Question

Помогите решить, пожалуйста

Answer 1

Ответ:

$log_3(5x+1)+log_{5x+1}3=\dfrac{17}{4}\ \ ,\\\\ODZ:\ \left\{\begin{array}{l}5x+1>0\\5x+1\ne 1\end{array}\right\ \ \left\{\begin{array}{l}x>-0,2\\x\ne 0\end{array}\right\ \ \ x\in (-0,2\ ;\ 0\ )\cup (\ 0\ ;+\infty \, )\\\\\\log_3(5x+1)+\dfrac{1}{log_3(5x+1)}-\dfrac{17}{4}=0\ \ ,\ \ \ \ t=log_3(5x+1)\ \ ,$

$t+\dfrac{1}{t}-\dfrac{17}{4}=0\ \,\ \ \ \dfrac{4t^2-17t+4}{4t}=0\ \ ,\ \ t\ne 0\ ,\\\\\\4t^2-17t+4=0\ \ ,\ \ D=225=15^2\ \ ,\ \ t_1=\dfrac{1}{4}=0,25\ \ ,\ \ \ t_2=4\ \ ,\\\\a)\ \ log_3(5x+1)=\dfrac{1}{4}\ \ ,\ 5x+1=3^{\frac{1}{4}}\ \ ,\ 5x=\sqrt[4]{3}-1\ \ ,\ x=\dfrac{\sqrt[4]{3}-1}{5}\approx 0,0632\\\\b)\ \ log_3(5x+1)=4\ \ ,\ 5x+1=3^4\ \ ,\ 5x=80\ \ ,\ \ x=16\\\\Otvet:\ \ x_1=\dfrac{\sqrt[4]{3}-1}{5}\ \ ,\ \ x_2=16\ .$