x = -9\)[/tex]:
[tex]\[
x^2 + 3x - 44 = (-9)^2 + 3(-9) - 44 = 81 - 27 - 44 = 10
\][/tex]
Check:
[tex]\[
\log (10) = 1 \quad \text{(True)}
\][/tex]
So, the solutions to the equation [tex]\(\log \left(x^2 + 3x - 44\right) = 1\)[/tex] are:
[tex]\[
x = 6 \quad \text{and} \quad x = -9
\][/tex]
