Search any question & find its solution
Question:
Answered & Verified by Expert
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.
Options:
Solution:
2564 Upvotes
Verified Answer
The correct answer is:
$3+\sqrt{15}$
$$
\begin{array}{l}
\log _{(3 x-1)}(x-2)=\log _{\left(i x^{2}-6 x+1\right)}\left(2 x^{2}-10 x-2\right) \\
\log _{(3 x-1)}(x-2)=\log _{(3 x-1)} \sqrt{2 x^{2}-10 x-2} \\
x-2=\sqrt{2 x^{2}-10 x-2}
\end{array}
$$
Square
$$
\begin{array}{l}
x^{2}-4-4 x=2 x^{2}-10 x-2 \\
x^{2}-6 x-6=0 \\
x=\frac{6 \pm \sqrt{36-4(-6)}}{2} \\
x=\frac{6 \pm 2 \sqrt{15}}{2} \\
x=3+\sqrt{15} \text { possible }
\end{array}
$$
\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}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.