If a,b,c are distinct positive numbers each different from 1 such that
$[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0$ then abc

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Solution

The correct option is A 1
Changing all the logarithms to base $α (α > 0, α \neq 1)$ given equation yields
$\sum [\frac{x}{y} . \frac{x}{z} - 1] = 0 w h e r e x = l o g_{α} α$ etc.
or $\frac{x^{2}}{y z} + \frac{y^{2}}{z x} + \frac{z^{2}}{x y} = 3$
or $x^{3} + y^{3} + z^{3} - 3 x y z = 0$
or $[x + y + z] [x^{2} + y^{2} + z^{2} - x y - y z - z x] = 0$ ... (1)
Since $x \neq y \neq z$ , we have
$x^{2} + y^{2} + z^{2} - x y - y z - z x$
$= \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}] \neq 0$
Hence we conclude from (1)that\\
(x+y+z)=0, that is
$l o g_{α} a + l o g_{α} b + l o g_{α} c = 0$
$l o g_{α}$ abc = 0 or abc =1

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Similar questions

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0, t h e n a b c =

[({log}_{b} a) ({log}_{c} b) ({log}_{a} c)]

If $\frac{log a}{b - c} = \frac{log b}{c - a} = \frac{log c}{a - b}$ then prove that $a^{a} . b^{b} . c^{c} = 1$

log (\frac{a}{b - c}) = log (\frac{b}{c - a}) = log (\frac{c}{a - b})

a^{a} \times b^{b} \times c^{c} = 1

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