Question

Решите уравнение номер 39

Answer 1

ОДЗ: x > 0. Заменим Log(5) x = t.

t^2 + t - 2 = 0.
D = 1 + 4 * 2 * 1 = 9.
t1 = (-1 + 3)/2 = 1.
t2 = -2.

Log (5) x = 1 или log (5) x = -2.
x = 5. x = 1/25

Answer 2

$log^2_5(x)+log_5(x)=2\\x\ \textgreater \ 0\\log_5(x)=t\\t^2+t=2\\t^2+t-2=0\\a+b+c=0\\t_1=1\\t_2=-2\\log_5(x)=1\\x=5\\log_5(x)=-2\\x=5^{-2}= \frac{1}{25}$