Where did I go wrong?

[Mandides]

My steps were:
y^0.5 . y⁴
= y^4.5

Help!

 

[Deleted member 4993]

View attachment 23490
My steps were:
y^0.5 . y⁴
= y^4.5

Help!

If 'x' indicates multiplication, then you are correct.

Are you saying that you "know the answer" and it does match? Please explain!!

 

[Harry_the_cat]

Is the radical (square root) sign meant to apply only to the first y?

 

[HallsofIvy]

View attachment 23490
My steps were:
y^0.5 . y⁴
= y^4.5

Help!

First are those "x"s multiplications or variables? It is a very bad idea to use "x" to mean multiplication in algebra!

Also does the square root apply to only the first "y" or to the entire expression. If yes to both of those then \(\displaystyle \sqrt{y(y^3)(y)}= \sqrt{y^{1 + 3+ 1}}= \sqrt{y^5}= (y^5)^{1/2}= y^{5/2}= y^{2.5}\)

IF the square root applied only to the first y THEN you would have \(\displaystyle \sqrt{y}(y^3)(y)= y^{1/2}(y^4)= y^{4.5}\).

So which is it?