Evaluate expression with help of properties of logarithms E(x)= lg(1/8)+lg(9/10)+lg(10/11)+...+lg(999/1000)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We use the relation for logarithm that states log a + log b = log (a*b).

E(x)= lg(1/8) + lg(9/10) + lg(10/11) +...+ lg(999/1000)

=> E(x) = lg [ (1/8)*(9/10)*(10/11)*...*(999/1000)]

we see that we can cancel terms which are present in the denominator as well as numerator.

=> E(x) = lg ( (1*9)/(8*1000))

=> E(x) = lg (9 / 8000)

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neela | High School Teacher | (Level 3) Valedictorian

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To evaluate expression with help of properties of logarithms:

E(x)= lg(1/8)+lg(9/10)+lg(10/11)+...+lg(999/1000).

Specially it is  noticed  that there is no term lg(8/9) on the right  side.

Solution:We use the property of logarithms: lga+lgb = lgab.

Therefore E(x) = lg(1/8)+lg(9/10)+(10/11)+lg(11/12)...+lg(998/999)+lg(999/1000).

E(x) = lg(1/8)+lg(9/10)(10/11)(11/12)(12/13)....(998/999)(999/1000)}, as lga+lgb = lgab.

E(x) = lg(1/8) + lg{9/1000) as other terms cancel.

So E(x) = lg{(9/8000), by property lga*lgb = lg ab.

So E(x) = lg9-log8 -lg1000

So E(x) = lg9-lg8 - 3.

If there was the 2nd term  lg(8/9), the sum E(x) = lg(8/9)+lg(9)-lg8-3 = lg8-lg9+lg8-lg9 3 = -3.

So E(x) = lg9-lg8 - 3 = -2.948847478.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply the quotient rule of logarithms:

lg(a/b) = lg a - lg b

According to this rule, we'll get:

lg(1/8) = lg 1 - lg 8 = - lg 8

lg(9/10) = lg 9 - lg 10

lg(10/11) = lg 10 - lg 11

......................................

lg(999/1000) = lg 999 - lg 1000

We'll add the terms from the left side and the terms from the right side:

E(x) = -lg8 + lg 9 - lg 10 + lg 10 - lg 11 + ... + lg 999 - lg 1000

We'll simplify and we'll get:

E(x) = lg (9/8) - 3

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