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Find the sum of all solutions to \begin{align*} (\log_2 x)(\log_3 x)(\log_4 x)(\log_5 x) &= (\log_2 x)(\log_3 x)(\log_4 x) + (\log_2 x)(\log_3 x)(\log_5 x) \\ &\quad + (\log_2 x)(\log_4 x)(\log_5 x) + (\log_3 x)(\log_4 x)(\log_5 x). \end{align*}

I have no idea hows to do this. Can someone help me?

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Hint: Write $$\log_{2}{x}=\frac{\ln(x)}{\ln(2)}$$ etc then it is $$\frac{(\ln(x))^4}{\ln(2)\ln(3)\ln(4)\ln(5)}=\frac{\ln(x)^3}{\ln(2)\ln(3)\ln(4)}+\frac{\ln(x)^3}{\ln(2)\ln(3)\ln(5)}+\frac{\ln(x)^3}{\ln(2)\ln(4)\ln(5)}+\frac{\ln(x)^3}{\ln(3)\ln(4)\ln(5)}$$

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  • $\begingroup$ The hint helps a lot. Thanks to you, I was able to solve this problem $\endgroup$ – sumi Apr 13 '19 at 4:30
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Hint:

If $$abcd=abc+bcd+cda+dab\ \ \ \ (1)$$

If $a=0,bcd=0$

Else if $abcd\ne0$

$(1)\implies a^{-1}+b^{-1}+c^{-1}+d^{-1}=1$

Now $\log_2x=\dfrac1{\log_x2}$

Finally use $\log (xyz\cdots)=\log x+\log y+\log z+\cdots$

where each of the logarithms remains defined

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  • $\begingroup$ $x=1\implies abcd=0$ otherwise $abcd\ne0$ $\endgroup$ – Peter Foreman Apr 12 '19 at 18:43

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