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If $a,b,c$ are positive integers such that $$\log_b a × \log_c a+ \log_c b × \log_a b+ \log_a c × \log_b c =3$$ find $a×b×c.$

When I tried to solve this problem I got last step as $\log a × \log a × \log a,$ which is not possible. Please help me.

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  • $\begingroup$ If $a=b=c$ then the equation is satisfied. $\endgroup$
    – lulu
    Commented Jun 9, 2018 at 16:42
  • $\begingroup$ Try using this $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$..The answer could be $1$ $\endgroup$
    – LM2357
    Commented Jun 9, 2018 at 16:50

2 Answers 2

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Since $\log_x(y)=\frac{\ln(y)}{\ln(x)}$ by definition, the condition can be rewritten as :

$$\frac{\ln^2(a)}{\ln(b)\ln(c)}+\frac{\ln^2(b)}{\ln(a)\ln(c)}+ \frac{\ln^2(c)}{\ln(a)\ln(b)}=3$$

Which is equivalent to :

$$\ln^3(a)+\ln^3(b)+\ln^3(c)-3\ln(a)\ln(b)\ln(c)=0$$

Now we can use the following algebraic identity :

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

The second factor in the rhs cannot be zero except if $x=y=z$, because it can be rewritten as :

$$\frac12\left[(x-y)^2+(y-z)^2+(z-x)^2\right]$$

Hence, there are only two possibilities :

$\ln(a)=\ln(b)=\ln(c)$ or $\ln(a)+\ln(b)+\ln(c)=0$

In other words :

$a=b=c$ or $abc=1$

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Use inequality between arithmetic in geometric mean: $$ {\log a\over \log c} {\log a\over \log b}+{\log b\over \log a} {\log b\over \log c}+{\log c\over \log a} {\log c\over \log b}\geq 3\sqrt[3]{ {\log a\over \log c} {\log a\over \log b}{\log b\over \log a} {\log b\over \log c}{\log c\over \log a} {\log c\over \log b}}=3$$

with equality iff $${\log a\over \log c} {\log a\over \log b}={\log b\over \log a} {\log b\over \log c}={\log c\over \log a} {\log c\over \log b}$$

Since we have equality we have $\log ^3a = \log ^3b = \log ^3c$ so $a=b=c$ and thus $abc =a^3$ for any positive integer $a\geq 2$.

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