- #1
antinerd
- 41
- 0
How can I solve:
(x^2)(e^x) - e^x = 0
and
2e^(2+x)=6
For the first one, tell me if this is right:
e^x(x^2-1)=0 ->
e^x = 0 and x^2 - 1 = 0
so x = 1 and 0? but 0 doesn't work when I plug it back in. so is 1 the solution for x?
(x^2)(e^x) - e^x = 0
and
2e^(2+x)=6
For the first one, tell me if this is right:
e^x(x^2-1)=0 ->
e^x = 0 and x^2 - 1 = 0
so x = 1 and 0? but 0 doesn't work when I plug it back in. so is 1 the solution for x?