## anonymous 5 years ago http://i1142.photobucket.com/albums/n608/nancylam67/Page31744.jpg

1. anonymous

never ?

2. anonymous

You have$x^y.x^z=x^{yz}$We know, by exponent laws, $x^y.x^z=x^{y+z}$so$x^{y+z}=x^{yz}$only if$y+z=zy$If you assume that it is always true, you are assuming$y+z=yz$for any y, z real. So choose y=1 and z=1. The LHS will be 2 and the RHS will be 1. This is a contradiction, so it's not true for all y and z. But, there will be cases when it is true. From the exponent equation above,$y+z-zy=0$and so$y(1-z)+z=0$That is,$y=\frac{z}{z-1}$So, if we choose a z, and then a y as given by the formula, we'll have that$x^y.x^z=x^{yz}$So your statement is *sometimes* true.

3. anonymous

thank, you get medal for this

4. anonymous

You're welcome.

5. anonymous