IIT JEE , AIEEE Forum - Mathematics , Algebra

debjyoti

how to find log of

log (1/3+1/9+1/27+1/81...............infinity)

2.5

(0.16)

so is their any derivation of formula through which we can find the log of this type of problem?

sachin_gupta1991

Important thing is that (1/3+1/9+1/27+1/81...............infinity) can be taken as 1/2.

log (1/3+1/9+1/27+1/81...............infinity)

2.5

(0.16)

(0.4)^2log1/2

(2/5)^log1/4{log has base2.5}

(5/2)^-log1/4

(5/2)^log4{log has base 2.5}

=4

Plzzz rate me for my answer.

dr.dane

log (1\3/2\3)

= log(1\2)

= - log2
= - 0.301
it is sum of infinite gp
do rate me

puneet

hey

if I understand the question correctly then sachin has given a right answer ..

cheers

spideyunlimited

yes sachin gupta is correct.. good work...