IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

iitkgp_bipin

1/4 + 1/8 + 1/16 + ....... forms an infinite GP with 1st term, a = 1/4 and common ratio, r = 1/2.

Sum of infinite GP = a / (1-r) = (1/4) / (1 - 1/2) = 1/2

Exponent is log_root5(1/4 + 1/8 + 1/16 + .......) = log_root5(1/2)

Now apply the property of logarithm : log_a^mn = (1/m)log_an

So, exponent = log_5^1/2(1/2) = 2log₅(1/2) = log₅(1/2)² = log₅(1/4)

So the expression becomes (0.2)^(log₅(1/4))

= (1/5)^(log₅(1/4))

= (5)^(-log₅(1/4))

= (5)^(log₅4)

= 4

Ask & Discuss Questions with Community & Experts