Show that $\frac 1{\log_2x}+\frac 1{\log_3x}+\cdots+\frac 1{\log_{43}x}=\frac 1{\log_{43!}x}$

Question

Show that $\frac 1{\log_2x}+\frac 1{\log_3x}+\cdots+\frac 1{\log_{43}x}=\frac 1{\log_{43!}x}$.I am just not able to get it.please help.

Answer 1

Hints $$\log_b x = \frac{\log x}{\log b} \qquad \log xy = \log x + \log y$$