Answer
(a) $x=\frac{1}{3}(lnk-1)$
(b) $x=\frac{1±\sqrt (1+4e)}{2}$
Work Step by Step
(a) Given: $e^{3x+1}=k$
ln$e^{3x+1}=$ln$k$
$(3x+1) =$ln$k$
This gives
$x=\frac{1}{3}(lnk-1)$
(b) Given: $lnx+ln(x-1)=1$
ln$[x(x-1)]=1$
$e^{ln[x(x-1)]}=e^{1}$
This implies
$x(x-1)=e$
$x^{2}-x=e$
$x^{2}-x-e=0$
Hence, $x=\frac{1±\sqrt (1+4e)}{2}$
