If x , y , R , then solve the equation :
$(3 y^{2} + 1) l o g_{3} x = 1$
$(2 y^{2} + 10) = l o g_{x} 27$

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Solution

By definition x > 0 and $x \neq 0$
$(3 y^{2} + 1) l o g_{3} x = 1$ ....(1)
and $(2 y^{2} + 10) = l o g_{x} 27 = 3 l o g_{x} 3 = \frac{3}{l o g_{3} x}$
or $(2 y^{2} + 10) l o g_{3} x = 3$ ....(2)
Dividing (1) and (2), we get
$\frac{2 y^{2} + 10}{3 y^{2} + 1} = \frac{3}{1}$
or $2 y^{2} + 10 = 9 y^{2} + 3$
$∴ y^{2} = 1 a n d f r o m (1), 4 l o g_{3} x = 1$
or $l o g_{3} x = \frac{1}{4} o r x = 3^{\frac{1}{4}}$
$∴ x = 3^{\frac{1}{4}}, y = \pm 1$

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