Question

If $ \log (m+n)=\log m+\log n, $ then what is the relation between m and n

Answer 1

Hint: Use the definition of logarithm
 $ \begin{align}
  & If\text{ }{{a}^{x}}=N \\
 & then\text{ x=lo}{{\text{g}}_{a}}N \\
\end{align} $
And then use the law of exponent $ {{a}^{x}}\times {{a}^{y}}={{a}^{x+y}} $ . Compare the given condition with standard formula of logarithm $ {{\log }_{a}}(m\times n)={{\log }_{a}}m+{{\log }_{a}}n $ , then find the relation between m and n.

Complete step-by-step answer:
Let us take L.H.S as P
 $ \Rightarrow P={{\log }_{a}}(m+n) $
Here we suppose that a is the base so, by definition of logarithm we can write
 $ {{a}^{p}}=(m+n)-----(1) $
Now we take R.H.S as $ Q+R $ , here we take base of logarithm is a, so we can write
 $ \begin{align}
  & Q={{\log }_{a}}m \\
 & R={{\log }_{a}}n \\
\end{align} $
So here we use the definition of logarithm and can write
\[\text{ }{{\text{a}}^{\text{Q}}}\text{=m------(2)}\]
and
 $ \text{ }{{\text{a}}^{\text{R}}}\text{=n------(3)} $
Now here we Multiply equation (2) and (3) so that we can find the relation among m, n, Q and R. Hence, we can write
 $ m\times n={{a}^{Q}}\times {{a}^{R}} $
As we know from law of exponents
 $ \begin{align}
  & \text{ }{{\text{a}}^{\text{x}}}\times {{\text{a}}^{\text{y}}}\text{=}{{\text{a}}^{\text{x+y}}} \\
 & \Rightarrow m\times n={{a}^{Q+R}} \\
\end{align} $
Here we logarithm both side with base a, we can write
 $ \begin{align}
  & {{\log }_{a}}(m\times n)=Q+R \\
 & \Rightarrow {{\log }_{a}}(m\times n)={{\log }_{a}}m+{{\log }_{a}}n \\
\end{align} $
Now in question it is given that
 $ \log (m+n)={{\log }_{a}}m+{{\log }_{a}}n----(a) $
And we get
 $ {{\log }_{a}}(m\times n)={{\log }_{a}}m+{{\log }_{a}}n-----(b) $
So, we can write by equating (a) and (b)
 $ \begin{align}
  & \log (m+n)={{\log }_{a}}mn \\
 & \therefore m+n=mn \\
\end{align} $
Subtracting both side mn we have
 $ m+n-mn=0 $
Here we take n common
 $ \begin{align}
  & \Rightarrow m+n(1-m)=0 \\
 & \Rightarrow n(1-m)=-m \\
 & \Rightarrow n=\dfrac{-m}{1-m} \\
 & \Rightarrow n=\dfrac{m}{m-1} \\
\end{align} $
Hence option C is correct.

Note: The logarithm of a number to a given base is the index of the power to which the base must be raised so that it may be equal to the given number. In this question instead of deriving the formula, we can use directly the formula $ {{\log }_{a}}(m\times n)={{\log }_{a}}m+{{\log }_{a}}n $ to find the relation. It should be noted that logarithm of negative number if not defined hence, $ m>0\text{ and n 0} $ also $ a>0 $ and $ a\ne 0 $