<s>[INST] How to evaluate $45^\frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm? Let $90^a=2$ and $90^b=5$, Evaluate  
$45^\frac {1-a-b}{2-2a}$ 
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
 [/INST] Let me try. 
$$10 = 90^{a+b} \Rightarrow 3^2 = 90^{1-a-b} \Rightarrow 3 = 90^{\frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{\frac{1}{1-a}} = 90 \Rightarrow 45^{\frac{1-a-b}{2(1-a)}} = 90^{\frac{1-a-b}{2}} = 3.$$
</s>