Question

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

Answer 1

Given,

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

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

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

Now we have

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c$ [ Using (1) ,(2) and (3)]

or, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\log }_{abc}abc$

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

or, $\frac{xy+yz+zx}{xyz}=1$

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