Answer
(a) $x=±\sqrt (e^{3}+1)$
(b) The calculated values for $x$ are ln$2$ and $0$.
Work Step by Step
(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 $0$.
