WHAT WILL BE THE VALUE OF
log e(e(e....)^1/2)^1/2)^1/2).....log base is e?
Choose the correct option.
A) 0  B) 1  C) 2  D) 3  E) 4
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)= 
<gadget id="calculator">(1/2)/(1-1/2)</gadget>
<output>1</output>
1   so e ^ 1   log(e^1) base e= 1   ANSWER:1
<result>B</result>
WHAT WILL BE THE VALUE OF
log e(e(e....)^1/2)^1/2)^1/2).....log base is e?