what is the value of log e e e 1 2 1 2 1 2 a 0 b 1 3

Answer is option 4

Let us take generalize it as log e^1/2n

Since log e=1,

Therefore, 1^1/2n=1, hence our answer is option d) 1.
12 years agoHelpfull: Yes(29) No(29)
e(e(e....)^1/2)^1/2)^1/2 = e ^ (1/2 + 1/4 + 1/8 +....)
it is in G.P(Geometric progression) then sum to infinite terms = (a)/(1-r)
(1/2 + 1/4 + 1/8 +....) = (1/2)/(1-1/2) = 1
so e ^ 1
log(e^1) base e = 1
answer is 1;
11 years agoHelpfull: Yes(23) No(2)
d-1
y=1/2[log(e(e(e(e...))^1/2)^1/2)]
=1/2[1+y]
=>2y=1+y
=>y=1
11 years agoHelpfull: Yes(14) No(0)
it follows like this...
1/2(log(e(e(e...)^1/2)^1/2)
logab=loga+logb
so similarly the series is as follows
(1/2)+(1/4)+(1/8)+.....=1
12 years agoHelpfull: Yes(8) No(6)
log(e(e(e.........)^1/2)^1/2)^1/2)
=log(e(e(e.....)))/2*2*2......
=log(e(e(e.....)))/infinity
=0
12 years agoHelpfull: Yes(6) No(35)
=1/2[loge+1/2[loge+1/2[loge+.......to infinite
=1/2[1+1/2[1+1/2[1+1/2[1+..........to infinite .'. loge base e =1
=1/2+1/4+1/8.......to infinite
now a=1/2, r=1/2
sum=a/(1-r)
=(1/2)/(1/2)
=1
Ans=d
10 years agoHelpfull: Yes(3) No(0)
infinite series
a/1-a;
(1/2)/(1-1/2);
=1; option is d;
11 years agoHelpfull: Yes(1) No(0)
simple property
10 years agoHelpfull: Yes(1) No(3)
let

x = log(e(e(e.....)^1/2)^1/2)^1/2)

so, x = (1/2)[log(e(e(e.....)^1/2)^1/2)^1/2)]

so, x= (1/2)* x itself

so, 2x-x = 0
so, x=0 is the ans
11 years agoHelpfull: Yes(0) No(11)
right option is c
9 years agoHelpfull: Yes(0) No(2)

what is the value of log(e(e(e.....)^1/2)^1/2)^1/2)?
a)0
b)1/3
c)1/2
d)1