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...

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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