anonymous
  • anonymous
http://i1142.photobucket.com/albums/n608/nancylam67/Page31744.jpg
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
never ?
anonymous
  • 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.
anonymous
  • anonymous
thank, you get medal for this

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anonymous
  • anonymous
You're welcome.
anonymous
  • anonymous
i know the answer its4x67xuyx=9

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