To solve the equation, [tex]\(\log \left(q^2 + 7q\right) = \log 18\)[/tex], follow these steps:
1. Remove the logarithms:
Since we have [tex]\(\log A = \log B\)[/tex], we can equate the arguments:
[tex]\[
q^2 + 7q = 18
\][/tex]
2. Set up the quadratic equation:
Rewrite the equation in standard form:
[tex]\[
q^2 + 7q - 18 = 0
\][/tex]
3. Solve the quadratic equation:
To solve [tex]\(q^2 + 7q - 18 = 0\)[/tex], factorize or use the quadratic formula [tex]\(q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = -18\)[/tex].
4. Find the solutions:
Solve the quadratic equation by factoring:
[tex]\[
(q + 9)(q - 2) = 0
\][/tex]
Set each factor to zero and solve for [tex]\(q\)[/tex]:
[tex]\[
q + 9 = 0 \quad \Rightarrow \quad q = -9
\][/tex]
[tex]\[
q - 2 = 0 \quad \Rightarrow \quad q = 2
\][/tex]
5. Verify the solutions:
Substitute [tex]\(q = -9\)[/tex] and [tex]\(q = 2\)[/tex] back into the original argument [tex]\(q^2 + 7q\)[/tex] to ensure the solutions are valid:
[tex]\[
\log((-9)^2 + 7(-9)) = \log(81 - 63) = \log(18) \quad \text{(valid)}
\][/tex]
[tex]\[
\log(2^2 + 7 \cdot 2) = \log(4 + 14) = \log(18) \quad \text{(valid)}
\][/tex]
Both solutions are valid. Thus, the exact solution set is:
[tex]\[
\boxed{-9, 2}
\][/tex]
