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If $\log _b a\cdot\log_c a +\log _a b\cdot\log_c b+\log _a c\cdot\log_b c=3$ and $a,b,c$ are different positive real numbers not equal to 1, then find the value of $abc$.

I tried to simplify this by different methods like using the identity $\log_b a=\frac{1}{\log_a b}$, but I couldn't get anywhere. Addition, subraction can't be used. I am not able to figure out a way to simplify the LHS. It would be great if someone could help me to proceed with this problem.

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    $\begingroup$ Hint: $\log_a b = \frac{\ln b }{\ln a}$. Apply this to all terms, then simplify using log rules. $\endgroup$ Commented Dec 15, 2016 at 12:43

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If you set $X=\log a$, $Y=\log b$ and $Z=\log c$ (any base you like), your condition becomes easily $$ \frac{X^3+Y^3+Z^3}{XYZ}=3 $$ Now you can use the identity $$ X^3+Y^3+Z^3=(X+Y+Z)^3-3(X+Y+Z)(XY+YZ+ZX)+3XYZ $$ to get that $$ (X+Y+Z)\bigl((X+Y+Z)^2-3(XY+YZ+ZX)\bigr)=0 $$ The second factor can be written $$ X^2+Y^2+Z^2-XY-YZ-ZX $$ and we should consider the quadratic form having as matrix $$ A=\begin{bmatrix} 1 & -1/2 & -1/2 \\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1 \end{bmatrix} $$ Since $$ \det[1]=1>0, \qquad \det\begin{bmatrix}1 & -1/2 \\ -1/2 & 1\end{bmatrix}=3/4>0 \qquad \det A=0 $$ the quadratic form is positive semidefinite. Its null space contains the vector $[1\;1\;1]^T$, so we can conclude that your condition implies $$ X+Y+Z=0\qquad\text{or}\qquad X=Y=Z $$ Since by assumption $a$, $b$ and $c$ are pairwise distinct, we end with $X+Y+Z=0$.


Without quadratic forms, you can reason about $X^2+Y^2+Z^2-XY-YZ-ZX=0$ as follows. Since $Z\ne0$ by assumption, we can set $X=uZ$ and $Y=vZ$, so the equation becomes $$ u^2-uv+v^2-u-v+1=0 $$ and, solving with respect to $v$, $v^2-v(u+1)+u^2-u+1=0$, the discriminant is $$ (u+1)^2-4(u^2-u+1)=-3(u-1)^2 $$ so the equation has a solution only for $u=1$, which gives $v=1$. Therefore $X=Y=Z$.


Another “elementary” approach. Suppose $(X+Y+Z)^2=3(XY+YZ+ZX)$. Set $s=X+Y+Z$ and $p=XYZ$; then $X$, $Y$ and $Z$ are the roots of the equation $$ t^3-st^2+\frac{s^2}{3}t-p=0 $$ by Viète's formulas. We can complete the cube getting $$ \left(t-\frac{s}{3}\right)^3=p-\frac{s^3}{27} $$ which should have three real roots. This is impossible unless the roots are coincident, so $p=s^3/27$ and $X=Y=Z$.

