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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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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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.
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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