# Finding the domain of $\log(\log(x))$

Why is the domain of $$f(x)=\log(x)$$ is $$x>0$$ but the domain of $$f(x)=\log\log(x)$$ is $$x>1$$.

Why is there a difference? (default value of base is 10)

• Try $\log(\log(1))$, then try $\log(\log(\frac{1}{10}))$. See what happens. Commented Jul 9, 2019 at 3:55
• Your function is not defined unless log(x) is positive. Commented Jul 9, 2019 at 3:59

If $$f(x)=\log(\log x)$$, then it has to be $$x>0$$, but the argument of the most external logarithm, argument which is $$\log x$$ also has to be greater than $$0$$.

So $$\log x>0$$ is equivalent to $$x>10^0$$, that is $$x>1$$. Then you need both $$x>0$$ and $$x>1$$ to be true, and you can just say that $$x>1$$. Then, $$(1,+\infty)$$ is the domain of $$f$$.

The input of the outer $$\log$$ needs to be greater than zero. Letting $$\log(x)=y$$:

$$\rightarrow\log(y)>0$$

This is only the case when the original argument is greater than one, so:

$$x>1$$

is the domain.

Since $$\log x$$ is defined only for $$x > 0$$, and $$\log x > 0$$ when $$x > 1$$, it follows that the domain of $$\log\log x$$ is all $$x > 1$$.

The domain of the function $$g(x)=\log_{10}(f(x))$$ is the set of all $$x$$ such that $$f(x)>0$$. In particular, if $$f=\log_{10}$$ then $$\log_{10}(x)>0 \quad \textrm{iff} \quad x>10^0 =1$$

The domain of logx is x>1 but when it comes to loglog(x) the function is defined when logx>0 which implies that x>e^0 or a^0 =1.(here a is any base)