Question

• logarithmic and exponential equations, applications 6 of 35 (1 point) | question attempt: 1 of unlimited 2 3 4 5 6 uation. write the solution set with the exact values give -6q)+2=14 is no solution, {}. xact solution set is: .

Explanation:

Step1: Isolate the exponential term

Assuming the equation is \( 3(-6q) + 2 = 14 \) (correcting possible typo, maybe \( 3e^{-6q}+2 = 14 \) or linear? Let's solve linear first: \( -18q + 2 = 14 \))
Subtract 2: \( -18q = 14 - 2 = 12 \)

Step2: Solve for q

Divide by -18: \( q = \frac{12}{-18} = -\frac{2}{3} \)
If it's exponential \( 3e^{-6q}+2 = 14 \):

Step1: Isolate exponential

\( 3e^{-6q} = 14 - 2 = 12 \)
\( e^{-6q} = 4 \)

Step2: Take natural log

\( -6q = \ln(4) \)
\( q = -\frac{\ln(4)}{6} = -\frac{2\ln(2)}{6} = -\frac{\ln(2)}{3} \)

Assuming linear (since the text is unclear, but solving linear):

Answer:

\( q = -\frac{2}{3} \) (or for exponential \( q = -\frac{\ln(2)}{3} \), but likely linear here)