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"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 $f'(t)\geq-1$ enter image description here

Graph when $f'(t)<-1$ enter image description here

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.

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  • $\begingroup$ Welcome to MathSE site! What have you tried so far? The point of asking this question is to help us understand what you are stuck in or offer some suggestions for your work. $\endgroup$ – NasuSama Feb 15 '14 at 3:10
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    $\begingroup$ Remember that if you were to prove "if and only if" statement, you need to prove both ways. For instance, assume the equation has only a root. We want to show the derivative $\geq -1$. Once you are done with that, assume the derivative $\geq -1$ and show the equation has only one root. In addition, you can prove this way vice versa. $\endgroup$ – NasuSama Feb 15 '14 at 3:12
  • $\begingroup$ Oh I have a proof that when derivate <-1 there become exist at least 3 roots. i think it's equivalent to the first one you suggested. $\endgroup$ – ElectronicKid Feb 15 '14 at 3:18
  • $\begingroup$ I added some photos to explain easier. $\endgroup$ – ElectronicKid Feb 15 '14 at 3:40

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