f(x) = ln(ln(lnx)) has the domain of (e, infinity). I am not sure how this occurs however. I know in the case of ln x, x must be greater than 0. However where does e come from. e is the base of log in the case of ln, but something is just not clicking for me.
since ln(u) needs u>0,
ln(ln(ln(x))) needs ln(ln(x)) > 0
That is, ln(x) > 1
That is, x>e
I am still confused. hm.
To understand why the domain of the function f(x) = ln(ln(lnx)) is (e, infinity), let's break it down step by step.
First, we know that the natural logarithm function, ln(x), is only defined for x greater than zero. This is because the logarithm function requires a positive argument.
Next, consider the innermost logarithm, ln(lnx). For this to be defined, the argument of the logarithm - lnx - must be greater than zero. Since we know that ln(x) is only defined for x greater than zero, it follows that lnx will also be greater than zero as long as x is in the domain of ln(x).
Now, let's move to the next level, ln(ln(lnx)). In order for this to be defined, the argument of the outermost logarithm - ln(lnx) - must be greater than zero. From what we concluded earlier, we know that ln(lnx) will be greater than zero as long as x is in the domain of ln(lnx).
Finally, we have the domain of f(x) as (e, infinity). Here's where the number e comes into play. The number e is defined as the base of the natural logarithm, ln. It is an irrational number approximately equal to 2.71828.
In the case of our function f(x) = ln(ln(lnx)), the domain of x is restricted by the nested logarithms. To ensure ln(lnx) is defined, x must be such that lnx is defined. Therefore, the domain of ln(lnx) is x > e.
As a result, for the entire function f(x) = ln(ln(lnx)) to be defined, the input x must satisfy the conditions x > e and the inner logarithms must be defined. Hence, the domain of f(x) is (e, infinity).