Question

(0.16)^log2.5(1)3+1/3^2+....+to infinity)?​

Answer 1

infinity

Step-by-step explanation:

the answer will be infinity

Answer 2

Step-by-step explanation:

Before solving this question , you have to be acquainted with certain identities:

a^ [log b (base a)]=b….(1)

a^ [log b (base a^n)]=b^(1/n)……..(2)

sum of infinite G.P, = a/(1-r) …(3)

Solving the exponent part: ⅓ +(⅓)²+..............(⅓)^∞ = (⅓)/ [1-⅓ ] =½…using(3)

0.16= 4/25

2.5=5/2

(4/25) ^ log_(5/2) [ 1/2]

(4/25) ^ log_[(4/25)^(-1/2)] [ 1/2]

Here [(4/25)^(-1/2) is the base =5/2

(4/25) ^ log_[(4/25)] [ 1/2]^(-2) …… using(2) and (1)

= 4 …………..Ans