How do I find out if this ecuation has any solutions?

Question

arctgx=lnx on I=(0,+infinite)
I can't apply the Cauchy-Bolzano theorem here ,because of ln0 which doesn't exist and the appearance of infinite..

Answer 1

Let`study teh function
y=Arctg x-lnx
At zero its limit is +infinity(Very near to 0 the function is positive)
At +infinity the limit is - infinity as Arctg x tends to pi/2.
So at first glance there is at least one root as the function is continuos in 0,+infinite

Let`s tahe the derivative
y'= 1/(1+x^2) -1/x = (x-1-x^2)/x(1+x^2)
y'=0 -(x^2-x+1)=0 No roots and always negative
So the function y is strictly decreasing and there is only ONE root

Answer 2

you can draw the scheme of each Arctgx and lnx, you can see that offcourse it has answer, there is an intersection,
and also you can do trial and error methods to fin out the answer.

Answer 3

I don't think there are any solutions because infinity is... infinite.

But I'm actually not sure, since I'm taking highschool algebra right now.