Question

Let $x=1+{\log }_{a}bc$, $y=1+{\log }_{b}ac$ and $z=1+{\log }_{c}ab$ then prove that $xy+yz+zx=xyz$.

Answer 1

Given $x=1+{\log }_{a}bc$.......(1),

$y=1+{\log }_{b}ac$.......(2)

and $z=1+{\log }_{c}ab$......(3).

From (1) we have,

$x={\log }_{a}a+{\log }_{a}bc$

or, $x={\log }_{a}abc\dots (\log a+\log b+\log c=\log (abc\left)\right)$.

Similarly from (2) and (3) we get, $y={\log }_{b}abc$ and $z={\log }_{c}abc$.

Now,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{{\log }_{a}abc}+\frac{1}{{\log }_{b}abc}+\frac{1}{{\log }_{c}abc}$

$={\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c\dots (\because {\log }_{a}b=\frac{1}{{\log }_{b}a})$

$={\log }_{abc}abc$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

So,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{y}=1$

or, $xy+yz+zx=xyz$.