# Determine lg(1/2) + lg(2/3) + lg(3/4) + ... + lg(99/100)

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### 3 Answers

lg(1/2)+lg(2/3)+...+lg(99/100)

We know that:

logx1+lgx2+...lg(xn)=lg(x1*x2*...*xn)

==> lg(1/2)+lg(2/3)+...+lg(99/100)=lg(1/2*2/3*...*99/100)

= lg(1/100) = lg(100)^-1

= -log(100)

=-lg(10)^2

=-2lg(10)=-2

To determine lg(1/2)+lg(2/3)+lg(3/4).....log(99/100)

Solution:

We know that lg(a/b)+lg(b/c)+lg(c/d) = lg[(a/b)(b/c(c/d)] = lg(a/d) = lga - lgd.

So lg(1/2)+lg(2/3)+kg(3/4)+....lg(99/100) = lg(1) - lg100 = 0- lg(10^2) = -2.

We'll use the product property of logarithmic function, which means that the sum of logarithms is the logarithm of product, we'll get:

lgx1+lgx2+lgx3+....+lgxn=lg(x1*x2*x3*...*xn)

In our case:

lg(1/2)+lg(2/3)+...+lg(99/100)=

lg((1/2)*(2/3)*(3/4)*...*(98/99)*(99/100))=lg(1/100)=-2

Now, we'll use the quotient property:

**lg(1/100)=lg1 - lg100 = 0 - lg(10)^2 = 0-2lg10 = 0-2 = -2**