A) If\[{{a}^{2}}+4{{b}^{2}}=12ab,\]then \[\log (a+2b)=\frac{1}{2}(\log a+\log b)\]
B) If\[\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b},\]then\[{{x}^{a}}.{{y}^{b}}.{{z}^{c}}=abc\]
C) \[\frac{1}{{{\log }_{xy}}xyz}+\frac{1}{{{\log }_{yz}}xyz}+\frac{1}{{{\log }_{zx}}xyz}=2\]
D) All are correct
Correct Answer: C
Solution :
[a]\[\log (a+2b)=\frac{1}{2}\log {{(a+2b)}^{2}}\] \[=\frac{1}{2}\log ({{a}^{2}}+4{{b}^{2}}+4ab)\] \[=\frac{1}{2}\log (12ab+4ab)\] \[=\frac{1}{2}\log ({{2}^{4}}.ab)\] \[=\frac{1}{2}(4\log 2+\log a+\log b)\] [b] Let\[\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k\] \[\Rightarrow \] \[\log x=k(b-c),\,\,\log y=k(c-a)\] \[\log z=k(a-b)\] \[\therefore \] \[{{x}^{a}}.{{y}^{b}}.{{z}^{c}}={{p}^{k[a(b-c)+b(c-a)+c(a-b)]}}\] \[={{p}^{k(0)}}=1\]where \[p\] is any arbitrary base of the log. [c] Given expression \[={{\log }_{xyz}}xy+{{\log }_{xyz}}yz+{{\log }_{xyz}}zx\] \[={{\log }_{xyz}}(xy.yz.zx)={{\log }_{xyz}}({{x}^{2}}.{{y}^{2}}.{{z}^{2}})\] \[=2{{\log }_{xyz}}(xyz)=2\times 1=2\].
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