2020/004) Prove that $\log_abc.\log_bca.\log_abc = 2 + \log_abc+\log_bca+\log_abc$

Solution
Let $\log_ab= x\cdots(1)$
$\log_bc=y\cdots(2)$
$\log_ca=z\cdots(3)$

So we have $xyz= \log_ab.\log_bc.\log_ca=1 \cdots(4)$
$\log_abc= \log_ab + \log_ac = x + \frac{1}{z}= x + xy= x(1+y)\cdots(5)$
Similarly
$\log_bca= y(1+z)\cdots(6)$
And
$\log_bca= z(1+x)\cdots(7)$


Now $LHS= \log_abc.\log_bca.\log_abc$
$=x(1+y) . y(1+z) . z (1+x)$ from (5),(6),(7)
$= xyz(1+y)(1+z)(1+x)$
$= (1+y)(1+z)(1+x)$ from (4) as xyz=1
$= 1 + y + z + x + xy+yz + zx + xyz$
$= 1 + y + z + x + xy+yz + zx + 1$ from (4) as xyz=1
$= 2 + y + yz + x + xy+ z + zx$ Rearrangement of terms
$=2 + y(1+z) + x(1+y) + z(1+x)$
$=2 +\log_bca + \log_abc + \log_cab$
$=2 + \log_abc + \log_bca + \log_cab$  Rearrangement of terms
$=RHS$


Hence Proved