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  4. If log(3x-1) (x-2)= log(9x2-6x+1) (2x2-10x-2) , then x equals

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

KVPYKVPY 2015

Solution:

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