pacifist85 wrote:
Hello,
I ended up in the same but in a more complicated way perhaps. I don't know why, as I always use this proposed way by bunuel. But, this is a good oportunity to ask something.
So, from [1] I ended up that x=1, k=1 or that x=2 and k=0 (or the opposite). So, NS.
From [2] I concluded that k should be zero, so x will be greater than k, like this:
9^x*3^k=81
3*(3^x*1^k)=3^4
3*3^x=3^4, so again not sufficient.
From [1 and 2], as 1 in whichever power is 1, and because in [2] we have 1^k, I accepted that k=0 and x=2.
Anyway, this could be wrong, but would like to ask about the part in red. Can you take the common factor 3 out like this?
Or perhaps you cannot do that due to the order of calculations? So, we would need to resolve the powers first, and then take the common factor out..