[tex]\bf y=ln(3x+2)^k\implies y=k\cdot ln(3x+2)\implies \cfrac{dy}{dx}=k\cdot \stackrel{chain~rule}{\cfrac{1}{3x+2}\cdot 3}
\\\\\\
\cfrac{dy}{dx}=\cfrac{3k}{3x+2}\impliedby \textit{we know that }y'(2)=3\quad or\quad
\begin{cases}
x=2\\
y'(x)=3
\end{cases}
\\\\\\
\textit{therefore}\qquad 3=\cfrac{3k}{3(2)+2}\implies 3=\cfrac{3k}{8}\implies \cfrac{3\cdot 8}{3}=k\implies 8=k[/tex]
