Evaulate :

$$ \frac{1}{\log_{xy} (xyz)} + \frac{1}{\log_{yz} (xyz)} + \frac{1}{\log_{zx} (xyz)} $$

I think that the following property of log will be used:

$$ \log_a (b) * \log_b (c) * log_c (a) = 1 $$

But I don't know how?


Use the fact that

$$\log_a b = \frac{\ln b}{\ln a}.$$


$$\frac{1}{\log_{xy}(xyz)} = \frac{\ln (xy)}{\ln(xyz)}$$



By using the identity $$ \log_b(a)=\frac{\ln{a}}{\ln{b}} $$ Your equation becomes: $$ \frac{\ln (x y)}{\ln (x y z)}+\frac{\ln (x z)}{\ln (x y z)}+\frac{\ln (y z)}{\ln (x y z)}$$ Which you could further simplify to (using the identity $\ln{(a\times b)}=\ln{a}+\ln{b}$:

$$ \frac{\ln (x y)+\ln (x z)+\ln (y z)}{\ln (x y z)}= \frac{2\ln (x)+2\ln (y)+2\ln (z)}{\ln (x y z)} =\frac{2(\ln (x)+\ln (y)+\ln (z))}{\ln (x y z)}$$ $$=\frac{2(\ln (x y z ))}{\ln (x y z)}=2$$


Hint :

$$\log (ab)=\log a + \log b$$

$$\log_a a=1$$

$$\log_a b = \frac{\log b}{\log a}$$

I am in hurry.. Please see if this can help you... If not i hope some one would help...

  • $\begingroup$ isn't $$ \log_a (b) = \frac {log_c (b)}{log_c (a)}$$ ?? $\endgroup$ – kartikeykant18 Apr 16 '14 at 12:18
  • $\begingroup$ @kool_kartikey : Both are one and the same... I should have been careful though! $\endgroup$ – user87543 Apr 17 '14 at 5:38

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