Solve $ \log_3x\log_4x(\log_5x-1)$=$\log_5x(\log_4x+\log_3x)$ Solve $ \log_3x\log_4x(\log_5x-1)$=$\log_5x(\log_4x+\log_3x)$ for $x&gt;0$.


The constants $3$, $4$ and $5$ are meant to be the bases of the logs.
The trick, as N.S.JOHN already commented,is to use a common base.

Let us make the problem even more complex, looking for the solutions of $$\log_a(x)\log_b(x)(\log_c(x)-d)-\log_e(x)(\log_f(x)+\log_g(x)=0$$ and apply for each term $$\log_a(x)=\frac{\log (x)}{\log (a)}$$ After simplification, this write $$\log ^2(x) \left(\frac{\log (x)-d \log (c)}{\log (a) \log (b) \log
   (c)}-\frac{\frac{1}{\log (f)}+\frac{1}{\log (g)}}{\log (e)}\right)=0$$ So, the first solution is $$\log(x)=0\implies x=1$$ and the second one (from the term in parentheses) is given by $$\log(x)=\log (c) \left(\frac{\log (a) \log (b) (\log (f)+\log (g))}{\log (e) \log (f) \log
   (g)}+d\right)\implies x= $$
$$\sum\dfrac{(\log x)^2}{\log3\log4}=\dfrac{(\log x)^3}{\log3\log4\log5}$$

If $\log x=0\iff x=1$

Else $$\sum_{\text{cyc}(3,4)}\dfrac1{\log3\log4}=\dfrac{\log x}{\log3\log4\log5}$$

$$\iff\log x=\sum\log3$$

See Change of Base of Logarithm