What is the minimum degree of x so that it is greater than or equal to ln(x)?

Question

I was thinking of this question and couldn't find it anywhere. I was trying to find a solution by finding the maximum of the function $f(x) = \frac{ln(ln(x))}{ln(x)}$,
yet I'm not sure if that's gonna work. Thanks in advance.

Answer 1

We can solve this by using calculus:

Let $$f(x) = \frac{\ln(\ln(x))}{\ln(x)}$$

Then, taking the derivative:

$$f'(x) = \frac{1-\ln(\ln(x))}{x\ln^2(x)}$$

Setting it equal to 0 to find the maximum value:

$$\frac{1-\ln(\ln(x))}{x\ln^2(x)} = 0$$ $$1-\ln(\ln(x)) = 0$$ $$\ln(x) = e$$ $$x = e^e$$

Our maximum value occurs when $x = e^e$ Plugging this back into our function: $$f(e^e) = \frac{\ln(\ln(e^e))}{\ln(e^e)}$$ $$f(e^e) = \frac{1}{e}$$