## Calculus 8th Edition

(a) $x=±\sqrt (e^{3}+1)$ (b) the calculated values for $x$ are ln$2$ and $0$
(a) Apply exponential functions to both sides of the equation: $e^{ln(x^{2}-1)}=e^{3}$ $(x^{2}-1)=e^{3}$ This implies $x=±\sqrt (e^{3}+1)$ (b) Given: $e^{2x}-3e^{x}+2=0$ Suppose $e^{x} = t$ Then $t^{2}-3t+2=0$ gives $t=2,1$ Thus, $e^{x} = 2,1$ Case 1: For $e^{x} = ln2$ ln$e^{x} =2$ means $x=$ln $2$ Case 2: For $e^{x} = 1$ ln$e^{x} =1$ means $x=$ln $1$ But ln$1=0$ Thus, $x=0$ Hence, the calculated values for $x$ are ln$2$ and $0$.