The set of positive real values of x for which the function #f(x) = x/(ln⁡x)# is a decreasing function is ?

Question

a) #x < e#
b) #x = 1#
c) #x < e^2#
d) #x > e#
e) empty space

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Answer 1 · 2016-09-26T16:09:29

#0 < x < e#

Calling #y = x/(ln x)#, #y# is decreasing for #forall x | dy/(dx) < 0#

but #dy/(dx) = 1/ln x(1-1/lnx)#. Now calling #z = 1/lnx# we have the condition

#z(1-z) < 0# and this happens for #0 < z < 1# so #y# is decreasing for

#0 < 1/lnx < 1# or #0 < x < e#