$$\frac{(0,01)^{x-1}}{(0,1)^{3-x}} = 10^{-4x-2}$$
I don't have any idea about where I'm going wrong and why.
$$\frac{(\frac{1}{10^2})^{x-1}}{(\frac{1}{10})^{3-x}} = 10^{-4x-2}$$
$$\frac{(\frac{1}{10^2})^{x-1}}{10^{-3+x}} = 10^{-4x-2}$$
The thing I want to know is If there's any strategy/tips to solve the questions which contain the terms like $x \cdot 10^y$ That's why it makes me confused.
On
You can proceed as follow
$$\frac{(\frac{1}{10^2})^{x-1}}{10^{-3+x}} = 10^{-4x-2}\iff(\frac{1}{10^2})^{x-1} = 10^{-4x-2}{10^{-3+x}}\iff10^{-2x+2}=10^{-3x-5}\\\iff-2x+2=-3x-5\iff x=-7$$
On
$$(0,01)^{x-1}=(10^{-2})^{x-1}=10^{-2 (x-1)}=10^{-2x+2} $$
$$(0,1)^{3-x}=(10^{-1})^{3-x}=10^{-3+x} $$
the quotient will be $$10^{(-2x+2)-(-3+x)}=10^{5-3x} =10^{-4x-2}$$
thus $$5-3x=-4x-2$$ and $$x=-7$$
On
The first thing you should have in mind is to convert everything to base $10$. Note that: $$a^{b+c}=a^ba^c$$
$$a^{b-c}=\dfrac{a^b}{a^c}$$ $$(0.1)^a=10^{-a}$$
This should help you.
Hint(s):
$$0.01=10^{-2}\implies (0.01)^{x-1}=(10^{-2})^{x-1}=10^{-2x+2}$$$$0.1^{3-x}=(10^{-1})^{3-x}=10^{x-3}$$ Using the second property of exponents I had mentioned: $$\dfrac{10^{-2x+2}}{10^{x-3}}=10^{-2x+2-(x-3)}$$
it is $$\frac{(10^{-2})^{x-1}}{(10^{-1})^{3-x}}=10^{--4x-2}$$ and this is $$\frac{10^{-2x+2}}{10^{-3+x}}=10^{-4x-2}$$ and you will get $$10^{-2x+2-(x-3)}=10^{-4x-2}$$ so $$-2x+2-(x-3)=-4x-2$$