Is #x^y*x^z=x^(yz)# sometimes, always, or never true?

Question

3688 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-27T22:50:49

#x^y*x^z = x^(yz)# is sometimes true.

If #x = 0# and #y, z > 0# then:

#x^y*x^z = 0^y*0^z = 0*0 = 0 = 0^(yz) = x^(yz)#

If #x != 0# and #y = z = 0# then:

#x^y*x^z = x^0*x^0 = 1*1 = 1 = x^0 = x^(0*0) = x^(yz)#

If #x = 1# and #y, z# are any numbers then:

#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)#

#color(white)()#
Footnote

The normal "rule" for #x^y*x^z# is:

#x^y*x^z = x^(y+z)#

which generally holds if #x != 0#