# The Center of Math Blog

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## Tuesday, May 17, 2016

### Problem of the Week

See what you can come up with for this week's irrationally cool Problem of the Week. Let us know how you do in the comments!
Solution below the break.

Solution video:

Solution transcript:

#### 1 comment:

1. This is an interesting problem. Thank you for posting it.

However, I feel there are a few things missing from the solution, which - if nothing more - increase a student's understanding and promote the application of necessary care when it comes to more complex problems.

1. It is important to clarify that the domain of f is the set of positive real numbers except 1.

2. Just after finding the local maximum at x = e, you write "Since f'(e) > 0", which is clearly an error; f'(e) = 0 and you have just located e as the solution to the equation f'(x) = 0.

3. It is not at all obvious (at least to me) that the graph that you show is the graph of f (apart from the discontinuity at x = 1). I think it would be better to invoke the second derivative test (a standard part of calculus). Calculate f''(x) = (1/x) . [(2-lnx)/(lnx)^3] and then f''(e) = 1/e > 0. From this, infer that f has a local minimum at x = e.

So, in the interval (1, +infinity), f(x) >= f(e) (*1)

Furthermore, for x > e:

lnx > lne or
lnx > 1 of
f'(x) > 0

Hence for x > e, f is strictly increasing and thus (*1) becomes a strict inequality. This means that f(x) > f(e) for x > e.

Since π > e, it follows that f(π) > f(e) or π/lnπ > e. QED