Question

The domain of the function $y = \underbrace{log_{10}log_{10}log_{10}...log_{10}}_{n \text{ times}}x$ is

• A. $[10^n, +\infty)$
• B. $(10^{n-1}, +\infty)$
• C. $(10^{n-2}, +\infty)$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

For the function

$y=\underbrace{\log_{10}\log_{10}\log_{10}\cdots\log_{10}}_{n\text{ times}}x$,

each logarithm is defined only if its argument is positive.

Let's denote:

$f_1(x)=\log_{10}x, \quad f_2(x)=\log_{10}(\log_{10}x), \quad f_3(x)=\log_{10}(\log_{10}(\log_{10}x)), \quad \text{etc.}$

Step-by-step conditions:

1. For $f_1(x)=\log_{10}x$:

   $$x>0.$$

2. For $f_2(x)=\log_{10}(\log_{10}x)$:

   The argument must be positive:

   $\log_{10}x>0 \implies x>10^0=1.$

3. For $f_3(x)=\log_{10}(\log_{10}(\log_{10}x))$:

   We require $\log_{10}(\log_{10}x)>0 \implies \log_{10}x>1 \implies x>10^1=10.$

4. For $f_4(x)=\log_{10}(\log_{10}(\log_{10}(\log_{10}x)))$:

   We need $\log_{10}(\log_{10}(\log_{10}x))>0$ which gives

   $\log_{10}(\log_{10}x)>1 \implies \log_{10}x>10 \implies x>10^{10}.$

In general, if we denote the lower bound for $x$ needed for $n$ logarithms as $a_{n-1}$, the recursion is:

$$a_0=0,\quad a_1=10^{a_0}=1,\quad a_2=10^{a_1}=10,\quad a_3=10^{a_2}=10^{10},\quad \ldots$$

So the domain is

$$x > \underbrace{10^{10^{\iddots^{10}}}}_{n-1\text{ times}}.$$

None of the options correctly represent the recursive tower of exponents obtained.