What relationship between a,b and c ?
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3Hint: $\log_xy=\dfrac{\ln x}{\ln y}$ and $\ln(xyz)=\ln x+\ln y+\ln z$. – Lucian Mar 01 '14 at 19:43
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Transform all the logarithms so they have all the same base, say $abc$. – alex Mar 01 '14 at 19:44
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Thanks to all. @Dietrich Burde can you publish your answer? – user2511140 Mar 01 '14 at 19:54
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I can't found relationship between a,b and c ! – user2511140 Mar 01 '14 at 20:08
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From Lucian's suggestion to transform $\log_a u=\frac{\ln u}{\ln a}$ etc. we find, using $x=\ln a, y=\ln b, z=\ln c$ and cancelling the common $\log u$ factor, that $$\frac1x+\frac1y+\frac1z-\frac{1}{x+y+z}=0.\tag{1}$$ The left side factors into $$\frac{(x+y)(x+z)(y+z)}{xyz(x+y+z)}.\tag{2}$$ Now since we are working with $a,b,c$ being the base of logs, we are assuming each is positive and not $1$, so that $x,y,z$ are nonzero. We also know $x+y+z \neq 0$ since it is the log of $abc$ which also appears as a logarithm base in the original equation.
So the top of $(2)$ must be zero, which means some two of the three logs add to zero, which in turn means that at least one of the three possibilities $ab=1,ac=1,bc=1$ must hold. It is then easy to check that on the other hand provided one of the three products is $1$ then the statement holds for all $u$.
coffeemath
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