Question

Срочно 3 и 4 !!!!!! Алгебра

Answer 1

$3)\ \ log^2_3x-log_3x-6=0\ \ ,\ \ \ \ ODZ:\ x>0\ ,\\\\t=log_3x\ \ ,\ \ t^2-t-6=0\ \ ,\ \ \ t_1=-2\ ,\ t_2=3\ \ (teorema\ Vieta)\\\\log_3x=-2\ ,\ \ x=3^{-2}\ \ ,\ \ x=\dfrac{1}{9}\\\\log_3x=3\ \ ,\ \ x=3^3\ \ ,\ \ x=27\\\\Otvet:\ \ x=\dfrac{1}{9}\ ,\ x=27\ .$

$4)\ \ log_5(2-x)=log_{25}x^4\ \ ,\ \ \ \ ODZ:\ \left\{\begin{array}{ccc}2-x>0\\x^4>0\end{array}\right\ \ \left\{\begin{array}{ccc}x<2\\x\ne 0\end{array}\right\\\\x\in (-\infty ;\, 0\, )\cup (0;2)\\\\log_5(2-x)=log_{5^2}x^4\\\\log_5(2-x)=\dfrac{4}{2}log_5x\\\\log_5(2-x)=log_5x^2\\\\2-x=x^2\\\\x^2+x-2=0\\\\x_1=-2\ ,\ x_2=1\ \ (teorema\ Vieta)\\\\Otvet:\ \ x=-2\ ,\ x=1\ .$