Sum to n terms and to infinity when |x| < 1 :
log(1+x)+log(1+x² ) + log(1+ x⁴ ) + log(1+x⁸ )...

log(1+x) can be written as a power series.
log(1+x) = x - 1/2 x² + 1/3 x - 1/4 x⁴ + ...
Try writing the power series to x⁸ for each of the terms in your list above.
I.e. log(1+x)+log(1+x² ) + log(1+ x⁴ ) + log(1+x⁸ )
and add the coeffiecients of x to form a sinle power series. Do you recognise it?

Have a go and message again if you get stuck

Geoff

Alternatively, you can prove directly that:

log(1+x)+log(1+x²)+¼+log(1+x^2n-1)=log((1-x²ⁿ)/(1-x))

for each natural n.

The result is trivial for n=1. And if it holds for n then upon adding log(1+x²ⁿ) to each side you get:

log(1+x)+log(1+x²)+¼+log(1+x²ⁿ)=log((1-x²ⁿ)/(1-x))+ log(1+x²ⁿ)=log((1-x²ⁿ)(1+x²ⁿ)/(1-x))=log((1-x²ⁿ⁺¹)/(1-x))

Hence result by induction. I'm sure you can figure out the result for the infinite product from here.

Dear Sir,
I don't know whether it is appropriate to ask you this...
I would like a brief history on the topic of logarithms; founder and origin, basically the background information.

Thank-you,
Levu

Levu,
You can find the things you wanted at this site.
http://forum.swarthmore.edu/dr.math/problems/temple.7.12.96.html

Hope this helps.

love arun