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If a,b,c are distinct positive real nos such that log b a.log c a-log a a + log a b.log c b-log b b + log a c.log b c-log c c = 0,then prove that abc=1.

If a,b,c are distinct positive real nos such that logba.logca-logaa + logab.logcb-logbb + logac.logbc-logcc = 0,then prove that abc=1.

Grade:9

1 Answers

gayatri
16 Points
5 years ago
logaa = logbb = logcc = 1 and logba = \frac{log a}{log b}
\frac{(loga)^{2}}{logb \times logc } - 1  + \frac{(logb)^{2}}{loga \times logc } -1 + \frac{(logc)^{2}}{loga \times logb } - 1 = 0   
By making the denominator common and bringing 3 to the LHS of the equation :
 
\frac{(loga)^{3}}{loga \times logb \times logc} + \frac{(logb)^{3}}{loga \times logb \times logc} + \frac{(logc)^{3}}{loga \times logb \times logc} = 3
 
(loga)^{3} + (logb)^{3} + \(logc)^{3} = 3 \times loga\times logb\times logc
\Rightarrow loga + logb + logc =0                    \textbf{}\textbf{[} a^{3} + b^{3}+c^{3} = 3 abc \Rightarrow a+b+c = 0 \textbf{]}}
\Rightarrow log (abc)=0                                       \textbf{[ loga + logb + log c = log (abc)]}
\Rightarrow abc= 1
 
 

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