Say we have $e^{e^{e^{e^e}}}$. Since exponents raised to exponents is the same as multiplying them, this is equivalent to $e^{4e}$:


Factoring out an $e^e$:


The left side now collapses to $e^{3e}$, leading to the equality $3e=4$.

Clearly my mistake was in the part where I divided both sides by $e^e$, but why am I not able to do this?

  • 3
    $\begingroup$ $e^{4e} = e^ee^ee^ee^e$. $\endgroup$ – chepner Jul 25 '18 at 18:24

It's because $e^{e^{e^{e^e}}}$ isn't $e^{4e}$ in the first place (and neither is $((((e^e)^e)^e)^e)$, which is $e^{e^4}$). The left expression is a power tower, where exponentiation happens right-to-left, not left-to-right as required for the power multiplication rule $(a^b)^c=a^{bc}$.

  • 5
    $\begingroup$ $(((e^e)^e)^e)^e$ also isn't $e^{4e}$ $\endgroup$ – Omnomnomnom Jul 25 '18 at 15:29
  • $\begingroup$ Also note the similarity in the MathJax: e^{e^{e^{e^e}}} $\endgroup$ – Simply Beautiful Art Aug 13 '18 at 2:27

Note that

$$e^{e^{e^{e^e}}}\neq e^{4e}$$


$$3^{3^3} = 3^{27} \neq 3^9$$

  • 2
    $\begingroup$ "Note that $$2^4 \ne 4^2$$ as $$9^3 = (3^2)^3 = 3^6 \ne 3^9$$ right?" Sorry, just teasing; I know how your answer is to be understood. You are saying that $x^{x^{x^{x^x}}} = x^{4x}$ is not a valid identity for all $x$ (because there is no power rule of the type the asker seems to use), so it would be strange if it was valid for $x=e$. However, isolated solutions like $x=1.69447\ldots$ exist. $\endgroup$ – Jeppe Stig Nielsen Jul 25 '18 at 20:23
  • $\begingroup$ @JeppeStigNielsen We have that $2^4=(2^2)^2=4^2=16$ and yes we are claiming that $x^{x^{x^{x^x}}} = x^{4x}$ is not true in general and in particular also for $x=e$. $\endgroup$ – gimusi Jul 25 '18 at 20:35

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