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.
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, $$45^{\frac{1}{1-a}} = 90 \Rightarrow 45^{\frac{1-a-b}{2(1-a)}} = 90^{\frac{1-a-b}{2}} = 3.$$
Given $90^a = 2$, $90^b = 5$

$$(90/2)^{(1-a-b)/(2–2a)}$$
$$=(90/90^a)^{(1-a-b)/(2–2a)}$$
$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$
$$=90^{(1-a-b)/2}$$
$$=\sqrt{\frac{90}{90^a 90^b}}$$
$$=\sqrt{\frac{90}{2\cdot 5}} = \sqrt{9} =3$$