Question

Решить неравенства: 1) log3 x * (5 - 2 log3 x) = 3 2) (log2 x)^2 + 3 log1/2 x + 2 = 0 3) (1/2log3 x - 6) * log9 x = 4 (2 - log9 x) 4)log2 x * log3 x = 4 log3 2 5)lg x + 4 = 2 lg 100 lg x

Answer 1

1) log₃x * (5 - 2log₃x) = 3
5log₃x - 2log₃²x = 3
2log₃²x - 5log₃x + 3 = 0
log₃x = t
2t² - 5t + 3 =0
t₁ = 1
t₂ = $\frac{3}{2}$
log₃x = 1
x = 3
log₃x = $\frac{3}{2}$
x = $\sqrt{27}$
ответ: 3 ; $\sqrt{27}$
2) log₂²x + 3log₁/₂x + 2 =0
   log₂²x + 3*$\frac{ log_{2}x }{ log_{2} \frac{1}{2} }$ + 2 = 0
log₂²x  - 3log₂x + 2 =0
log₂x = t
t² - 3t + 2 =0
t₁ = 1
t₂ = 2
log₂x = 1
x = 2
log₂x = 2
x = 4
ответ: 2; 4
3) (1/2log₃x - 6 )*log₉x = 4(2-log₉x)
(1/2log₃x - 6) * $\frac{ log_{3}x }{2} = 4(2- \frac{ log_{3}x }{2} )$
$\frac{ log_{3} ^{2}x }{4}- 3 log_{3}x = 8-2 log_{3}x$
log₃x = t
t² - 12t = 32 - 8t
t² - 4t - 32 = 0
D₁ = 4+32=36
t₁ = 8
t₂ = -4
log₃x = 8
x = $3^{8}$
log₃x = -4
x = $3^{-4} = \frac{1}{81}$
4) log₂x * log₃x = 4log₃2
$\frac{ log_{3}x }{ log_{3}2 } * log_{3}x = 4 log_{3}2$
$log_{3} ^{2}x = 4 log_{3} ^{2}2$
log₃x = 2log₃2              log₃x = -2log₃2
log₃x = log₃4                log₃x = log₃$\frac{1}{4}$
x=4                              x = $\frac{1}{4}$
5) lg²x + 4 = 2*2*lgx
lg²x - 4lgx + 4 = 0
lgx = t
t² - 4t + 4=0
D₁ = 4-4=0
t = 2
lgx = 2
x = 100