A) \[{{\log }_{a}}C={{\log }_{b}}a\]
B) \[{{\log }_{b}}a={{\log }_{c}}b\]
C) \[{{\log }_{c}}b={{\log }_{a}}c\]
D) none of these
Correct Answer: B
Solution :
Since, \[x,y,z\]are in GP \[\Rightarrow \] \[{{y}^{2}}=xz\] ?(i) Now, \[{{a}^{x}}={{b}^{y}}={{c}^{z}}=m\,(say)\] \[\Rightarrow \]\[x{{\log }_{e}}a=y{{\log }_{e}}=z{{\log }_{e}}c={{\log }_{e}}m\] \[\Rightarrow \]\[x{{\log }_{e}}a={{\log }_{e}}m,y\,{{\log }_{e}}b={{\log }_{e}}m,\] \[z{{\log }_{e}}C={{\log }_{e}}m\] \[\Rightarrow \]\[x={{\log }_{a}}m,y={{\log }_{b}}m,z={{\log }_{c}}m\] From Eq. (i) \[{{y}^{2}}=xz\] \[\Rightarrow \] \[{{({{\log }_{b}}m)}^{2}}={{\log }_{a}}m{{\log }_{c}}m\] \[\Rightarrow \] \[\frac{{{\log }_{b}}m}{{{\log }_{a}}m}=\frac{{{\log }_{c}}m}{{{\log }_{b}}m}\] \[\Rightarrow \] \[{{\log }_{b}}\alpha ={{\log }_{c}}b\]
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