Question

If a,b,c are distinct positive numbers each different from 1 such that
$[lo{g}_{b}alo{g}_{c}a-lo{g}_{a}a]+[lo{g}_{a}blo{g}_{c}b-lo{g}_{b}b]+[lo{g}_{a}clo{g}_{b}c-lo{g}_{c}c]=0$ then abc

• A. 1
• B. 2
• C. 3
• D. None

Answer 1

The correct option is A 1

Changing all the logarithms to base $\alpha (\alpha >0,\alpha \ne 1)$ given equation yields
$\sum [\frac{x}{y}.\frac{x}{z}-1]=0wherex=lo{g}_{\alpha }\alpha$ etc.
or $\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3$
or $x^3+y^3+z^3-3xyz=0$
or $[x+y+z][x^2+y^2+z^2-xy-yz-zx]=0$ ... (1)
Since $x\ne y\ne z$, we have
$x^2+y^2+z^2-xy-yz-zx$
$=\frac{1}{2}\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]\ne 0$
Hence we conclude from (1)that\\
(x+y+z)=0, that is
$lo{g}_{\alpha }a+lo{g}_{\alpha }b+lo{g}_{\alpha }c=0$
$lo{g}_{\alpha }$ abc = 0 or abc =1