Answer
(a) $x =\dfrac{1}{3}[ \ln (k)-1]$
(b) $x = \dfrac{c^2}{m}$
Work Step by Step
(a) Since
\begin{align*}
e^{3 x+1}&=k\\
\ln e^{3 x+1}&=\ln k\\
3 x+1&=\ln k\\
3x&= \ln (k)-1\\
x&=\frac{1}{3}[ \ln (k)-1]
\end{align*}
(b) Since
\begin{align*}
\log _{2}(m x)&=c\\
\frac{\ln (mx)}{\ln 2}&=c\\
\ln(mx) &= 2\ln c \\
mx &= e^{\ln c^2}\\
x&= \frac{c^2}{m}
\end{align*}
