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calculate lg(7/8) + lg(6/7)+... +lg(1/2)

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lexijuly | Student, Undergraduate | (Level 1) Honors

Posted August 18, 2010 at 12:48 AM via web

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calculate lg(7/8) + lg(6/7)+... +lg(1/2)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted August 18, 2010 at 12:52 AM (Answer #1)

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First, we'll use the product property of logarithms: the sum of logarithms is the logarithms of the product:

lg x1+lg x2+lg x3+....+lg xn = lg(x1*x2*x3*...*xn)

In our case:

lg(7/8) + lg(6/7)+... +lg(1/2) = lg [(7/8)*(6/7)*...*(2/3)*(1/2)]

We'll cancel like terms and we'll get:

lg [(7/8)*(6/7)*...*(2/3)*(1/2)] = lg (1/8) = lg (1/2^3)

lg (1/2^3) = lg (2^-3) = -3*lg 2

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 18, 2010 at 12:55 AM (Answer #2)

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Let S = lg(7/8) + lg(6/7) + ...+lg(1/2)

We know that : lg a/b = lg a - lg b

==> lg 7 - lg 8 + lg 6 - lg 7 + lg 5 - lg 6 + ... + lg 2 - lg 3 + lg 1 - lg 2

Now eliminate si,ilar:

=>  S = - lg 8 + lg 1

==> S = lg 1 - lg8

But we now that lg 1 = 0

==> S = -lg 8 = - lg (2^3) = -3lg 2 = 3*lg (1/2)

==> S = 3*log(1/2)

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neela | High School Teacher | (Level 3) Valedictorian

Posted August 18, 2010 at 1:40 AM (Answer #3)

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log (7/8)+log(6/7)+log(5/6)....(1/2) . To find the sum.

 We use the property of logarithm, loga+logb = logab.

So log(7/8)+log(6/7)+log(5/6).... log(1/2)

= log { (7/8)(6/7)(5/6)(4/5)(3/4) (2/3)(1/2)}

= log{(7!/8!) = log(1/8) = log1- log8 = 0- log8 =

= - log8

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jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted August 18, 2010 at 11:52 AM (Answer #4)

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lg(7/8) + lg(6/7)+... +lg(1/2)

=lg[7/8*6/7*.......*1/2]

=lg(1/8)

=lg(1/2^3)

=lg(2^-3)

=(-3)lg(2)

=(-3)*0.3010

=(-0.9030)

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