Evaluate the log expression

Question

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?

Answer 1

Use the fact that

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

Thus

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

etc.

Answer 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...

Answer 3

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 $\ln{(a\times b)}=\ln{a}+\ln{b}$:

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