We need to find:

-Find the interval where the function is decreasing.

-Find the inflection point.

-Find the interval where the function is concave up/down

$\displaystyle F(x) = ln(4-ln(x))$

$\displaystyle F'(x) = \frac{1}{4-ln(x)} * \frac{-1}{x} = \frac{-1}{4x-xln(x)}$

I thought this was right, but that gave me a decreasing interval of $\displaystyle (e^3,\infty)$

$\displaystyle F'(x) = \frac{-1}{4x-xln(x)}$

$\displaystyle F''(x) = \frac{0-(5-ln(x))(-1)}{(4x-xln(x))^2} = \frac{5-ln(x)}{(4x-xln(x))^2}$

Which is pretty much where I've gotten stuck.

Is it one of my derivatives that's gone askew? Or did I do those right and I'm just getting the wrong intervals?