Question

Prove that: ${\log }_{a}N\cdot {\log }_{b}N+{\log }_{b}N\cdot {\log }_{c}N+{\log }_{c}N\cdot {\log }_{a}N=\frac{{\log }_{a}N\cdot {\log }_{b}N\cdot {\log }_{c}N}{{\log }_{abc}N}$

Answer 1

${\log }_{a}N.{\log }_{b}N+{\log }_{b}N.{\log }_{c}N+{\log }_{c}N.{\log }_{a}N$

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

$=\frac{{\log }_{N}c+{\log }_{N}a+{\log }_{N}b}{{\log }_{N}a.{\log }_{N}b.{\log }_{N}c}$

$=\frac{{\log }_{N}abc}{{\log }_{N}a{\log }_{N}b{\log }_{N}c}$ ($\because \log mnp=\log m+\log n+\log p$)

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

$=\frac{{\log }_{a}N.{\log }_{b}N.{\log }_{c}N}{{\log }_{abc}N}$

Hence, ${\log }_{a}N.{\log }_{b}N+{\log }_{b}N.{\log }_{c}N+{\log }_{c}N.{\log }_{a}N=\frac{{\log }_{a}N.{\log }_{b}N.{\log }_{c}N}{{\log }_{abc}N}$