$\displaystyle solve \frac{3^n}{49}=1029^n$ without using logarithms
$\displaystyle \frac{3^n}{49} = 1029^n$
$\displaystyle \frac{3^n}{1029^n} = 49$
$\displaystyle \left(\frac{3}{1029}\right)^n = 49$
$\displaystyle \left(\frac{1}{343}\right)^n = 49$
$\displaystyle \frac{1^n}{343^n} = 7^2$
$\displaystyle \frac{1}{(7^{3})^n} = 7^2$
$\displaystyle \frac{1}{7^{3n}} = 7^2$
$\displaystyle 7^{-3n} = 7^2$
$\displaystyle -3n = 2$
$\displaystyle n = -\frac{2}{3}$.
Hint: 1029 = 3*7*7*7. Rewrite RHS in a way that allows you to cancel something out.
Edit: Too slow, so might as well show all the steps I had in mind.
$\displaystyle \frac{3^n}{49}=1029^n$
$\displaystyle 3^n7^{-2}=3^n7^{3n}$
$\displaystyle 7^{-2}=7^{3n}$
$\displaystyle -2=3n$
$\displaystyle n=-\frac{2}{3}$