If #log_8(y*(x+2))=z-1/3# and #log_2((x-2)/y)=2z+1#, how to show that #x^2=32^z+4#?

Question

923 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-20T06:28:48

#log_8(y*(x+2))=z-1/3#

#=>y*(x+2)=8^(z-1/3)=8^zxx8^(-1/3)#

#=>y*(x+2)=8^z/2=2^(3z-1)........[1]#

Again

#log_2((x-2)/y)=2z+1#

#=>((x-2)/y)=2^(2z+1).......[2]#

Multiplying [1] and [2] we get

#=>cancely*(x+2)xx((x-2)/cancely)=2^(3z-1)xx2^(2z+1)#

#=>x^2-4=2^(5z)=32^z#

#=>x^2=32^z+4#

Proved