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 :
\[x,\ y,\ z\] are in G.P., then \[{{y}^{2}}=x\ .\ z\] Now \[{{a}^{x}}={{b}^{y}}={{c}^{z}}=m\] \[\Rightarrow \] \[x{{\log }_{e}}a=y{{\log }_{e}}b=z{{\log }_{e}}c={{\log }_{e}}m\] \[\Rightarrow \]\[x={{\log }_{a}}m,\ y={{\log }_{b}}m,z={{\log }_{c}}m\] Again as \[x,\ y,\ z\] are in G.P., so \[{{(p-r)}^{2}}={{(p+r)}^{2}}-4pr=16{{K}^{2}}-16{{K}^{2}}=0\] \[\Rightarrow \]\[\frac{{{\log }_{b}}m}{{{\log }_{a}}m}=\frac{{{\log }_{c}}m}{{{\log }_{b}}m}\]\[\Rightarrow \]\[{{\log }_{b}}a={{\log }_{c}}b\].
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