Solve (sic) for #z#: #y^z/x^4 = y^3/x^z# ?

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Answer 1 · 2017-01-05T05:19:30

#z=(4lnx+3lny)/(lnx+lny)#

Clearly you can't "solve" for #z# here but you can express #z# in terms of #x# and #y# as follows:

#y^z/x^4 = y^3/x^z#

Cross multiply #-> x^z*y^z = x^4*y^3#

Take logs of both sides #-> zlnx+zlny = 4lnx+3lny#

#z(lnx+lny)= 4lnx+3lny#

#z=(4lnx+3lny)/(lnx+lny)#

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