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  • $\begingroup$ Can't be understood If you don't know linear algebra, I suggest to replace it with a non-negative discriminant argument. $\endgroup$
    – i9Fn
    Commented Dec 15, 2016 at 14:05
  • $\begingroup$ @i9Fn Agreed, but first of all I wanted to write a correct answer, which the others aren't. $\endgroup$
    – egreg
    Commented Dec 15, 2016 at 14:10
  • $\begingroup$ I cannot express how much I appreciate this answer. I tried to write it myself, and realized I didn't know how. It definitely fills out the interesting (and more challenging) part of the original problem. $\endgroup$ Commented Dec 15, 2016 at 14:59
  • $\begingroup$ @JohnHughes Quadratic forms are a big fun. ;-) $\endgroup$
    – egreg
    Commented Dec 15, 2016 at 15:39
  • $\begingroup$ Oh, yeah. Esp. over $\Bbb Z/ 2\Bbb Z$, where there's the Arf invariant. :) $\endgroup$ Commented Dec 15, 2016 at 15:42
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We can write $\log_{b}a = \frac{\log a}{\log b}$ and similarly $\log_{c}a = \frac{\log a}{\log c}$. Expanding like this, we have, $$\log_ba\log_ca + \log_ab\log_cb + \log_ac\log_bc =3$$ $$\Rightarrow \frac{(\log a)^2}{\log b\log c} + \frac{(\log b)^2}{\log a\log c} + \frac{(\log c)^2}{\log a\log b} =3$$ $$\Rightarrow \frac{(\log a)^3 + (\log b)^3 + (\log c)^3}{\log a\log b\log c} =3$$ $$\Rightarrow (\log a)^3 + (\log b)^3 + (\log c)^3 = 3\log a\log b\log c...(1)$$ Let $\log a =x, \log b=y, \log c =z$. Then equation $(1)$ transforms to $$x^3+y^3+z^3 = 3xyz \Leftrightarrow x+y+z=0$$ Then, $$\log a+\log b+\log c=0$$ $$\Rightarrow \log abc =0$$ $$\Rightarrow abc =10^0=1$$ Hope it helps.

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    $\begingroup$ Also: the fact that $x+y+z = 0$ solves the equation doesn't mean that every solution has the property that $x +y+z = 0$. $\endgroup$ Commented Dec 15, 2016 at 13:02
  • $\begingroup$ That double-implication is not the result of transforming the previous equation as claimed; it's a separate result, and requires justification. $\endgroup$ Commented Dec 15, 2016 at 13:47
  • $\begingroup$ @JohnHughes Also see Roman83's answer below. $\endgroup$
    – user371838
    Commented Dec 15, 2016 at 13:49
  • $\begingroup$ Right: Roman83 skips the same step that you do. That doesn't make it right. :) $\endgroup$ Commented Dec 15, 2016 at 13:51
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    $\begingroup$ Also note that your conclusion that $x^3 + y^3 + z^3 = 3xyz$ is equivalent to $x+y+z = 0$ is simply false: consider the case $x = y = z = 1$. $\endgroup$ Commented Dec 15, 2016 at 15:52
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Use $\log_a b=\frac{\ln a}{\ln b}$ $$\log _b a\cdot\log_c a +\log _a b\cdot\log_c b+\log _a c\cdot\log_b c=3$$ $$\frac{\ln a}{\ln b}\cdot \frac{\ln a}{\ln c}+\frac{\ln b}{\ln a}\cdot\frac{\ln b}{\ln c}+\frac{\ln c}{\ln a}\cdot\frac{\ln c}{\ln b}=3$$ $$\ln^3a+\ln^3b+\ln^3c=3(\ln a\ln b\ln c)$$ But $x^3+y^3+z^3=3xyz \Leftrightarrow x+y+z=0$

Then $\ln a+\ln b+\ln c=0$ $$\ln abc=0$$ Then $$abc=1$$

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  • $\begingroup$ Your conclusion that $x^3 + y^3 + z^3 = 3xyz$ is equivalent to $x+y+z = 0$ is simply false: consider the case $x = y = z = 1$. (This possibility, in the problem, is ruled out by the assumption that $a,b,c$ are distinct, an assumption that your solution never uses.) $\endgroup$ Commented Dec 15, 2016 at 15:53
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Got it!
$\log _b a\cdot\log_c a +\log _a b\cdot\log_c b+\log _a c\cdot\log_b c=3$
$$\frac{{(\log a)}^2}{\log b\cdot\log c} +\frac{{(\log b)}^2}{\log a\cdot\log c} +\frac{{(\log c)}^2}{\log a\cdot\log b} =3$$

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

$${(\log a)}^3+{(\log b)}^3+{(\log c)}^3=3\log a\log b\log c$$

This means $ \log a+\log b+\log c=0 \Rightarrow \log{abc}=0$
Hence, $abc=1$

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  • $\begingroup$ The jump to "this means that..." could use a little justification; otherwise, you're good. $\endgroup$ Commented Dec 15, 2016 at 13:03
  • $\begingroup$ See egreg's answer to fill in the justification for the "this means that". $\endgroup$ Commented Dec 15, 2016 at 15:00

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