How to evaluate $45^\frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm?

Question

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.

Answer 1

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

Answer 2

$$2=90^a=(2\cdot5\cdot3^2)^a\iff5^a3^{2a}=2^{1-a}$$

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 $3$