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Hint- In this question, we use the concept of logarithm. We use the property of logarithm like ${\log _x}\left( {{a^b}} \right) = b{\log _x}\left( a \right)$ and ${\log _x}\left( x \right) = 1$ . In this question we use the base of log is 10 for better understanding and \[{\log _{10}}\left( 2 \right) = 0.301\] and ${\log _{10}}\left( {10} \right) = 1$ .

__Complete step-by-step answer:__

Now, we have two numbers ${100^2}$ and ${2^{100}}$ . We have to find which one is greater ${100^2}$ or ${2^{100}}$ .

So, we use the concept of logarithm.

Let, $p = {100^2}$

Take a log on both sides and the base of the log is 10.

$ \Rightarrow {\log _{10}}\left( p \right) = {\log _{10}}\left( {{{100}^2}} \right)$

We know, ${\log _x}\left( {{a^b}} \right) = b{\log _x}\left( a \right)$

$ \Rightarrow {\log _{10}}\left( p \right) = 2{\log _{10}}\left( {100} \right)$

We can write as $100 = {10^2}$

$

\Rightarrow {\log _{10}}\left( p \right) = 2{\log _{10}}\left( {{{10}^2}} \right) \\

\Rightarrow {\log _{10}}\left( p \right) = 4{\log _{10}}\left( {10} \right) \\

$

As we know, ${\log _{10}}\left( {10} \right) = 1$

$

\Rightarrow {\log _{10}}\left( p \right) = 4 \times 1 \\

\Rightarrow {\log _{10}}\left( p \right) = 4 \\

$

We have to use this property, ${\log _x}\left( a \right) = b \Rightarrow a = {x^b}$

$ \Rightarrow p = {10^4}...............\left( 1 \right)$

Let, $q = {2^{100}}$

Take a log on both sides and the base of the log is 10.

${\log _{10}}\left( q \right) = {\log _{10}}\left( {{2^{100}}} \right)$

We know, ${\log _x}\left( {{a^b}} \right) = b{\log _x}\left( a \right)$

$ \Rightarrow {\log _{10}}\left( q \right) = 100{\log _{10}}\left( 2 \right)$

We know the value of \[{\log _{10}}\left( 2 \right) = 0.301\]

$

\Rightarrow {\log _{10}}\left( q \right) = 100 \times 0.301 \\

\Rightarrow {\log _{10}}\left( q \right) = 30.1 \\

$

We have to use this property, ${\log _x}\left( a \right) = b \Rightarrow a = {x^b}$

\[ \Rightarrow q = {10^{30.1}}..............\left( 2 \right)\]

Now, we can easily compare (1) and (2) equations.

$

\Rightarrow p < q \\

\Rightarrow {100^2} < {2^{100}} \\

$

So, ${2^{100}}$ is greater than ${100^2}$ .

Note- We can use another method to solve the above question in an easy way. Let’s take an example, if we compare ${\left( 2 \right)^2}{\text{ and }}{\left( 3 \right)^3}$ so we can easily identify ${\left( 3 \right)^3}$ is greater because base and power of ${\left( 3 \right)^3}$ is also greater than ${\left( 2 \right)^2}$.

Now, we take ${\left( 2 \right)^{100}}$ and we can express it in form of ${\left( {{2^{10}}} \right)^{10}}$

As we know, ${2^{10}} = 1024$

Now, ${\left( 2 \right)^{100}} = {\left( {1024} \right)^{10}}$

If we compare ${\left( {1024} \right)^{10}}{\text{ and }}{\left( {100} \right)^2}$ so we can easily identify \[{\left( {1024} \right)^{10}}\] is greater than ${\text{ }}{\left( {100} \right)^2}$ because its base 1024 greater than 100 and also power 10 greater than 2. So, ${\left( 2 \right)^{100}}$ is greater than ${\text{ }}{\left( {100} \right)^2}$ .

Now, we have two numbers ${100^2}$ and ${2^{100}}$ . We have to find which one is greater ${100^2}$ or ${2^{100}}$ .

So, we use the concept of logarithm.

Let, $p = {100^2}$

Take a log on both sides and the base of the log is 10.

$ \Rightarrow {\log _{10}}\left( p \right) = {\log _{10}}\left( {{{100}^2}} \right)$

We know, ${\log _x}\left( {{a^b}} \right) = b{\log _x}\left( a \right)$

$ \Rightarrow {\log _{10}}\left( p \right) = 2{\log _{10}}\left( {100} \right)$

We can write as $100 = {10^2}$

$

\Rightarrow {\log _{10}}\left( p \right) = 2{\log _{10}}\left( {{{10}^2}} \right) \\

\Rightarrow {\log _{10}}\left( p \right) = 4{\log _{10}}\left( {10} \right) \\

$

As we know, ${\log _{10}}\left( {10} \right) = 1$

$

\Rightarrow {\log _{10}}\left( p \right) = 4 \times 1 \\

\Rightarrow {\log _{10}}\left( p \right) = 4 \\

$

We have to use this property, ${\log _x}\left( a \right) = b \Rightarrow a = {x^b}$

$ \Rightarrow p = {10^4}...............\left( 1 \right)$

Let, $q = {2^{100}}$

Take a log on both sides and the base of the log is 10.

${\log _{10}}\left( q \right) = {\log _{10}}\left( {{2^{100}}} \right)$

We know, ${\log _x}\left( {{a^b}} \right) = b{\log _x}\left( a \right)$

$ \Rightarrow {\log _{10}}\left( q \right) = 100{\log _{10}}\left( 2 \right)$

We know the value of \[{\log _{10}}\left( 2 \right) = 0.301\]

$

\Rightarrow {\log _{10}}\left( q \right) = 100 \times 0.301 \\

\Rightarrow {\log _{10}}\left( q \right) = 30.1 \\

$

We have to use this property, ${\log _x}\left( a \right) = b \Rightarrow a = {x^b}$

\[ \Rightarrow q = {10^{30.1}}..............\left( 2 \right)\]

Now, we can easily compare (1) and (2) equations.

$

\Rightarrow p < q \\

\Rightarrow {100^2} < {2^{100}} \\

$

So, ${2^{100}}$ is greater than ${100^2}$ .

Note- We can use another method to solve the above question in an easy way. Let’s take an example, if we compare ${\left( 2 \right)^2}{\text{ and }}{\left( 3 \right)^3}$ so we can easily identify ${\left( 3 \right)^3}$ is greater because base and power of ${\left( 3 \right)^3}$ is also greater than ${\left( 2 \right)^2}$.

Now, we take ${\left( 2 \right)^{100}}$ and we can express it in form of ${\left( {{2^{10}}} \right)^{10}}$

As we know, ${2^{10}} = 1024$

Now, ${\left( 2 \right)^{100}} = {\left( {1024} \right)^{10}}$

If we compare ${\left( {1024} \right)^{10}}{\text{ and }}{\left( {100} \right)^2}$ so we can easily identify \[{\left( {1024} \right)^{10}}\] is greater than ${\text{ }}{\left( {100} \right)^2}$ because its base 1024 greater than 100 and also power 10 greater than 2. So, ${\left( 2 \right)^{100}}$ is greater than ${\text{ }}{\left( {100} \right)^2}$ .