Latest Elitmus Aptitude Question SOLUTION: log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3 opt a)0 b)1 c)1/3 d) not remember

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(#M40152588) ELITMUS Numerical Ability question Keep an EYE

log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3

opt
a)0 b)1 c)1/3 d) not remember

Asked In Elitmus Manish Agrawal (8 years ago)

Unsolved Read Solution (10)

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kumar saurav  (8 years ago)

ans-1/2
log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3
loge^1/3+1/9+1/27+....
since 1/3+1/9+1/27+.... is in GP so
sum= (1/3)/(1-1/3)
1/2
so
loge^1/2
=1/2

Is this solution Helpfull? Yes (17) | No (1)  

Abhishek kumar keshri  (8 years ago)

let e(e(e(e(e...)^1/3)^1/3)^1/3....=x
then (e*x)^1/3=x
e^1/3=x^2/3
e=x^2
e^1/2=x
taking log both side then log x=1/2log e
then log x=1/2
answer is 1/2.

Is this solution Helpfull? Yes (8) | No   

Jyothi Prakash Thaduri  (8 years ago)

log(e(e(e(e....)^1/3)^1/3)^1/3)^1/3
lets take from inner most 3 values as
(e(e(e^1/3)^1/3)^1/3)
e^1/3(e(e^1/3)^1/9)
e^1/3*e^1/9*e^1/27.....
e^(1/3+1/9+1/27+......)
e^((1/3)/(1-1/3)) as from GP infinite sum....
e^(1/2)
now loge^1/2 with base e
1/2 is answer.......
if log(e(e(e...)^1/2)^1/2)^1/2 with base e will give.. 1
e^(1/2)(1-(1/2))
e^1
loge^1 with base e gives 1

Is this solution Helpfull? Yes (5) | No   

Dilraj vyas  (8 years ago)

let log(e(e(e....)^1/3)^1/3)^1/3 = x
=> x = (1/3)*[(loge) + log(e(e(e....)^1/3)^1/3)^1/3 ]
=> x = (1/3)*(1 + x)
=> 3x = 1 + x
=> x = 1/2

Is this solution Helpfull? Yes (4) | No   

Mohammadzaki  (8 years ago)

Let, loge(e(e(e...)1/3)1/3))1/3 = x
=> (1/3)loge(e(e(e...)1/3)1/3)) = x
=> (1/3)[logee + loge(e(e(e...)1/3)1/3))1/3] = x
=> (1/3)[1 + x] = x
=> 1 + x = 3x
=> x = 1/2

Is this solution Helpfull? Yes (3) | No   

Amit Singh Deopa  (8 years ago)

ans = 1/d
log(e(e(e.....)^1/3)^1/3)^1/3) = z
1/3 log(e(e(e(.....)^1/3^1/3) = z
1/3 [loge + log(e(e(e.....)^1/3)^1/3)^1/3)] = z
1/3 [1 + z] = z
z = 1/2

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Abhishek Kakad  (8 years ago)

Ans is b)1.
bcoz log e^anything = 1.

Is this solution Helpfull? Yes | No (8)  

jojo  (8 years ago)

answer is 2 ...this question is very common and repeated many a time in elitmus

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jitendra  (8 years ago)

log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3=1/3(loge+1/3(log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3))
let log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3=p
logp=1/3(loge+1/3(logp))
logp -1/9logp=1/3*1
8/9logp=1/3
p=e^(3/8)
log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3=logp=3/8

Is this solution Helpfull? Yes | No (2)  

jitendra  (8 years ago)

let logp=log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3
logp=1/3(1+log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3))
logp=1/3(1+logp)
logp=1/2
log(e(e(e(e.........)^1/3)^1/3)^1/3)^1/3)^1/3=1/2

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