WHAT is the domain of defination of #log_4 (-log_1/2 (1+ 6/root(4)x) -2)#?

1 Answer
John D.
May 30, 2017

#x in (16, oo)#

Explanation:

I'm assuming this means #log_4(-log_(1/2)(1+6/root(4)(x))-2)#.

Let's start by finding the domain and range of #log_(1/2)(1+6/root(4)(x))#.

The log function is defined such that #log_a(x)# is defined for all POSITIVE values of #x#, as long as #a > 0 and a != 1#

Since #a = 1/2# meets both of these conditions, we can say that #log_(1/2)(x)# is defined for all positive real numbers #x#. However, #1+6/root(4)(x)# cannot be all positive real numbers. #6/root(4)(x)# must be positive, since 6 is positive, and #root(4)(x)# is only defined for positive numbers and is always positive.

So, #x# can be all positive real numbers in order for #log_(1/2)(1+6/root(4)(x))# to be defined. Therefore, #log_(1/2)(1+6/root(4)(x))# will be defined from:

#lim_(x->0)log_(1/2)(1+6/root(4)(x))# to #lim_(x->oo)log_(1/2)(1+6/root(4)(x))#

#lim_(x->0)log_(1/2)(oo)# to #(log_(1/2)(1))#

#-oo to 0#, not inclusive (since #-oo# is not a number and #0# is only possible when #x=oo#)

Finally, we check the outer log to see if it requires us to narrow down our domain even more.

#log_4(-log_(1/2)(1+6/root(4)(x))-2)#

This meets the requirements for the same log domain rule as listed above. So, the inside must be positive. Since we have already shown that #log_(1/2)(1+6/root(4)(x))# must be negative, we can say that the negative of it must be positive. And, in order for the entire inside to be positive, the log with base 1/2 must be less than #-2#, so that its negative is greater than #2#.

#log_(1/2)(1+6/root(4)(x)) < -2#

#1+6/root(4)(x) < (1/2)^-2#

#1+6/root(4)(x) < 4#

#6/root(4)(x) < 3#

#2 < root(4)(x)#

#16 < x#

So #x# must be greater than 16 in order for the entire log to be defined.

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