Latest Elitmus Aptitude Question SOLUTION: loge(e(e(e....)^1/2)^1/2)^1/2?

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loge(e(e(e....)^1/2)^1/2)^1/2?

Asked In Elitmus ajay mathela (11 years ago)

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vishal raj  (11 years ago)

log(e(e(e...)^1/2)^1/2)^1/2
=log(e^1/2(e(e...)^1/2)^1/4
=log(e^1/2.e^1/4(e...)^1/8
=log(e^1/2.e^1/4.e^1/8(e...)^1/16)
=1/2loge+1/4loge+1/8loge+....
since loge =1,so
=1/2+1/4+1/8+....
this expression is in G.P.
so here,a=1/2 and r=1/2 and n->infinite,so using formula
s=a/(1-r)
s=(1/2)/(1-(1/2)
s=1
so the answer is 1.

Is this solution Helpfull? Yes (22) | No   

Dolly  (11 years ago)

I think mp is right.

let loge(e(e(e....)^1/2)^1/2)^1/2 = x
=>(1/2) log (e) + log(e(e(e....)^1/2)^1/2)^1/2 = x
=> 1/2 + (1/2)loge(e(e(e....)^1/2)^1/2)^1/2 = x
=> 1/2 + (x/2) = x
=> x = 1

Is this solution Helpfull? Yes (13) | No (2)  

MP   (11 years ago)

loge(e(e(e....)^1/2)^1/2)^1/2 =1

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ajay mathela  (11 years ago)

ans this question ASAP......

Is this solution Helpfull? Yes | No (8)  

ajay mathela  (11 years ago)

@MP-can you solve it plz????//

Is this solution Helpfull? Yes | No (7)  

glory agarwal  (11 years ago)

@mp please explain it.

Is this solution Helpfull? Yes | No (8)  

Vinay Sipani   (11 years ago)

let loge(e(e(e....)^1/2)^1/2)^1/2 = x
=> log (e) + log(e(e(e....)^1/2)^1/2)^1/2 = x
=> 1 + (1/2)loge(e(e(e....)^1/2)^1/2)^1/2 = x
=> 1 + (x/2) = x
=> x = 2

Is this solution Helpfull? Yes | No (28)  

SHREYASH ANAND  (7 years ago)

log e base e is 1 ....so 1^anything is 1..

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