Question
If $${x^{\ln \left( {\frac{y}{z}} \right)}} \cdot {y^{\ln{{\left( {xz} \right)}^2}}} \cdot {z^{\ln\left( {\frac{x}{y}} \right)}} = {y^{4\,\ln\,y}}$$ for any $$x > 1, y > 1$$ and $$z > 1,$$ then which one 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 $$\ln\,x, \ln\,x$$ and $$\ln\,z$$
D.
$$\ln\,y$$ is the AM of $$\ln, \ln\,x, \ln\,z$$ and $$\ln\,z$$
Answer :
$$\ln\,y$$ is the AM of $$\ln\,x, \ln\,x, \ln\,x$$ and $$\ln\,z$$
Solution :
$$\eqalign{
& {x^{\ln \left( {\frac{y}{z}} \right)}} \cdot {y^{\ln {{\left( {xz} \right)}^2}}} \cdot {z^{\ln \left( {\frac{x}{y}} \right)}} = {y^{4\,\ln \,y}} \cr
& \Rightarrow \ln \,\left[ {{x^{\ln \left( {\frac{y}{z}} \right)}}} \right] + \ln \left[ {{y^{\ln {{\left( {xz} \right)}^2}}}} \right] + \ln \left[ {{z^{\ln \left( {\frac{x}{y}} \right)}}} \right] = \ln \left[ {{y^{4\,\ln \,y}}} \right] \cr
& \Rightarrow \left[ {\ln \left( {\frac{y}{z}} \right)\ln \,x} \right] + \left[ {2\,\ln \,\left( {xz} \right)\ln \,y} \right] + \left[ {\ln \left( {\frac{x}{y}} \right)\ln \,z} \right] = 4{\left[ {\ln \,y} \right]^2} \cr
& \Rightarrow \ln \,x\left[ {\ln \,y - \ln \,z} \right] + 2\,\ln \,y\left[ {\ln \,x + \ln \,z} \right] + \ln \,z\left[ {\ln \,x - \ln \,y} \right] = 4{\left[ {\ln \,y} \right]^2} \cr
& \Rightarrow 3\,\ln \,x + \ln \,z = 4\,\ln \,y \cr
& \Rightarrow \frac{{\ln \,x + \ln \,x + \ln \,x + \ln \,z}}{4} = \ln \,y \cr} $$
∴ $$\ln\,y$$ is the AM of $$\ln\,x, \ln\,x, \ln\,x\, $$ & $$\ln\,z$$