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)
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2$\begingroup$ Try $\log(\log(1))$, then try $\log(\log(\frac{1}{10}))$. See what happens. $\endgroup$– quasiCommented Jul 9, 2019 at 3:55
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1$\begingroup$ Your function is not defined unless log(x) is positive. $\endgroup$– Chris LearyCommented Jul 9, 2019 at 3:59
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5 Answers
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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 $f$.
answered Jul 9, 2019 at 4:00
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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.
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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 $x > 1$.
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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 $$\log_{10}(x)>0 \quad \textrm{iff} \quad x>10^0 =1$$
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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)