Step-by-step explanation:
Using the following property of logarithms,
[tex]\ln{\left(\dfrac{a}{b}\right)} = \ln{a} - \ln{b}[/tex]
we can write the given equation as
[tex]\ln{(x^2 + 1)} - \ln{(x^2 - 1)} = \ln{\left(\dfrac{x^2 + 1}{x^2 - 1}\right)}[/tex]
[tex]\:\:\:\:\:\:\:\:\:= \ln{\dfrac{x(x + 1)}{x(x - 1)}} = \ln{\left(\dfrac{x + 1}{x - 1}\right)}[/tex]
But recall that [tex]\ln{e} = 1[/tex] so we can write our original equation as
[tex]\ln{\left(\dfrac{x + 1}{x - 1}\right)} = \ln{e}[/tex]
or simply
[tex]\dfrac{x + 1}{x - 1} = e[/tex]
Multiplying both sides by [tex](x - 1),[/tex] we get
[tex]x + 1 = e(x - 1)[/tex]
Solving for x, we get
[tex]x = \dfrac{e + 1}{e - 1}[/tex]
