A) \[\ell n\,y\] is the GM of \[\ell n\,x,\,\ell n\,x,\,\ell n\,x\] and \[\ell n\,z\]
B) \[\ell n\,y\] is the AM of \[\ell n\,x,\,\ell n\,x,\,\ell n\,x\]and \[\ell n\,z\]
C) \[\ell n\,y\] is the HM of \[\ell n\,x,\,\ell n\,x,\,\ell n\,x\] and \[\ell n\,z\]
D) \[\ell n\,y\] is the AM of \[\ell n,\,\,In\,\,x,\,\,\ell n\,z\] and \[\ell n\,z\]
Correct Answer: B
Solution :
[d] \[{{x}^{ln\left( \frac{y}{z} \right)}}.\,\,{{y}^{ln{{(xz)}^{2}}}}.\,\,{{z}^{ln\left( \frac{x}{y} \right)}}={{y}^{4\,\,ln\,\,y}}\] \[\Rightarrow \,\,ln\left[ {{z}^{ln\left( \frac{y}{z} \right)}} \right]+ln\,\left[ {{y}^{ln{{(xz)}^{2}}}} \right]+ln\left[ {{z}^{ln\left( \frac{x}{y} \right)}} \right]=ln\,\,\left[ {{y}^{4\,\,ln\,\,y}} \right]\] \[\Rightarrow \,\,\left[ ln\left( \frac{y}{z} \right)ln\,x \right]+[2\,\,ln\,\,(xz)ln\,\,y]+\left[ ln\left( \frac{x}{y} \right)ln\,\,z \right]=4\,{{[ln\,\,y]}^{2}}\]\[\Rightarrow \,\,ln\,\,x[ln\,\,y-ln\,\,z]+2\,\,ln\,\,y[ln\,\,x+\,\,ln\,\,z]\] \[+\,ln\,\,z[ln\,\,x-ln\,\,y]=4\,\,{{[ln\,\,y]}^{2}}\] \[\Rightarrow \,\,\,\,3\,\,ln\,\,x+ln\,\,z=4\,\,ln\,\,y\] \[\Rightarrow \frac{ln\,\,x+ln\,\,x=ln\,\,x+ln\,\,z}{4}\,\,=\,\,ln\,\,y\] \[\therefore \] lny is the AM of lnx, lnx, lnx & lnz.
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