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.

Final Answer

Related questions
Impact of this question
1128 views around the world
You can reuse this answer
Creative Commons License