Solving ln(x)=b x^2 for Exactly One Solution

[fyziky]

the question is simple but i can't seem to think of a solution.
for what b>0 does ln(x)=b x^{2} have exactly one solution, and not 0.
I've tried playing with ln rules but can't seem to think of a solution

 

[hgfalling]

This is a pretty good calculus problem, but it's definitely not as simple as fiddling with rules of logarithms. (Or maybe it is and I'm not seeing it, there's always that possibility!)

Here are some hints on a method for solving it:

1) Consider the function f(x) = bx^2 - \ln x. Finding a solution to the original problem is the same as finding a zero of this function.
2) Analyze the behavior of this function. For what x is it defined? What happens near zero? As x goes to infinity? Where are the local maxima/minima, etc?
3) The analysis in 2) will lead you to the special circumstances (ie, value of b) where the function has only one zero. That's your answer. The answer is kind of elegant.

If you get stuck, post in this thread. I or others will help you.

 

[fyziky]

thank you very much for the hints but unfortunately I am still stuck. I still am unable to solve for the zeros, and when i use maple the answer i get is dependent on the lambert function. when i try different plots varying my b i get one zero when b=1/2e. maybe that is the solution but i still don't know how i would solve it independently.

 

[hgfalling]

That is the answer. But you don't need the Lambert W function to find it; you can do it with first semester calculus.

Let me follow up on some of my hints:

f(x) = bx^2 - \ln x

f'(x) = 2bx - \frac{1}{x}

2bx - \frac{1}{x} = 0

\frac{1}{2b} = x^2

So we know that we have just one local minima/maxima, at \frac{1}{\sqrt{2b}}.

Also,

\lim_{x \rightarrow 0} f(x) = \infty

\lim_{x \rightarrow \infty} f(x) = \infty

So what can we say about the zeros of f based on these facts?

 

[fyziky]

thank you hgfalling you are wise.
Since f(x) approaches infinity from both directions and we do have a max/min value it must be a min. also since it approaches infinity from both sides if the min is negative it will cross the X axis twice so to have a unique solution thi min must be the zero. using the min value you showed earlier:
ln((2b)^(1/2))=b((2b)^(-1/2))^2=(1/2)
(1/2)ln(2b)=(1/2)
ln(2b)=-1
2b=e^-1
b=1/2e
!
thanks again