Question

Prove this logarithmic equation .




If log18 ( base 12 ) = alpha .
log54 ( base 24 ) = beta .

Then prove that
alpha × beta + 5(alpha - beta ) = 1

Answer 1

Heya.......!!!!


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$given \: in \: the \: question \: \\ = > log_{12}(18) = \alpha \: \: and \: log_{24}(54) = \beta \\ \\ = > \: log_{12}(18) = \alpha = \frac{ log(18) }{ log(12) } \\ = > \frac{ log(3 {}^{2} \times 2 ) }{ log(3 \times 2 {}^{2} ) } \\ = > \frac{2 log(3) + log(2) }{ log(3) + 2 log(2) } \\ \\ \beta = \frac{ log(54) }{ log 24} \\ = > \frac{ log(3 {}^{3} \times 2 ) }{ log(3 \times 2 {}^{3} ) } \\ = > \frac{ 3log(3) + \: log(2) }{ log(3 ) + \: 3 log(2) } \\ \\ now \: let \: x \: = log(2) \: \: and \: y \: = log(3) \\ then \\ \\ \alpha = \frac{x + 2y}{2x + y} \\ \beta = \frac{x + 3y}{3x + y} \\ \\ we \: have \: to \: prove \: that \: \: \alpha \beta \: + 5( \alpha - \beta ) = 1 \\ \\ putting \: the \: vaues \: we \: get \\ \\ = > \frac{x + 2y}{2x + y} \times \frac{x + 3y}{3x + y} \: + 5( \frac{(x + 2y)(3x + y) - (x + 3y)(2x + y)}{(2x + y)(3x + y)} \\ \\ its \: long \: calcultion....... \\ \\ = > \frac{(x {}^{2} \times 5xy + 6y {}^{2}) + 5(x {}^{2} - y {}^{2} )}{6x {}^{2} + 5xy + y {}^{2} } \\ that \: is \: equal \: to \: \\ = > \frac{6x {}^{2} + 5xy + y {}^{2} }{6x {}^{2} + 5xy + y {}^{2} } = 1 \\ \\ \\ hence \: proved$


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