Show that 1-:logx to the base 2 +1-:logx to the base 3 +........+1-:logx to the base 43 =1-:log x to the base 431 ?

Question

2432 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-03T14:34:54

The result is not $1-:log_431x$ but $1:-log_(43!)x$

To simplify $1-:log_2x+1-:log_3x+1-:log_4x........+1-:log_43x$

we can use the identity that $log_abxxlog_ba=1$ or $1/log_ab=log_ba$

Hence $1-:log_2x=log_x2$ , $1-:log_3x=log_x3$ and so on.

Hence $1-:log_2x+1-:log_3x+1-:log_4x........+1-:log_43x$

= $log_x2+log_x3+log_x4+....+log_x43$

= $log_x(2xx3xx4xx.....xx43)$

= $log_x43!$

= $1-:log_43!x$

It is obvious that answer is not $1-:log_431x$ but $1-:log_(43!)x$