Question

solve. simplify your answer(s). if there are multiple answers, separate them with commas.\\(\log \left(r^{2}+4r + 12\
ight)=\log \left(-2r + 3\
ight)\\)\\(r = \square\\)

Explanation:

Step1: Use the property of logarithms

If \(\log(a)=\log(b)\), then \(a = b\) (for the same base and \(a>0\), \(b>0\)). So we set \(r^{2}+4r + 12=-2r + 3\).

Step2: Rearrange into quadratic equation

\(r^{2}+4r+2r + 12 - 3=0\)
\(r^{2}+6r + 9 = 0\)

Step3: Factor the quadratic equation

The quadratic \(r^{2}+6r + 9\) is a perfect square, \((r + 3)^{2}=0\)

Step4: Solve for \(r\)

Taking square root on both sides, \(r+3 = 0\), so \(r=-3\)

Step5: Check the domain

For \(\log(-2r + 3)\), when \(r=-3\), \(-2(-3)+3=6 + 3=9>0\)
For \(\log(r^{2}+4r + 12)\), when \(r=-3\), \((-3)^{2}+4(-3)+12=9-12 + 12=9>0\)
So \(r=-3\) is valid.

Answer:

\(-3\)