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

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 $$C^x-A^x=\frac{\ln(\frac{\ln(C)}{\ln(A)})}{\ln(B)}$$ Mar 12, 2013 at 6:36
• @Ethan Yes. Thanks for pointing that out! I apologize for my carelessness. Mar 12, 2013 at 6:40

\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.

• Is there a typo in your 4th line with $e^{x\ln A}-e^{x\ln A}$? Mar 12, 2013 at 7:10
• @alex.jordan Yes, thanks for catching that. Mar 12, 2013 at 7:11
• Is there a typo in 3rd line? ($\frac{\ln{C}}{\ln{A}})/\ln B$ Mar 13, 2013 at 13:00
• oh, yep, I dropped all the logs. sloppy. Mar 13, 2013 at 20:34

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.