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

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Question

(b)

We have,

$\log _{3 x-1}(x-2)$

$=\log _{\left(9 x^2-6 x+1\right)}\left(2 x^2-10 x-2\right)$

$\log _{(3 x-1)}(x-2)$

$=\log _{(3 x-1)^

Vedclass · Accepted Answer

$3+\sqrt{15}$

Vedclass · Answer

(b)

We have,

$\log _{3 x-1}(x-2)$

$=\log _{\left(9 x^2-6 x+1\right)}\left(2 x^2-10 x-2\right)$

$\log _{(3 x-1)}(x-2)$

$=\log _{(3 x-1)^2}\left(2 x^2-10 x-2\right)$

$\log _{3 x-1}(x-2)$

$=\frac{1}{2} \log _{3 x-1}\left(2 x^2-10 x-2\right)(x-2)^2$

$(x-2)^2=2 x^2-10 x-2$

$x^2-4 x+4=2 x^2-10 x-2$

$x^2-6 x-6=0$

$x=3 \pm \sqrt{15}$

$x=3-\sqrt{15}, 3+\sqrt{15}$

$x=3-\sqrt{15} \Rightarrow x-2 < 0$

$x=3+\sqrt{15}$

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

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