Question

If a,b,c are in H.P. ,then the value of $ {{\log }_{e}}(a+c)+{{\log }_{e}}(a-2b+c) $ is equal to ..........
A. $ 2{{\log }_{e}}\left( c-a \right) $
B. $ 2{{\log }_{e}}\left( b-a \right) $
C. $ 2{{\log }_{e}}\left( b-c \right) $
D. $ 2{{\log }_{e}}\left( a-c \right) $

Answer 1

Hint: It is given that a,b,c are in HP. Therefore, b will be the harmonic mean of a and c and the harmonic mean (b) of two numbers (a and c) is given by the formula $ b=\dfrac{2ac}{a+c} $ . We will first write ‘b’ in the terms of ‘a’ and ‘c’ and then we will try to solve both of the logarithm functions so that we will get the value of the required sum. To simplify the calculation, we will be using the property $ \log \dfrac{m}{n}=\log m-\log n $ .

Complete step-by-step answer:
It is given that a,b and c are in HP.
Therefore, b will be the harmonic mean of a and c
Therefore, b can be written as:
 $ b=\dfrac{2ac}{a+c} $ .......(1)
We have to find $ {{\log }_{e}}(a+c)+{{\log }_{e}}(a-2b+c) $ ......(2)
Replacing the value of ‘b’ from (1) in (2) we will get:
 $ \begin{align}
  & \Rightarrow {{\log }_{e}}(a+c)+{{\log }_{e}}(a-2.\dfrac{2ac}{a+c}+c) \\
 & \Rightarrow {{\log }_{e}}(a+c)+{{\log }_{e}}(a+c-\dfrac{4ac}{a+c}) \\
 & \Rightarrow {{\log }_{e}}(a+c)+{{\log }_{e}}\left( \dfrac{{{\left( a+c \right)}^{2}}-4ac}{a+c} \right) \\
 & \Rightarrow {{\log }_{e}}(a+c)+{{\log }_{e}}\left( \dfrac{{{a}^{2}}+{{c}^{2}}+2ac-4ac}{a+c} \right) \\
 & \\
\end{align} $
 $ \Rightarrow {{\log }_{e}}(a+c)+{{\log }_{e}}\left( \dfrac{{{a}^{2}}+{{c}^{2}}-2ac}{a+c} \right) $
 $ \Rightarrow {{\log }_{e}}\left( a+c \right)+{{\log }_{e}}\left( \dfrac{{{\left( a-c \right)}^{2}}}{a+c} \right) $ ........(3)
Now applying the property of $ \log \left( \dfrac{m}{n} \right)=\log m-\log n $ we will get:
 $ \begin{align}
  & \Rightarrow {{\log }_{e}}(a+c)+{{\log }_{e}}{{\left( a-c \right)}^{2}}-{{\log }_{e}}\left( a+c \right) \\
 & \Rightarrow {{\log }_{e}}{{(a-c)}^{2}} \\
\end{align} $
Now we will apply the property $ \log {{a}^{m}}=m\log a $
Applying this property we will get:
 $ \Rightarrow {{\log }_{e}}{{\left( a-c \right)}^{2}}=2{{\log }_{e}}\left( a-c \right) $
Therefore, the answer to our question is $ 2{{\log }_{e}}\left( a-c \right) $
So, the correct answer is “Option D”.

Note: We may also try to solve the question by applying the property $ \log \left( mn \right)=\log m+\log n $ in the first step itself and try to solve the term within then obtained logarithm function but it will lead to a very complicated solution and it is not guaranteed that everyone will be able to solve that very tedious terms inside the logarithm function.
Even if some kind of answer is obtained, it is highly likely to be a wrong one due to its very complicated and confusing calculation hence we should avoid applying this property in the first step itself and try to solve both the terms separately before combining them.
Also, we can apply the inverse of this property after obtaining equation 3. It will give us the same answer as:
 $ \begin{align}
  & {{\log }_{e}}\left( a+c \right)+{{\log }_{e}}\left( \dfrac{{{\left( a-c \right)}^{2}}}{a+c} \right)={{\log }_{e}}\left( \left( a+c \right).\dfrac{{{\left( a-c \right)}^{2}}}{a+c} \right) \\
 & \Rightarrow {{\log }_{e}}{{\left( a-c \right)}^{2}} \\
\end{align} $