How do you solve the equation: $e^{t/\!\ln(x)}=x$ for $x$? Here $t=1,2,3,\dotsc$
This is what I did:
$\ln(e^{t/\!\ln(x)})=\ln(x), $
$t/\!\ln(x)=\ln(x),$
$t=(\ln(x))^2,$
$x=e^\sqrt{t},e^{-\sqrt{t}}.$
Is this correct? What kinds of numbers are the solutions?
You did perfect (apart from not noting that solutions must satisfy $x>0$ and $x\ne1$).
It might be easier if you set $\ln x=y$, so $x=e^y$. Then the equation becomes $$ e^{t/y}=e^y $$ that yields $$ \frac{t}{y}=y $$ and so $y^2=t$. Hence $y=\sqrt{t}$ or $y=-\sqrt{t}$ (assuming $t>0$, or the equation has no solution) and $x=e^{\sqrt{t}}$ or $x=e^{-\sqrt{t}}$.