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My lecturer gave us a potential bonus to the grade if we can fully solve those equations: $y^x=9 $ $x^y=8$

easy to see that the value of $x$ and $y$ are $2$ & $3$ but what is the correct full way of showing the solution for that?

appreciate your help geniuses <3

dmtri
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niRo
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1 Answers1

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$y^x=9 \Rightarrow x=\frac{ln9}{lny}$ , $x^y=8 \Rightarrow y=\frac{ln8}{lnx}$ Substitute to get $x=\frac{ln9}{ln(\frac{ln8}{lnx})}=\frac{ln9}{ln(ln8)-ln(lnx)} \Rightarrow x(ln(ln8)-ln(lnx))=ln9 \Rightarrow xln(ln8)-xlnln(x)=ln9$ And then well, using Wolframe Alpha to solve the above equation, we get $x=2$. This then implies $2^y=8 \Rightarrow y=3$.

Hope that help!

user 6663629
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    I don't think you need Wolframe Alpha to prove uniqueness at this point. The function $x(\ln(\ln(8)-\ln(\ln(x))$ is strictly increasing. Hence there is at most one solution and we already know the solution $x=2$. – quarague Mar 27 '19 at 13:26
  • Ah right! Thank you! But the function is strictly decreasing, right? – user 6663629 Mar 27 '19 at 13:35
  • You're right, it's decreasing. The argument still works the same though :-) – quarague Mar 27 '19 at 13:41
  • I appreciate your effort alot, that helped me much I just didnt get the last step you did to find x=2 is that some common form ? xln(ln8)−xlnln(x)=ln9 the alpha site doesnt show me full solution from that stage without paying them ^^ – niRo Mar 27 '19 at 17:34
  • if you can screen the solution from wolframe alpha site it would be much appreciated!!! TY – niRo Mar 27 '19 at 17:36