"For two positive number a, b which satisfies the condition $\ln a \ln b <0$. equation $a^{b^{a^{b^x}}}=x$ has only one root if and only if ${\frac{d}{dx}a^{b^x}}_{x=t}\geq-1$, where $a^{b^t}=t$"
It's the last unsolved proposition on my personal reserch.
Please help me prove this, or if it is false, let me know it. You can give a lot of help to this newbie at writing paper, just throwing several hints to him..
== added ==
some pictures to help you understand:
(before picture: $a^{b^{a^{b^x}}}=x$ is equivalent to $a^{b^x}=\log_b{\log_ax}$. and letting $f(x)=a^{b^x}$, it is same with $f(x)=f^{-1}(x)$)
Blue is $f$, red is $f^{-1}$
Graph when $f'(t)\geq-1$
Graph when $f'(t)<-1$
it is SO clear on the graphs. But hard to prove.
Q. Where did you get so far?
A. I've tried many ways. and the best result i got so far is: It is easily proved if I show that $f(f(x))''>0$ for $x<p$, $f(f(x))''=0$ for $x=p$, $f(f(x))''>0$ for $x>p$ for a number p. but... it is also hard.