Q33 / 90 · KVPY 2015 (SB/SX) · Mathematics
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.
- A$9-\sqrt{15}$
- Correct answerB$3+\sqrt{15}$
- C$\mathrm 2+\sqrt{15}$
- D$6-\sqrt{5}$
Solution
$$
\begin{array}{l}
\log _{(3 x-1)}(x-2)=\log _{\left(i x^{2}-6 x+1\right)}\left(2 x^{2}-10 x-2\right) \\
\log _{(3 x-1)}(x-2)=\log _{(3 x-1)} \sqrt{2 x^{2}-10 x-2} \\
x-2=\sqrt{2 x^{2}-10 x-2}
\end{array}
$$
Square
$$
\begin{array}{l}
x^{2}-4-4 x=2 x^{2}-10 x-2 \\
x^{2}-6 x-6=0 \\
x=\frac{6 \pm \sqrt{36-4(-6)}}{2} \\
x=\frac{6 \pm 2 \sqrt{15}}{2} \\
x=3+\sqrt{15} \text { possible }
\end{array}
$$
Hence, the correct answer is option: 'B'.