Processing math: 100%

Thursday, January 17, 2019

Change of base

Suppose that  \displaystyle a,b\ \text{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)}} .

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)}} 

\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)}} 

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

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

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

\displaystyle =3

No comments:

Post a Comment