Question

Решите уравнение

2+log2(7x²+1)-log корень из 2 (корень из 9х²+7)=0

Answer 1

ОДЗ

$\left \{ {{7x^2+1 > 0} \atop {9x^2+7 > 0}} \right.$

$x \in (-\infty; \infty)$

$2+log_{2}(7x^2+1)-log_{\sqrt{2}}\sqrt{9x^2+7}=0$

$2\cdot log_{2}2+log_{2}(7x^2+1)-log_{2^{\frac{1}{2} }}(9x^2+7)^{\frac{1}{2} }}=0$

$log_{2}2^2+log_{2}(7x^2+1)-\frac{\frac{1}{2}}{\frac{1}{2}}log_{2}(9x^2+7)=0$

$log_{2}4+log_{2}(7x^2+1)-log_{2}(9x^2+7)=0$

$log_{2}4\cdot (7x^2+1)=log_{2}(9x^2+7)$

$4\cdot (7x^2+1)=9x^2+7$

$28x^2+4=9x^2+7$

$28x^2-9x^2=7-4$

$19x^2=3\\\\x=\pm\sqrt{\frac{3}{19}}$