Latest Elitmus Aptitude Question SOLUTION: log(e(e(e...)^1/3)^1/3)^1/3 what will be its value?

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(#M40040068) ELITMUS Category question Keep an EYE

log(e(e(e...)^1/3)^1/3)^1/3 what will be its value?

Asked In Elitmus Jayesh Hazari (9 years ago)

Solved subodh khandelwal Read Solution (3)

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subodh khandelwal  (9 years ago) Best Solution

let us assume [log(e(e(e..)^1/3)^1/3)^1/3] = x
now as log m^n = n log(m)
we can write it as 1/3[log(e(e(e...)^1/3)^1/3)] = x
now 1/3[log(e)+log(e(e..)^1/3)^1/3)] = x
now we can see that it is again our question so we can replace it as x
so 1/3(1+x)= x
=> 3x = 1+x
=> x= 1/2

Is this solution Helpfull? Yes (21) | No (4)  

DINKAR BHARTI  (9 years ago)

e(e(e..)^1/3)^1/3)^1/3=e^(1/3+1/9++1/27+......infinte series)=e^(1/3)/(1-1/3)=e^1/2
now,
loge^1/2=1/2loge=1/2*1=1/2 since loge base e=1

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Ashutosh Ranjan  (9 years ago)

Let x=(e(e(e.........)^1/3)^1/3)^1/3)
now squaring both side we get
x^3=e(e(e(e.........)^1/3)^1/3)^1/3)
i.e., x^3=e*x [since x=(e(e(e.........)^1/3)^1/3)^1/3)]
therefore x=e [since x can not be zero]
Finally, log(x)=log(e^1/2)=1/2

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