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UrbanXrisis
Sep11-04, 11:16 PM
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
Sep11-04, 11:20 PM
Was that supposed to be \cos \frac {\pi}{3}?

UrbanXrisis
Sep11-04, 11:22 PM
yes sorry, that is 2cos(pi/3)=LnX

Tide
Sep11-04, 11:35 PM
In that case, good job!

UrbanXrisis
Sep11-04, 11:47 PM
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
Sep12-04, 12:02 AM
Just find the derivative of the function and set it to zero.

UrbanXrisis
Sep12-04, 12:17 AM
for the domain or for the max value?

Tide
Sep12-04, 12:24 AM
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.