Why ##a^0=1##?

[Roberto Pavani]

Equation 1 has no solution: $2^x > 0$ for all $x \in \mathbb{R}$. The only "value" satisfying $2^x = 0$ is $x = -\infty$, which is not a real number, and certainly cannot equal $0$.

 

[weirdoguy]

I decided to solve following equations:

Where in the world did you even get this from?

 

[sophiecentaur]

I decided to solve following equations:

1) Let 2^x =0

Your reality check should kick in here, at this stage. Your equation 1) has to be dodgy. We've all agreed that 2⁰ = 1 and all the various 'justifications / proofs are up there.
Any magnitude less than 1 has to be 2 raised to the power of a number less than zero (a negative number). Smaller would be more negative. Think of a log scale graph (just to get pictorial); smaller numbers need more and more negative powers of 2, which means further and further left along the X axis. Your piece of paper would need to be infinitely wide, to the left.
If the idea came from someone else than check before you believe anything else they say. If it came from you then just give your head a wobble. and 'Look before you leap.'.etc.