Question

If ${\log }_4\left({\log }_{3}x\right)+{\log }_{\frac{1}{4}}\left({\log }_{\frac{1}{3}}y\right)=0$ and $x^2+y^2=\frac{17}{4}$, then find the values of $x$ and $y$

Answer 1

Given ${\log }_4\left({\log }_{3}x\right)+{\log }_{1/4}\left({\log }_{1/3}y\right)=0$

$\Rightarrow {\log }_4\left({\log }_{3}x\right)+{\log }_{1/4}\left({\log }_3\frac{1}{y}\right)=0$ $(\because {\log }_{a}b=\frac{1}{{\log }_{b}a}$)

$\Rightarrow {\log }_4\left({\log }_{3}x\right)-{\log }_4\left({\log }_3\frac{1}{y}\right)=0$ $(\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x$)

$\Rightarrow {\log }_4\left({\log }_{3}x\right)={\log }_4\left({\log }_3\frac{1}{y}\right)$

$\Rightarrow x=\frac{1}{y}$ .....(1)

Also given $x^2+y^2=\frac{17}{4}$

$\Rightarrow x^2+\frac{1}{x^2}=\frac{17}{4}$ (by (1))
$\Rightarrow \frac{x^4+1}{x^2}=\frac{17}{4}$

$\Rightarrow 4x^4-17x^2+4=0$

$\Rightarrow 4x^4-16x^2-x^2+4=0$

$\Rightarrow 4x^2(x^2-4)-1(x^2-4)=0$

$\Rightarrow (4x^2-1)(x^2-4)=0$

$\Rightarrow 4x^2-1=0$ or $x^2-4=0$

$\Rightarrow x=\pm \frac{1}{2}$ or $x=\pm 2$

Since, log of negative numbers is not defined .

Hence, $x=2,\frac{1}{2}$

When $x=2\Rightarrow y=\frac{1}{2}$

When $x=\frac{1}{2}\Rightarrow y=2$