# exponents and logarithms question

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?

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

• The hint helps a lot. Thanks to you, I was able to solve this problem
– sumi
Commented Apr 13, 2019 at 4:30

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

• $x=1\implies abcd=0$ otherwise $abcd\ne0$ Commented Apr 12, 2019 at 18:43