Answer
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Hint: use the basic definition of logarithm that is if we have ${{a}^{x}}=N$, then by taking log on both sides, we get ${{\log }_{a}}N=x$. And vice versa is also true. Use this to solve the given problem
We have equation given;
${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0$………………. (1)
Here, we need to apply the basic definition of logarithm function i.e. if we have the expression ${{a}^{x}}=N$, we can convert it into log by taking log to both sides with base ‘a’.
Now, taking log to both sides with base ‘a’, we get;
${{\log }_{a}}{{a}^{x}}={{\log }_{a}}N$…………………. (2)
Now, we know identity of logarithm function as;
${{\log }_{c}}{{m}^{n}}=n{{\log }_{c}}m$
Rewriting the equation (2), with the help of above equation, we get;
$x{{\log }_{a}}a={{\log }_{a}}N$
As we know ${{\log }_{c}}C=1$ i.e. value of any log on the same base is 1.
Hence, we get;
$x={{\log }_{a}}N$
Therefore, if ${{a}^{x}}=N$, then we can write this equation in logarithmic form as;
$x={{\log }_{a}}N$and vice – versa is also true i.e. if $x={{\log }_{a}}N$then ${{a}^{x}}=N$.
Now, using the given property with equation (1) we get
Since,
$\begin{align}
& {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0 \\
& {{\log }_{10}}{{\log }_{10}}x=10{}^\circ \\
\end{align}$
Now, we know that $a{}^\circ =1,$ therefore; above equation can be written as;
${{\log }_{10}}{{\log }_{10}}x=1$………………… (3)
Now, we can use same property as explained initially in the solution i.e. if ${{\log }_{a}}N=x$, then ${{a}^{x}}=N$
Using the same property with equation (3), we get,
Since, we have;
$\begin{align}
& {{\log }_{10}}{{\log }_{10}}x=1 \\
& {{\log }_{10}}x={{10}^{1}}=10 \\
\end{align}$
Hence,
${{\log }_{10}}x=10$……………… (4)
Now, we can use the same property again with equation (4), we get
Since,
$\begin{align}
& {{\log }_{10}}x=10 \\
& x={{10}^{10}} \\
\end{align}$
Hence, on solving ${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0$, we get $x={{10}^{10}}$
Note: One can go wrong while doing conversion of ${{\log }_{a}}N=x$to $N={{a}^{x}}$. One can write $a={{N}^{x}}\text{ or }N={{x}^{a}}$ which is wrong. One can give answer x = 1, as log 1= 0, which is wrong, because this will be true for ${{\log }_{10}}x=0$ but we equation as ${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0$. Hence x = 1 will not be the correct solution to the given equation.
We have equation given;
${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0$………………. (1)
Here, we need to apply the basic definition of logarithm function i.e. if we have the expression ${{a}^{x}}=N$, we can convert it into log by taking log to both sides with base ‘a’.
Now, taking log to both sides with base ‘a’, we get;
${{\log }_{a}}{{a}^{x}}={{\log }_{a}}N$…………………. (2)
Now, we know identity of logarithm function as;
${{\log }_{c}}{{m}^{n}}=n{{\log }_{c}}m$
Rewriting the equation (2), with the help of above equation, we get;
$x{{\log }_{a}}a={{\log }_{a}}N$
As we know ${{\log }_{c}}C=1$ i.e. value of any log on the same base is 1.
Hence, we get;
$x={{\log }_{a}}N$
Therefore, if ${{a}^{x}}=N$, then we can write this equation in logarithmic form as;
$x={{\log }_{a}}N$and vice – versa is also true i.e. if $x={{\log }_{a}}N$then ${{a}^{x}}=N$.
Now, using the given property with equation (1) we get
Since,
$\begin{align}
& {{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0 \\
& {{\log }_{10}}{{\log }_{10}}x=10{}^\circ \\
\end{align}$
Now, we know that $a{}^\circ =1,$ therefore; above equation can be written as;
${{\log }_{10}}{{\log }_{10}}x=1$………………… (3)
Now, we can use same property as explained initially in the solution i.e. if ${{\log }_{a}}N=x$, then ${{a}^{x}}=N$
Using the same property with equation (3), we get,
Since, we have;
$\begin{align}
& {{\log }_{10}}{{\log }_{10}}x=1 \\
& {{\log }_{10}}x={{10}^{1}}=10 \\
\end{align}$
Hence,
${{\log }_{10}}x=10$……………… (4)
Now, we can use the same property again with equation (4), we get
Since,
$\begin{align}
& {{\log }_{10}}x=10 \\
& x={{10}^{10}} \\
\end{align}$
Hence, on solving ${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0$, we get $x={{10}^{10}}$
Note: One can go wrong while doing conversion of ${{\log }_{a}}N=x$to $N={{a}^{x}}$. One can write $a={{N}^{x}}\text{ or }N={{x}^{a}}$ which is wrong. One can give answer x = 1, as log 1= 0, which is wrong, because this will be true for ${{\log }_{10}}x=0$ but we equation as ${{\log }_{10}}{{\log }_{10}}{{\log }_{10}}x=0$. Hence x = 1 will not be the correct solution to the given equation.
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