IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Anupamagrawal13a

Question-_{[ x]}

_{[ infinity]} (0.2)^log[

5]^{(1/4+1/8+1/16+................to n terms)_{is equal to}}

_^a)2

_^b)4

_^c)8

_^d)0

_{^{Answer is b}}

_{^{Note that}}

5 is base of logarithm.

Please solve it

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

kishan12

Hey i have a small dout.....
the above expreesion has summation upto n terms and not inf..
so should we not simplify things and then apply limits????or one can apply limits directly??

venkat_tatolu

While doing limit problems in descriptive type exam.,first simplify the function and apply limit.

In objective type exam , follow as in above solution.

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