Question

solve. simplify your answer(s). if there are multiple answers, separate them with commas.\\(\log\left(v^{2}+2v + 6\
ight)=\log(4v + 14)\\)\\(v = \square\\)

Explanation:

Step1: Use the property of logarithms

If \(\log(a)=\log(b)\), then \(a = b\) (for the same base and valid domains). So we set \(v^{2}+2v + 6=4v + 14\).

Step2: Rearrange into quadratic equation

Subtract \(4v\) and \(14\) from both sides: \(v^{2}+2v-4v + 6 - 14=0\), which simplifies to \(v^{2}-2v - 8=0\).

Step3: Factor the quadratic equation

We need two numbers that multiply to \(- 8\) and add to \(-2\). The numbers are \(-4\) and \(2\). So \((v - 4)(v+2)=0\).

Step4: Solve for \(v\)

Set each factor equal to zero: \(v - 4=0\) gives \(v = 4\); \(v+2=0\) gives \(v=-2\).

Step5: Check the domain

For \(\log(4v + 14)\), when \(v=-2\), \(4v + 14=4\times(-2)+14=-8 + 14 = 6>0\), and \(v^{2}+2v + 6=(-2)^{2}+2\times(-2)+6=4-4 + 6=6>0\). When \(v = 4\), \(4v+14=4\times4 + 14=16 + 14=30>0\), and \(v^{2}+2v + 6=16+8 + 6=30>0\). Both are valid.

Answer:

\(4, - 2\)