- #1
UrbanXrisis
- 1,196
- 1
How do I solve for x in the interval [0,2pi]:
2cos(x/3)=LnX
Here's what I got:
2(.5)=LnX
1=LnX
e^(1)=x
x=2.718
is this correct?
If x=e^(Ln3)+Lne^2-5Ln1
what is x?
x=3+2-0=5
is this correct?
2cos(x/3)=LnX
Here's what I got:
2(.5)=LnX
1=LnX
e^(1)=x
x=2.718
is this correct?
If x=e^(Ln3)+Lne^2-5Ln1
what is x?
x=3+2-0=5
is this correct?