Question

e^ { 2 x } - 6 e^ { x } +8 = 0
please help find the 2 solutions

Answer 1

solve the quadratic equation
\[z^2-6z+8=0\] for z and then replace z by
\[e^x\]

Answer 2

how did you get that quadratic equation?

Answer 3

get
\[(z-2)(z-4)=0\]
\[z=2,z=4\] so
\[e^x=4,x=\ln(4)\] or
\[e^x=2,x=\ln(2)\]

Answer 4

i get it because
\[e^{2x}=(e^x)^2\]

Answer 5

so i am thinking that this is
\[(e^x)^2-6e^x+8\] which is a quadratic equation in
\[e^x\]

Answer 6

i think i understand how you got the equation now, thank you!