Question 1087381
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Next step notice that

{{{3/4*4/5*5/6*6/7*7/8*8/9}}} = {{{3/9}}} = {{{1/3}}}    (after canceling the terms in the numerator and denominator)


{{{4/5*5/6*6/7*7/8*8/9*9/10}}} = {{{4/10}}}       (after canceling the terms in the numerator and denominator)


Then {{{log(3, (3/4*4/5*5/6*6/7*7/8*8/9))}}} = {{{log(3, (1/3))}}} = -1   and


{{{log((4/5*5/6*7/8*8/9*9/10))}}} = {{{log(10, (4/5*5/6*7/8*8/9*9/10))}}} = {{{log(10,(4/10))}}} = {{{log(10,(4))-1}}} = {{{2*log(10,(2))-1}}},


so your answer is 

. . . = {{{(-1)/(2*log(10,(2))-1)}}} = {{{1/(1-2*log(10,(2)))}}}.
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