Question

solve. simplify your answer(s). if there are multiple answers, separate them with commas. \\(\log \left(t^{2} - 4t + 12\
ight) = \log \left(-3t + 14\
ight)\\) \\(t = \square\\)

Explanation:

Step1: Use the property of logarithms

If \(\log(a)=\log(b)\), then \(a = b\) (for the same base and valid domain). So we set \(t^{2}-4t + 12=-3t + 14\).

Step2: Rearrange into quadratic equation

\(t^{2}-4t+3t + 12 - 14 = 0\) simplifies to \(t^{2}-t - 2=0\).

Step3: Factor the quadratic equation

We factor \(t^{2}-t - 2\) as \((t - 2)(t+1)=0\).

Step4: Solve for t

Setting each factor equal to zero: \(t - 2=0\) gives \(t = 2\); \(t + 1=0\) gives \(t=-1\).

Step5: Check the domain

For \(\log(-3t + 14)\), we need \(-3t+14>0\).

• For \(t = 2\): \(-3(2)+14=-6 + 14 = 8>0\), valid.
• For \(t=-1\): \(-3(-1)+14 = 3+14 = 17>0\), valid.

Answer:

2, -1