If log _{(3x-1)}(x-2) = log _{(9x^2-6x+1)}(2x | vedclass

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(B) Given the equation: $\log _{(3x-1)}(x-2) = \log _{(9x^2-6x+1)}(2x^2-10x-2)$.Note that $9x^2-6x+1 = (3x-1)^2$.Using the property $\log_{a^n}(b) = \

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$3+\sqrt{15}$

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(B) Given the equation: $\log _{(3x-1)}(x-2) = \log _{(9x^2-6x+1)}(2x^2-10x-2)$.Note that $9x^2-6x+1 = (3x-1)^2$.Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have:$\log _{(3x-1)}(x-2) = \frac{1}{2} \log _{(3x-1)}(2x^2-10x-2)$.This implies $\log _{(3x-1)}(x-2)^2 = \log _{(3x-1)}(2x^2-10x-2)$.Equating the arguments: $(x-2)^2 = 2x^2-10x-2$.$x^2-4x+4 = 2x^2-10x-2$.$x^2-6x-6 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{6 \pm \sqrt{36+24}}{2} = 3 \pm \sqrt{15}$.For the logarithm to be defined,the base $3x-1 > 0$ and $3x-1 
eq 1$,and the argument $x-2 > 0$.If $x = 3-\sqrt{15} \approx 3-3.87 = -0.87$,then $x-2 If $x = 3+\sqrt{15} \approx 6.87$,then $3x-1 > 0$ and $x-2 > 0$,which is valid.Thus,$x = 3+\sqrt{15}$.

If $\log _{(3x-1)}(x-2) = \log _{(9x^2-6x+1)}(2x^2-10x-2)$,then $x$ equals

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