WHAT is the domain of defination of #log_4 (-log_1/2 (1+ 6/root(4)x) -2)#?
1 Answer
Explanation:
I'm assuming this means
Let's start by finding the domain and range of
The log function is defined such that
Since
So,
#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)) < -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
Final Answer