## Calculus 8th Edition

(a) $x=\frac{1}{3}(lnk-1)$ (b) $x=\frac{1±\sqrt (1+4e)}{2}$
(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, $x=\frac{1±\sqrt (1+4e)}{2}$