SOLUTION: 1/log2a+1/log4a+1/log8a+......upto n term

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Question 991818: 1/log2a+1/log4a+1/log8a+......upto n term
Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
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1%2Flog%282%2Ca%29 + 1%2Flog%284%2Ca%29 + 1%2Flog%288%2Ca%29 + . . . . . upto n term
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Notice that

1%2Flog%28b%2Ca%29 = log%28a%2Cb%29.

It is the consequence of the  "base change formula"  for logarithms  (see any  Algebra  textbook or the lesson  Change of Base Formula for logarithms  in this site).

Therefore,  we can re-write the formula in the form

1%2Flog%282%2Ca%29 + 1%2Flog%284%2Ca%29 + 1%2Flog%288%2Ca%29 + . . . + 1%2Flog%282%5En%2Ca%29 = log%28a%2C2%29 + log%28a%2C4%29 + log%28a%2C8%29 + . . . + log%28a%2C2%5En%29

Next,  notice that   log%28a%2C4%29 = 2%2Alog%28a%2C2%29,   log%28a%2C8%29 = 3%2Alog%28a%2C2%29, . . . ,   log%28a%2C2%5En%29 = n%2Alog%28a%2C2%29.

Thus you have,  actually,  the sum

log%28a%2C2%29 + 2%2Alog%28a%2C2%29 + 3%2Alog%28a%2C2%29 + . . . + n%2Alog%28a%2C2%29 = log%28a%2C2%29.(1 + 2 + 3 + . . . +n),

which is equal to   log%28a%2C2%29.%28%28n%28n%2B1%29%29%2F2%29.

Answer.  The sum is equal to  log%28a%2C2%29.%28%28n%28n%2B1%29%29%2F2%29 = %281%2Flog%282%2Ca%29%29.%28%28n%28n%2B1%29%29%2F2%29.

Thank you for submitting so exciting problem.