# Solution for an exponential expression without using logarithms, with two defined variables

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

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I didn't understand this so 12 = 5/60 = 60^(1-b), could you help me? – user202981 Dec 24 '14 at 2:23
@WanderleyTiowann It should be mentioned that this post was over two years old. Had I not noticed you ask this question from the review queues, it would have likely gone unanswered. As it so happens, there was a typo. It should have been $12=\frac{60}{5}$, which by the condition that $60^b = 5$ becomes $\frac{60}{5} = \frac{60}{60^b} = 60^{1-b}$. I have improved the formatting of the answer so it is hopefully easier to read. – JMoravitz Dec 24 '14 at 2:59

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 $12^{\frac{1-a-b}{2-2b}} = (60^{1-b})^{\frac{1-a-b}{2-2b}} = 60^{\frac{(1-a-b)(1-b)}{2-2b}} = 60^{\frac{1-a-b}{2}} = (\frac{60}{60^a60^b})^{1/2} = (\frac{60}{3\cdot 5})^{1/2} = 4^{1/2} = 2$