If A is not 0 then A^0 = 1 whatever A is. In 3 x^0, which means (3 multiplied by x^0), I deduce that x^0 = 1 so substitute 1 for x^0 and I get (3 multiplied by 1) In (3x)^0, I deduce that (3x)^0 = 1 (I put A = 3x) I get 1.

According to the order of operations 3x^0 = 3 since you take x to the power of 0 first, then multiply that times three, and anything to the power of 0 = 1; now when the 3x are in parenthesis (3x)^0 the answer will be 1 because you take 3 times x first, then you will take that to the power of 0 which is how the answer comes out to be 1

[^^^] = [ Y = X^2 ] [^^^] = [ Y' = 2X^2-1 = 2X^1 = 2X ] [^^^] = [ Y'' = 2(1 * X^1-1) = 2(X^1-1) = 2(X^0) = 2(1) = 2 ] [^^^] = [ X^0 ] = [ X^(0_) ] = [ X^{ (+1) + (-1) ] = [ X^(1-1) ] = [ { X^1 } / { X^1 } ] = [ 1 ] [^^^] = [ X^0 ] = [ X^(_0) ] = [ X^(Nothing) ] = [ Nothing ] = [ 0 ] [ 0 ÷ 0 ] = [ (Nothing) ÷ (Nothing) ] = [ (Nothing) ] = [ 0 ] [ 0 ÷ 0 ] = [ { (Money $1) + (Debt paper in $1) } ÷ (Nothing) ] = [ (Nothing) ] = [ 0 ] [ 0 ÷ 0 ] = [ (Nothing) ÷ { (Money $1) + (Debt paper in $1) } ] = [ (Nothing) ] = [ 0 ] [ 0 ÷ 0 ] = [ { (Money $1) + (Debt paper in $1) } ÷ { (Money $1) + (Debt paper in $1) } ] = [ 1 ] [ 0^0 ] = [ (Nothing)^(Nothing) ] = [ (Nothing) ] = [ 0 ] [ 0^0 ] = [ (Nothing)^{ (Money $1) + (Debt paper in $1) } ] = [ (Nothing) ] = [ 0 ] [ 0^0 ] = [ { (Money $1) + (Debt paper in $1) }^(Nothing) ] = [ (Nothing) ] = [ 0 ] [ 0^0 ] = [ { (Money $1) + (Debt paper in $1) }^{ (Money $1) + (Debt paper in $1) } ] = [ 1 ] http://www.flickr.com/photos/trapassing/3577686187 http://www.flickr.com/photos/trapassing/3577686187/sizes/o http://www.flickr.com/photos/trapassing/3718926763 http://www.flickr.com/photos/trapassing/3718926763/sizes/o