Question

If \[{{x}^{\ln \left( \dfrac{y}{x} \right)}}\cdot {{y}^{\ln {{\left( xz \right)}^{2}}}}\cdot {{z}^{\ln \left( \dfrac{x}{y} \right)}}={{y}^{4\ln y}}\] for any \[x>1,y>1\] and \[z>1\], then which of the following is correct?
A. \[\ln y\] is the GM of \[\ln x,\ln x,\ln x\] and \[\ln z\]
B. \[\ln y\] is the AM of \[\ln x,\ln x,\ln x\] and \[\ln z\]
C. \[\ln y\] is the HM of \[\ln x,\ln x,\ln x\] and \[\ln z\]
D. \[\ln y\] is the AM of \[\ln ,Inx,\ln z\] and \[\ln z\]

Answer 1

Hint: In this question, we have to find the correct statement. By simplifying the given expression, we get the required statement. Applying logarithms on both sides we can able to simplify the given expression.

Formula Used: Some of the important logarithmic formulae:
$\begin{align}
  & \log a\cdot \log b=\log (a+b) \\
 & \dfrac{\log a}{\log b}=\log (a-b) \\
 & \log {{a}^{n}}=n\log a \\
\end{align}$
Some of the important progressions and their means:
The arithmetic mean of $x$ and $y$ is $AM=\dfrac{x+y}{2}$
The geometric mean of $x$ and $y$ is $GM=\sqrt{ab}$
The harmonic mean of $x$ and $y$ is $HM=\dfrac{2ab}{a+b}$

Complete step by step solution: Given expression is
\[{{x}^{\ln \left( \dfrac{y}{x} \right)}}\cdot {{y}^{\ln {{\left( xz \right)}^{2}}}}\cdot {{z}^{\ln \left( \dfrac{x}{y} \right)}}={{y}^{4\ln y}}\]
Applying logarithms on both sides of the given expression,
\[\begin{align}
  & {{x}^{\ln \left( \dfrac{y}{x} \right)}}\cdot {{y}^{\ln {{\left( xz \right)}^{2}}}}\cdot {{z}^{\ln \left( \dfrac{x}{y} \right)}}={{y}^{4\ln y}} \\
 & \Rightarrow \ln \left( {{x}^{\ln \left( \dfrac{y}{z} \right)}}\cdot {{y}^{\ln {{\left( xz \right)}^{2}}}}\cdot {{z}^{\ln \left( \dfrac{x}{y} \right)}} \right)=\ln \left( {{y}^{4\ln y}} \right) \\
\end{align}\]
On simplifying, we get
\[\begin{align}
  & \Rightarrow \ln {{x}^{\ln \left( \dfrac{y}{z} \right)}}+\ln {{y}^{\ln {{\left( xz \right)}^{2}}}}+\ln {{z}^{\ln \left( \dfrac{x}{y} \right)}}=\ln \left( {{y}^{4\ln y}} \right) \\
 & \Rightarrow \ln \left( \dfrac{y}{z} \right)\cdot \ln x+\ln {{\left( xz \right)}^{2}}\cdot \ln y+\ln \left( \dfrac{x}{y} \right)\cdot \ln z=4\ln y\cdot \ln y \\
 & \Rightarrow \left( \ln y-\ln z \right)\ln x+2\left( \ln x+\ln z \right)\cdot \ln y+\left( \ln x-\ln y \right)\ln z=4\ln y\cdot \ln y \\
 & \Rightarrow \ln y\cdot \ln x-\ln z\cdot \ln x+2\ln x\cdot \ln y+2\ln z\cdot \ln y+\ln x\cdot \ln z-\ln y\cdot \ln z=4\ln y.\ln y \\
\end{align}\]
Simplifying further, we get
\[\begin{align}
  & 3\ln y\cdot \ln x+\ln z\cdot \ln y=4\ln y.\ln y \\
 & \ln y\left( 3\ln x+\ln z \right)=4\ln y.\ln y \\
 & 3\ln x+\ln z=4\ln y \\
 & \Rightarrow \ln y=\dfrac{3\ln x+\ln z}{4} \\
\end{align}\]
Thus, we can write the obtained expression as
\[\Rightarrow \ln y=\dfrac{\ln x+\ln x+\ln x+\ln z}{4}\]
Since the obtained expression is the average of four terms that are $\ln x,\ln x,\ln x,\ln z$. So, by this, we can say that $\ln y$ is an arithmetic mean of the series with the four terms $\ln x,\ln x,\ln x,\ln z$.

Option ‘B’ is correct

Note: Here we need to remember that the average of all the terms or the ratio of the sum of all the terms of the series to the total number of terms is the definition of arithmetic mean. But here the term of the series is logarithmic terms.