Question

Please answer this Question......​

Answer 1

Answer:

GIVES THAT :–

$log_{1 - 2x}(6 {x}^{2} - 5x + 1 ) - log_{1 - 3x}(4 {x}^{2} - 4x + 1) = 2$

USING BASE CHANGING THEOREM :–

$\frac{ log_{2}(6 {x}^{2} - 5x + 1) }{ log_{2}(1 - 2x) } - \frac{ log_{2}(4 {x}^{2} - 4x + 1 ) }{ log_{2}(1 - 3x) } = 2$

$log_{2}(6 {x}^{2} - 3x) - log_{2} (4 {x}^{2} - x) = 2$

$log_{2} ( \frac{6 {x}^{2} - 3x }{4 {x}^{2} - x} ) = 2$

$log_{2} ( \frac{6x - 3}{4x - 1} ) = 2$

$\frac{6x - 3}{4x - 1} = {2}^{2}$

$6x - 3 = 4(4x - 1)$

$10x = 1$

$x = \frac{1}{10}$

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