# Thread: A summation with logs

1. ## A summation with logs

What is the simplified numerical value of the sum:

$\log_{2} (1-\frac{1}{2}) + \log_{2} (1-\frac{1}{3}) + ... \log_{2} (1-\frac{1}{64})$

I'm not sure how to go about solving this problem.

Would it be

$\log_{2} ((1-\frac{1}{2})(1-\frac{1}{3}) etc)$

I can see how other properties of logs would apply as well, but I don't know the best way to go about this, and I'm not sure how to correctly arrive at a final answer

2. Originally Posted by mikedwd
What is the simplified numerical value of the sum:

$\log_{2} (1-\frac{1}{2}) + \log_{2} (1-\frac{1}{3}) + ... \log_{2} (1-\frac{1}{64})$

I'm not sure how to go about solving this problem.

Would it be

$\log_{2} ((1-\frac{1}{2})(1-\frac{1}{3}) etc)$

I can see how other properties of logs would apply as well, but I don't know the best way to go about this, and I'm not sure how to correctly arrive at a final answer
write 1 - (1/2) as 1/2, 1 - (1/3) as 2/3, and so on

now what do you see?

3. I see the pattern:

$\log_{2} (\frac{1}{2}) + \log_{2} (\frac{2}{3}) + \log_{2} (\frac{3}{4}) ... \log_{2} (\frac{63}{64})$

But I don't know how to go about the summation...I feel a bit lost

(0-1) + (1- $\log_{2} 3$)...I really don't know

4. wait is it

$\log_{2} (\frac{1}{64})$

alright well thanks for the tip, it helped me out

But how should I write this conclusion in good mathematics? The way I got to the answer is just as important as the solution.

Should I write
(1/2)(2/3)=1/3
(1/2)(2/3)(3/4)(4/5)=4/5
or something like that? is that sufficient?

5. Is the correct answer $\log_{2} (\frac{1}{64})$

which would make the sum -6

?

6. Hello, mikedwd!

Absolutely correct!

7. Originally Posted by mikedwd
Is the correct answer $\log_{2} (\frac{1}{64})$

which would make the sum -6

?
yes

do you see how? rewriting and then putting them as a product causes a lot of things to cancel out, leaving you with $\frac 1{64} = 2^{-6}$