Solving equation $A^{(B^{(C^x)})} = C^{(B^{(A^x)})}$ for $x$

Question

Recently I came across the equation

$$A^{(B^{(C^x)})} = C^{(B^{(A^x)})}$$

where $A \neq B \neq C$, and if $A, B, C > 1$ or if $0 < A,B,C < 1$, there exists a unique solution for $x$.

Here is my attempt:

$$A^{(B^{(C^x)})} = C^{(B^{(A^x)})}$$ $$B^{(C^x)} = \log_A{C^{(B^{(A^x)})}}$$ $$C^x-A^x = \log_B{\log_A{C}}$$

$$C^x-A^x = \frac{\ln{(\frac{\ln{C}}{\ln{A}}})}{\ln B}$$

And I was stuck at $C^x-A^x$..

It should be $$C^x-A^x=\frac{\ln(\frac{\ln(C)}{\ln(A)})}{\ln(B)}$$ — Ethan Splaver, Mar 12 '13 at 06:36
@Ethan Yes. Thanks for pointing that out! I apologize for my carelessness. — Archy Will He 何魏奇, Mar 12 '13 at 06:40

Josephine Moeller · Accepted Answer · 2013-03-13T20:35:50.790

4

$$\begin{align*} A^{B^{C^x}}&=C^{B^{A^x}} \\{B^{C^x-A^x}}&=\frac {\ln C}{\ln A} \\{C^x-A^x}&=\ln\left(\frac {\ln C}{\ln A}\right)/\ln B \\e^{x\ln(CA)/2}\left(e^{x\ln(C/A)/2}-e^{-x\ln(C/A)/2}\right)&=\ln\left(\frac {\ln C}{\ln A}\right)/\ln B \\2\sinh({x\ln (C/A)/2})&=\ln\left(\frac {\ln C}{\ln A}\right)\frac{e^{-x\ln (CA)/2}}{\ln B} \end{align*}$$

If there's a solution, it mostly comes down to signs of the logs and whether $A > C$ or not.

I'll leave you to figure out the details, but this should give you a general approach.

edited Mar 13 '13 at 20:35

answered Mar 12 '13 at 06:57

Josephine Moeller

2,601

Is there a typo in your 4th line with $e^{x\ln A}-e^{x\ln A}$? – 2'5 9'2 Mar 12 '13 at 07:10
@alex.jordan Yes, thanks for catching that. – Josephine Moeller Mar 12 '13 at 07:11
Is there a typo in 3rd line? ($\frac{\ln{C}}{\ln{A}})/\ln B$ – Archy Will He 何魏奇 Mar 13 '13 at 13:00
oh, yep, I dropped all the logs. sloppy. – Josephine Moeller Mar 13 '13 at 20:34

score 1 · Answer 2 · answered Mar 12 '13 at 07:09

By symmetry in $A\leftrightarrow C$, we may assume that either $0<C<A<1$ or $1<A<C$. So far, by taking logarithms and reordering
$$\tag0A^{(B^{(C^x)})} = C^{(B^{(A^x)})}$$ $$B^{(C^x)}\ln A = B^{(A^x)}\ln C$$ $$C^x\ln B + \ln\ln A = A^x\ln B +\ln\ln C$$ $$\tag1C^x-A^x = \frac{\ln\ln C-\ln\ln A}{\ln B}$$ where the right hand side is constant. We rewrite the left hand side as $$ \tag2C^x-A^x = A^x\left(\left(\frac CA\right)^x-1\right).$$ If $C>A>1$, the first factor on the right os strictly positive and strictly increasing, while the second factor is strictly increasing (but might be negative). However, the product is not monotonuous for all $x$. But from $B>1$ we infer that the right hand side in $(1)$ is positive, hence we can restrict to $x$ where the second factor in $(2)$ is positive, that is $x>0$. In that case, both factors in $(2)$ are positive and increasing, hence so is their product. This shows that at most one solution $x$ exists. At $x=0$, we obtain $C^x-A^x=1-1=0$, whereas each factor $A^x$ and $\left(\frac CA\right)^x-1$ goes $\to+\infty$ as $x\to+\infty$. Therefore $x\mapsto C^x-A^x$ is a bijection $[0,\infty)\to[0,\infty)$ and there exists a unique solution.

The same discussion works in the case $0<A<C<1$, $0<B<1$ apart from changed signs and monotonicity.

Solving equation $A^{(B^{(C^x)})} = C^{(B^{(A^x)})}$ for $x$

2 Answers2