Elitmus
Exam
Numerical Ability
Log and Antilog
loge(e(e(e....)^1/4)^1/4)^1/4
Read Solution (Total 1)
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- let us assume [log(e(e(e..)^1/4)^1/4)^1/4] = x
now as log m^n = n log(m)
we can write it as 1/4[log(e(e(e...)^1/4)^1/4)] = x
now 1/4[log(e)+log(e(e..)^1/4)^1/4)] = x
now we can see that it is again our question so we can replace it as x
so 1/4(1+x)= x
=> 4x = 1+x
=> x= 1/3 - 7 years agoHelpfull: Yes(12) No(0)
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