A) \[\log \left( \frac{m}{n} \right)\]
B) \[\log (m\times n)\]
C) \[log{{(m)}^{2}}\]
D) \[\log ({{n}^{m}})\]
Correct Answer: B
Solution :
\[{{\log }_{e}}m=a\]and \[{{\log }_{e}}n=b\] \[\because \] \[m={{e}^{e}}\]and \[n={{e}^{b}}\] \[\because \] \[(m\times n)={{e}^{a}}\times {{e}^{b}}={{e}^{a+b}}\] \[\Rightarrow \] \[a+b={{\log }_{e}}(m\times n)\] \[\therefore \]\[{{\log }_{e}}m+{{\log }_{e}}n={{\log }_{e}}(m\times n)\]You need to login to perform this action.
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