Question

if x=logbc base a, y=logca base b, z=logab base c then prove That x+y+z=xyz-2

Answer 1

I hope this helps.
You will not get the answer if you try it one go by solving only the LHS. In the rough try to solve the RHS and back trace the question till you get an expression in the LHS and just write the steps in reverse in the RHS side

Answer 2

Using  property:- $p=\log_qr\implies q^{p}=r$

$x=\log_abc\implies a^{x}=bc\\\;\\y=\log_bca\implies b^{x}=ca\\\;\\z=\log_cab\implies c^{x}=ab$

Now,

$a^{xyz}=(a^{x})^{yz}\\\;\\=(bc)^{yz}\\\;\\=(b^{y})^{z}\times(c^{z})^{y}\\\;\\=(ca)^{z}\times(ab)^{y}\\\;\\=c^{z}.a^{z}\times a^{y}.b^{y}\\\;\\=ab.a^{z+y}.b^{y}\\\;\\=a^{1+z+y}\times b.b^{y}\\\;\\=a^{1+z+y}\times b.ca\\\;\\=a^{1+z+y+1}\times bc\\\;\\=a^{2+z+y}.a^{x}\\\;\\a^{xyz}=a^{x+y+z+2}\\\;\\xyz=x+y+z+2\\\;\\xyz-x-y-z=2$