## anonymous 5 years ago Solve for x>0, e^x^e^x^e^x^... = 2

1. anonymous

$e^{x^{e^{x}}}$ till infinity =2 = y (say) $=e^{x^{y}}$ $=e^{x^{2}}$ =2 Therefore, $x^2 = \ln2$ x = ln2 Greetings!

2. anonymous

Oh wow, thanks, I didn't think about that approach!

3. anonymous

welcome :-)

4. anonymous

Sorry, I know its obvious but its $\sqrt{\ln2}$