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.

$$2=90^a=(2\cdot5\cdot3^2)^a\iff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^b\iff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$\implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}\iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2\cdot5)^{1-a-b}=3^{2-2(a+b)}\cdot5^{1-a-b}=3^{2-2(a+b)}\cdot3^{2b}$$
$$\implies45^{1-a-b}=3^{2(1-a)}$$
$\implies$ one of the values of $$45^{\frac{1-a-b}{2(1-a)}}$$ is $3$
