Is #x^y*x^z=x^(yz)# sometimes, always, or never true?
1 Answer
Jul 27, 2017
Explanation:
If
#x^y*x^z = 0^y*0^z = 0*0 = 0 = 0^(yz) = x^(yz)#
If
#x^y*x^z = x^0*x^0 = 1*1 = 1 = x^0 = x^(0*0) = x^(yz)#
If
#x^y*x^z = 1^y*1^z = 1*1 = 1 = 1^(yz) = x^(yz)#
It does not hold in general.
For example:
#2^3*2^3 = 2^6 != 2^9 = 2^(3*3)#
Footnote
The normal "rule" for
#x^y*x^z = x^(y+z)#
which generally holds if