Question

Решите пожалуйста 4..........

Answer 1

$\displaystyle\bf\\ODZ: \ x-2 > 0 \ \ \ \Rightarrow \ \ \ x > 2\\\\\log_{5} ^{2} (x-2)+3\log_{5} (x-2)-4=0$

Сделаем замену :

$\displaystyle\bf\\\log_{5} (x-2)=m\\\\m^{2} +3m-4=0\\\\Teorema \ Vieta \ :\\m_{1} + m_{2} =-3\\\\m_{1} \cdot m_{2} =-4\\\\m_{1} =1 \ \ , \ \ m_{2} =-4\\\\1) \ \log_{5} (x-2)=1\\\\x-2=5\\\\\boxed{x_{1} =7}\\\\2) \ \log_{5}(x-2)=-4\\\\x-2=0,0016\\\\\boxed{x_{2} =2,0016}$