Question 1089581
{{{ log(2,(3))log(3,(4))log(4,(5))}}} … {{{log(1023,(1024)) }}}

Using 'log' to denote log base 10 (i.e. the traditional log function):
= {{{ (log((3))/log((2)))*(log((4))/log((3)))*(log((5))/log((4))) }}} … * {{{ (log((1023))/log((1022)))* (log((1024))/log((1023))) }}}

Notice how log(3) cancels with log(3) of the 2nd factor, and then log(4) cancels, etc. leaving us with
= {{{ log((1024)) / log((2)) = log((2^10)) / log((2)) = 10 * log((2))/log((2)) = highlight(10) }}}