[tex]e^{2x}-e^x=6[/tex]
[tex](e^x)^2-e^x-6=0[/tex]
[tex](e^x-3)(e^x+2)=0[/tex]
[tex]e^x=3[/tex]
[tex]e^x=-2[/tex]
therefore [tex]x=\ln3[/tex] and [tex]x=\ln-2[/tex]
because [tex]\ln-2[/tex] doesn't exist, the only solution to your equation is [tex]x=\ln3[/tex]
