Question

Solve e^(2x) -2e^x + 1 = 0.

Answer 1

hint:
\[(a-b)^2=a^2-2ab+b^2\]

Answer 2

using above identity u can write
\[e^{2x}-2e^x+1=(e^x-1)^2\]
so
\[e^x-1=0\]
u can do it from here

Answer 3

Do you get him?

Answer 4

I need a full answer please

Answer 5

\[e^x-1=0\rightarrow e^x=1\]
now take log on both sides
\[\log[e^x]=\log1\rightarrow x=0\]