3

I have tried to use $\ln$, but couldn't solve:

\begin{equation} \ln x^{x^{x^{2017}}}=x^{x^{2017}}\ln x=\ln 2017. \end{equation}

I found that $x=\sqrt[2017]{2017}$ is a solution, and it is easy to check it. But how to find that solution without guessing and how to prove if it is the only solution?

Lee
  • 1,910
  • 12
  • 19

1 Answers1

3

$$x^{x^{x^{2017}}}=2017$$

Raise $x$ to the power of both sides:

$$x^{x^{x^{x^{2017}}}}=x^{2017}$$ Let $y=x^{2017}$. Then the equality becomes: $$x^{x^{x^{y}}}=y$$ Since this is in the form of the first equation, $y$ can equal 2017. Therefore: $$y=2017=x^{2017}$$ Which gives us our one real solution $2017^{\frac{1}{2017}}$. By the fundamental theorem of algebra, there are 2016 more complex solutions, however it will take you a while if you don't have a calculator. Hope this helps!

JC12
  • 1,040
  • 1
    You claim that from $x^{x^{x^{y}}}=y$ follows $y$ must be $2017$. But why is 2017 the only solution of $x^{x^{x^{y}}}=y$? – miracle173 Jul 06 '20 at 08:49
  • It isn't the only solution, however, I chose the solution $y=2017$ to arrive at the answer to OP was asking about. – JC12 Jul 06 '20 at 08:56
  • 1
    But the argument "Since this is in the form of the first equation, y '''must''' equal 2017" is only valid if the equation has only one solution – miracle173 Jul 06 '20 at 09:23
  • I realised the mistake and have edited the sentence from using the word "must" to "can", so that the substitution is one of many valid ones though the one used gives the OPs desired answer. Thanks for clearing things up! – JC12 Jul 06 '20 at 09:53