[tex]e ^{2x} -e ^x=6
[/tex] ----(i)
Let, [tex]e ^x =t[/tex]
Therefore, [tex]e^{2x} = t^{2} [/tex]
Therefore, from (i),
t²-t=6
t²-3t+2t-6=0
t(t-3)+2(t-3)=0
Therefore, (t-3)(t+2)=0
Therefore, t=3 or t = -2
But,
[tex]e^x =t [/tex]
Therefore,
[tex]e^x = 3 [/tex] or [tex]e^x = -2[/tex]
Taking ln on both sides,
Therefore,
[tex]ln(e^x)=ln(3)[/tex] or [tex]ln(e^x) = ln(-2)[/tex]
But natural log of negative numbers does not exist since negative numbers are not in the domain of ln(x)
Therefore,
[tex]ln(e^x) = ln(-2)[/tex] is discarded
Therefore, x = ln(3) is the only solution.
