Value of $\frac{1}{\log_aabc}+\frac{1}{\log_babc}+\frac{1}{\log_cabc}$

Question

How to find the value of $\frac{1}{\log_aabc}+\frac{1}{\log_babc}+\frac{1}{\log_cabc}$?

I guess the answer will be $1$. But I don't know how to evaluate it. Can someone give me some tips?

Answer 1

Hint:

$$ \log_x y = \frac{\ln y}{\ln x}$$ Thus $$ \log_a abc = \frac{\ln abc}{\ln a} = \frac{\ln a + \ln b + \ln c}{\ln a}$$

Answer 2

I don't know whether am i correct or not.

$\frac{1}{\log_aabc}+\frac{1}{\log_babc}+\frac{1}{\log_cabc}=\frac{1}{\frac{\ln abc}{\ln a}}+\frac{1}{\frac{\ln abc}{\ln b}}+\frac{1}{\frac{\ln abc}{\ln c}}$

$=\frac{\ln a}{\ln abc}+\frac{\ln b}{\ln abc}+\frac{\ln c}{\ln abc}$

$=\frac{\ln a+\ln b+\ln c}{\ln abc}=\frac{\ln (abc)}{\ln abc}=1$

Answer 3

The reciprocal logarithm identity $$\frac{1}{\log_ab} = \log_ba$$ follows from the change of base identity, or quite directly from the modern definition $\log_ab = \frac{\ln b}{\ln a}$. It's handy to know. A glance at the expression with this identity in mind yields $$\log_{abc}a + \log_{abc}b + \log_{abc}c$$ which equals 1 by basic properties of logarithms.