Solving 2cos(x/3) = ln x for x in [0,2π]

[UrbanXrisis]

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?

 

[Tide]

Was that supposed to be $\cos \frac {\pi}{3}$?

 

[UrbanXrisis]

yes sorry, that is 2cos(pi/3)=LnX

 

[Tide]

In that case, good job!

 

[UrbanXrisis]

how do I find the relative maximum value of the function y=(LnX)/X and the domain of the function Ln(X^2-1)?

 

[Tide]

Just find the derivative of the function and set it to zero.

 

[UrbanXrisis]

for the domain or for the max value?

 

[Tide]

That will give you the location of the relative maximum (you can verify that it is a maximum as opposed to a minimum). Then use that value of x that you find to determine the value of the function at that point.