Question

If the equation $x^{{\log }_a{x}^2}=\frac{x^{k-2}}{a^k},a\ne 0$, has exactly one solution for $x$, then the value of $k$ is/are

• A. $6+4\sqrt{2}$
• B. $2+6\sqrt{3}$
• C. $6-4\sqrt{2}$
• D. $2-6\sqrt{3}$

Answer 1

Answer: (A), (C)

$6+4\sqrt{2}$
$6-4\sqrt{2}$

$x^{{\log }_a{x}^2}=\frac{x^{k-2}}{a^k}$

Taking logarithm with base $a$ on both sides, we get

${\log }_a{x}^2\left({\log }_{a}x\right)=(k-2){\log }_{a}x-k[\because \log a^m=m\log a\text{\&}\log \frac{a}{b}=\log a-\log b\text{\&}{\log }_{a}a=1]$

$\Rightarrow 2{\left({\log }_{a}x\right)}^2-(k-2){\log }_{a}x+k=0$
Since, given equations have only one solution

Therefore, Discriminant, $D=0$

$\Rightarrow {(k-2)}^2-4\left(2k\right)=0$

$\Rightarrow k^2-4k+4-8k=k^2-12k+4=0$
$\Rightarrow k=\frac{12\pm \sqrt{144-16}}{2}=\frac{12\pm \sqrt{128}}{2}=\frac{12\pm \sqrt{64\times 2}}{2}=\frac{12\pm 8\sqrt{2}}{2}$

$\Rightarrow k=6\pm 4\sqrt{2}$