If x = log[a]bc y = log[b]ca z = log[c]ab ptove that x+y+z+2 = xyz

x = log a (bc)  

y= log b
(ca)

z= log c (ab)

Prove that
:

x+ y + z + 2 = xyz

We will
start from left side:

x+ y + z + 2 = log a (bc) + log b
(ca) + log c (ab) + 2

We know that: log b (x) = log a (x) +
log a (b)

Let us the base 10 as
a:

==>  (log bc/ log a)+(log ca/log b) +(log ab)/
log c) + 2 

We know that log ab = log a + log
b

==> (log c + log b)/ log a +(log c + log a)/log b
+(log a + log b)/log c  + 2 .........(1)

Now let us
calculate xyz:

xyz= log a (bc)* log b (ac)* log c
(ab)

      = log bc/ log a * log ac / log b * log ab/ log
c

      =(log b + log c)(log a + log c)(log a + log b)/ log
a*logb*logc

=[ logb(loga)^2 + logb*logc*loga + logc(loga)^2
+ loga(logc)^2+ loga(logb)^2 + logc(logb)^2 + loga*logb*logc +
logb(logc)^2]/loga*logb*logc

= loga/logc + 1 + loga/logb +
logc/logb + logb/logc + logb/loga + 1 + logc/loga

=
(loga+logb)/logc + (loga+logc)/logb + (logb+logc)/loga + 2
.....(2)

Now we can see that (1) and (2) are
identical.

Then we conclude
that:

x+y+z+2 = xyz