If $60^a=3$ and $60^b=5$, what is the result of $12^{\frac{1-a-b}{-2-2b}}$?
This has to be done without logarithms. The past four hours were helpless to me. Any hint, solution is welcome, I just want to learn it, or it will continue to bug my head
Thanks
Are you sure it's not $12^\frac{1-a-b}{2-2b}$? Because that works out nicely, but the posted version doesn't. I'll show the working for this version.
$60^b = 5$, so $12 = \frac{60}{5} = 60^{1-b}$.
So $12^{\frac{1-a-b}{2-2b}} = (60^{1-b})^{\frac{1-a-b}{2-2b}} = 60^{\frac{(1-a-b)(1-b)}{2-2b}} = 60^{\frac{1-a-b}{2}} = (\frac{60}{60^a60^b})^{1/2} = (\frac{60}{3\cdot 5})^{1/2} = 4^{1/2} = 2$