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

x=log(a)bc, y = log(b)ca and z = log(c)ab

To prove x+y+z+2 =xyz.

Solution:

log(a)bc =logbc/loga = (logb+logc)/loga, log(b) ca = (logc+loga)/logb and log (c) ab = (loga+logb)/logc.

Therefore,

x+y+z+2 = (logb+logc)/loga +(logc+loga)/logb+(loga+logb)/logc + 2

xyz = {(logb+logc)(logc+loga)(loga+logb)}/{(loga)(logb)(logc)}

Also we know that (m+n)/l + (n+l)/m+((l+m)/n +2 = (l+m)(m+n)(n+l)/(lmn) is an identity. Using this we can write:

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

Threfore,

x+y+z+2 = xyz