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

1 Answer
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