Typesetting math: 100%

Thursday, January 17, 2019

Change of base

Suppose that  \displaystyle a,b\ \text{and}\ ca,b and c are real numbers greater than 1 .Find the value of \displaystyle \frac{1}{{1+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}11+loga2b(ca)+11+logb2c(ab)+11+logc2a(bc) .

Solution

\displaystyle \frac{1}{{1+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}11+loga2b(ca)+11+logb2c(ab)+11+logc2a(bc) 

\displaystyle =\frac{1}{{{{{\log }}_{{{{a}^{2}}b}}}{{a}^{2}}b+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{{{{\log }}_{{{{b}^{2}}c}}}{{b}^{2}}c+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{{{{\log }}_{{{{c}^{2}}a}}}{{c}^{2}}a+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}=1loga2ba2b+loga2b(ca)+1logb2cb2c+logb2c(ab)+1logc2ac2a+logc2a(bc) 

\displaystyle =\frac{1}{{{{{\log }}_{{{{a}^{2}}b}}}abc}}+\frac{1}{{{{{\log }}_{{{{b}^{2}}c}}}abc}}+\frac{1}{{{{{\log }}_{{{{c}^{2}}a}}}abc}}=1loga2babc+1logb2cabc+1logc2aabc 

\displaystyle ={{\log }_{{abc}}}{{a}^{2}}b+{{\log }_{{abc}}}{{b}^{2}}c+{{\log }_{{abc}}}{{c}^{2}}a=logabca2b+logabcb2c+logabcc2a

\displaystyle ={{\log }_{{abc}}}{{(abc)}^{3}}=logabc(abc)3 

\displaystyle =3=3

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