Jameson Insights Author Gold Member MHB Messages 4,533 Reaction score 13 Thread starter Jan 19, 2014 #1 Solve for x. x^(x^(x^(x^(x^(x^(x^(x^(x…)))…) = 2 --------------------
MarkFL Gold Member MHB Messages 13,284 Reaction score 12 Jan 27, 2014 #2 I am standing in for Jameson this week. He has been working on various coding projects around the clock and so I am giving him a well-deserved break from his POTW duties. Congratulations to the following members for their correct solutions: 1) MarkFL 2) magneto 3) anemone 4) topsquark 5) mente oscura 6) Pranav 7) eddybob123 8) jacobi Honorable mention goes to springfan25 for having only made a minor but critical error in the last step of the logarithmic method. There were basically two methods used by those who submitted solutions. One was to use logarithms, as illustrated by topsquark: Spoiler [math]x^{x^{x^x...}} = 2[/math] [math]\ln \left ( x^{x^{x^x...}} \right ) = \ln(2)[/math] [math]x^{x^{x^x...}} \cdot \ln(x) = \ln(2)[/math] From the original problem statement [math]x^{x^{x^x...}} = 2[/math] so [math]2 \ln(x) = \ln(2)[/math] etc., so [math]x = \sqrt{2}[/math]. -Dan The other was to use a substitution, as illustrated by magneto: Spoiler Let $p := x^{x^{x^{x^{\cdots}}}}$. We can rewrite the equation as $x^p = x^2 = 2$. Therefore, $x = \sqrt{2}$.
I am standing in for Jameson this week. He has been working on various coding projects around the clock and so I am giving him a well-deserved break from his POTW duties. Congratulations to the following members for their correct solutions: 1) MarkFL 2) magneto 3) anemone 4) topsquark 5) mente oscura 6) Pranav 7) eddybob123 8) jacobi Honorable mention goes to springfan25 for having only made a minor but critical error in the last step of the logarithmic method. There were basically two methods used by those who submitted solutions. One was to use logarithms, as illustrated by topsquark: Spoiler [math]x^{x^{x^x...}} = 2[/math] [math]\ln \left ( x^{x^{x^x...}} \right ) = \ln(2)[/math] [math]x^{x^{x^x...}} \cdot \ln(x) = \ln(2)[/math] From the original problem statement [math]x^{x^{x^x...}} = 2[/math] so [math]2 \ln(x) = \ln(2)[/math] etc., so [math]x = \sqrt{2}[/math]. -Dan The other was to use a substitution, as illustrated by magneto: Spoiler Let $p := x^{x^{x^{x^{\cdots}}}}$. We can rewrite the equation as $x^p = x^2 = 2$. Therefore, $x = \sqrt{2}$.