Question

2log(основание X)27-3log(основание 27)x=1

Answer 1

$2log_{x}2 - 3log_{27}x = 1 \\ \\ \dfrac{2}{log_{27}x} - 3log_{27}x = 1$
Пусть $t = log_{27}x$
$\dfrac{2}{t} - 3t = 1 \\ \\ 3t + 1 - \dfrac{2}{t} = 0 \\ \\ \dfrac{3t^2 + t - 2 }{t} = 0 \\ \\ 3t^2 + t - 2 = 0 \\ \\ D = 1 + 4 \cdot 2 \cdot 3 = 1 + 24 = 25 = 5^2 \\ \\ t_1 = \dfrac{-1 + 5 }{6} = \dfrac{2}{3} \\ \\ t_2 = \dfrac{-1 - 5}{6} = -1$
Обратная замена:
$log_{27}x = \dfrac{2}{3} \\ \\ 27^ \frac{2}{3}} = x \\ \\ 3^{2} = x \\ \\ \boxed{x = 9} \\ \\ log_{27}x = -1 \\ \\ \boxed{x = \dfrac{1}{27}}$

Answer 2

ОДЗ
x>0 ,x≠1
log(a)b=1/log(b)a
log(a)b-log(a)c=log(a)(b/c)
2log(x)27-3/log(x)27=1
2log²(x)27-log(x)27-3=0
log(x)27=a
2a²-a-3=0
D=1+24=25
a1=(1-5)/4=-1⇒log(x)27=-1⇒1/x=27⇒x=1/27
a2=(1+5)/4=3/2⇒log(x)27=3/2⇒x^(3/2)=27⇒x=9