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

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?

• $y=\frac{\ln a}{\ln abc}+\frac{\ln b}{\ln abc}+\frac{\ln c}{\ln abc}$ – Vikram Jun 19 '15 at 6:46
• Just curious, how did you correctly guess the answer to be 1. Wonder where your intuition is coming from. – Indominus Jun 19 '15 at 6:48
• If it has a constant as its solution, it's easiest to see that when $a=b=c$ we have $\frac{1}{3}$ for each fraction that sum to 1. When there are a lot of variables and you want to experiment or get familiar with an unknown expression you usually test cases like $a=0, b=0, c=0, a=b=c$, etc. These cases are the extreme cases and might shed some light on the behavior of the function. (Just plugging in values and seeing what happens does not, of course, constitute a proof. They just give a good background for you to work with, to make conjectures like OP did, etc.) – Linus S. Jun 19 '15 at 6:53
• @Indominus My six sense. – MathLOL Jun 19 '15 at 6:56

Hint:

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

• I posted my answer. Can you help me check it? Thanks – MathLOL Jun 19 '15 at 6:52

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

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

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

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

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.