Question

The value of \[\left( \log m+\log {{m}^{2}}+\log {{m}^{3}}+....+\log {{m}^{n}} \right)\] is equal to
\[\left( a \right)\dfrac{n\left( n+1 \right)}{2}\]
\[\left( b \right)\dfrac{mn}{2}\]
\[\left( c \right)\dfrac{n\left( n+1 \right)}{2}\log m\]
\[\left( d \right)n\left( n+1 \right).\log {{m}^{2}}\]

Answer 1

Hint: To solve this question we will use some properties of logarithm given as log m + log n = log (mn) and then we will use the property stated as the sum of n numbers is given by \[1+2+.....+n=\dfrac{n\left( n+1 \right)}{2}.\]

Complete step by step answer:
We have to find the value of the term \[\log m+\log {{m}^{2}}+\log {{m}^{3}}+.....+\log {{m}^{n}}.\] First of let us assume a variable to be as L.
\[\Rightarrow L=\log m+\log {{m}^{2}}+\log {{m}^{3}}+.....+\log {{m}^{n}}\]
Now we will use the property of log stated as \[\log m+\log n=\log \left( mn \right).\] Using this property of log in L we have,
\[\Rightarrow L=\log m+\log {{m}^{2}}+\log {{m}^{3}}+.....+\log {{m}^{n}}\]
\[\Rightarrow L=\log \left( m.{{m}^{2}}.{{m}^{3}}....{{m}^{n}} \right)\]
Now, any term of the form \[{{x}^{1}}{{x}^{2}}{{x}^{3}}.....{{x}^{n}}\] can be written as
\[{{x}^{1}}{{x}^{2}}{{x}^{3}}.....{{x}^{n}}={{x}^{1+2+....+n}}\]
Using this in the above obtained equation, we have,
\[L=\log \left( m.{{m}^{2}}.....{{m}^{n}} \right)\]
\[\Rightarrow L=\log {{m}^{1+2+....+n}}\]
Now the formula of the sum up to n numbers is given by
\[1+2+.....+n=\dfrac{n\left( n+1 \right)}{2}\]
Using this formula in the above obtained equation of L, we have,
\[\Rightarrow L=\log {{m}^{1+2+....+n}}\]
\[\Rightarrow L=\log {{m}^{\dfrac{n\left( n+1 \right)}{2}}}\]
Now there is a property of log as \[\log {{m}^{n}}=n\log m.\] Using this in the above equation, we get,
\[\Rightarrow L=\log {{m}^{\dfrac{n\left( n+1 \right)}{2}}}\]
\[\Rightarrow L=\dfrac{n\left( n+1 \right)}{2}\log m\]
Hence, the value of \[\log m+\log {{m}^{2}}+....+\log {{m}^{n}}=\dfrac{n\left( n+1 \right)}{2}\log m.\]

So, the correct answer is “Option c”.

Note: You can also neglect some options by observing that ‘log’ cannot be removed from the given equations, as no value of m is given. Therefore, options (a) and (b) are never possible. Also, option (d) would have been correct if it was of the form \[n\left( n+1 \right)\log {{m}^{\dfrac{1}{2}}}\] as we have \[m\log n=\log {{n}^{m}}.\]
\[\Rightarrow \dfrac{n\left( n+1 \right)}{2}\log m=n\left( n+1 \right)\log {{m}^{\dfrac{1}{2}}}\]