Question

Find the domain of $f\left( x \right)=\underbrace{{{\log }_{10}}{{\log }_{10}}......{{\log }_{10}}x}_{10times}$.

Answer 1

Hint : To solve the given question, we need to find the interval in which x lies. This interval becomes the domain of the function, f(x). Also, we need to know that ${{10}^{{{\log }_{10}}x}}=x$ .

Complete step-by-step answer :
We have been given in the question that we have to find the domain of the function, $f\left( x \right)=\underbrace{{{\log }_{10}}{{\log }_{10}}......{{\log }_{10}}x}_{10times}$. So, we can write the function as,
$f\left( x \right)={{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x \right]$
Since the values following ${{\log }_{10}}$ are positive, we can write the function as follows.
${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x>0$
This is because we know that log takes only positive values.
Now, on raising the power to 10 on both the sides, we will get,
${{10}^{{{\log }_{10}}}}\underbrace{\left( {{\log }_{10}}{{\log }_{10}}......{{\log }_{10}}x \right)}_{8times}>{{10}^{0}}$
Taking the property, ${{10}^{{{\log }_{10}}x}}=x$ into consideration, we have,
${{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x \right]>1$
Again, raising the power to 10 on both the sides, we have,
${{10}^{{{\log }_{10}}}}\underbrace{\left( {{\log }_{10}}{{\log }_{10}}......{{\log }_{10}}x \right)}_{7times}>{{10}^{1}}$
This can be written as,
${{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x \right]>10$
Again, raising the power to 10 on both the sides, we have,
${{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x \right]>{{10}^{10}}$
Again, raising the power to 10 on both the sides, we have,
${{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x \right]>{{10}^{{{10}^{10}}}}$
Again, raising the power to 10 on both the sides, we have,
${{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x \right]>{{10}^{{{10}^{{{10}^{10}}}}}}$
Again, raising the power to 10 on both the sides, we have,
${{\log }_{10}}\left[ {{\log }_{10}}{{\log }_{10}}x \right]>{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}$
Again, raising the power to 10 on both the sides, we have,
${{\log }_{10}}{{\log }_{10}}x>{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}}}$
Again, raising the power to 10 on both the sides, we have,
${{\log }_{10}}x>{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}}}}}$
This means that, $x>{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}}}}}}}$
Thus, we have the value of x to be greater than ${{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}}}}}}}$.
Also, since x was plugged into the log function, it can take only positive values.
Therefore we have, $x\in \left[ {{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}}}}}}},\infty \right]$

Note : Here, the value of x varies in the interval $\left[ {{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{{{10}^{10}}}}}}}}}}}}}},\infty \right]$ because x can take positive values only. If x was not plugged in the log function, the interval changes. Also, while doing problems like this, keep the count of terms after each step. If you miss out any one of the terms, then the final answer will be wrong.