Question

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The value of
${(0.16)}^{ log(2.5) } \times ( \frac{1}{3} + \frac{1}{ {3}^{2} } + \frac{1}{ {3}^{3} }..... \infty )$
is :

(a) 1
(b) 3
(c) 4
(d) None of these.



I need well explained answer.

Answer 1

Before solving this question , you have to be acquainted with certain identities:
These are basic formulas

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

hope this helps you

Answer 2

Step-by-step explanation:

Answer: 4 because once check in photo and if u like join our school NUM 6305510459