Question

What is the greater: \[{{1000}^{1000}}\] or \[{{1001}^{999}}\]?

Answer 1

Hint: This type of question depends on the basic rules of logarithm. Here instead of natural log we use logarithm to the base 10. We have to use the rule \[\log \left( {{a}^{m}} \right)=m\log a\] . Also in this question we use the value of \[\log 1000=\log \left( {{10}^{3}} \right)=3\log 10=3\], \[\log 1001=3.0004\] . Also we have to take into consideration that as that of \[\ln e=1\] the value of \[\log \left( 10 \right)=1\] . Also we have to use the term of rounding off to the nearest integer.

Complete step-by-step answer:
Now, we have to check which one is larger \[{{1000}^{1000}}\] or \[{{1001}^{999}}\]
For this, consider, \[{{1000}^{1000}}\]
Applying logarithm to the base 10 we can write,
\[\Rightarrow \log \left( {{1000}^{1000}} \right)=1000\log \left( 1000 \right)\]
But, 1000 can be expressed as \[{{10}^{3}}\] Hence,
\[\Rightarrow \log \left( {{1000}^{1000}} \right)=1000\log \left( {{10}^{3}} \right)\]
Now, by the rule of logarithm \[\log \left( {{a}^{m}} \right)=m\log a\] we get,
\[\Rightarrow \log \left( {{1000}^{1000}} \right)=1000\times 3\log \left( 10 \right)=1000\times 3=3000\]
Then consider, \[{{1001}^{999}}\]
Applying logarithm to the base 10 we can write,
\[\Rightarrow \log \left( {{1001}^{999}} \right)=999\log \left( 1001 \right)\] By the rule of logarithm\[\log \left( {{a}^{m}} \right)=m\log a\]
But \[\log 1001=3.0004\] Hence,
\[\Rightarrow \log \left( {{1001}^{999}} \right)=999\times 3.0004=2997.3996\approx 2997\]
Now, as \[3000>2997\]we can write,
\[\Rightarrow \log \left( {{1000}^{1000}} \right)>\log \left( {{1001}^{999}} \right)\]
As logarithm is an increasing function we can write,
\[\Rightarrow \left( {{1000}^{1000}} \right)>\left( {{1001}^{999}} \right)\] ,
\[\left( {{1000}^{1000}} \right)\] is greater than \[\left( {{1001}^{999}} \right)\].

Note: While solving this question student may make mistake in using logarithm to the base 10 instead they use natural log though the rules are same but \[\ln e=1\] while \[\log \left( 10 \right)=1\] . Also students have to take care when they use the rules of logarithm and at the time especially when they multiply by 1000 and 999 if they forgot to do so which may lead to totally reverse answer. Students have to take care when they rounded off the value 2997.3996 to the nearest integer.