вправа 6.54 гдз 11 клас алгебра Істер 2019
Вправа 6.54
Знайдіть множину коренів рівняння:
1) хlog0,5х = 1/16; 2) хlgх-2 = 1000; 3) х2-(log4х)/2 = 8;
4) хlg3х-5lgх = 0,0001; 5) (√х)log5х-1 = 5; 6) 10lg2х + 9хlgх = 100.
1) хlog0,5х = 1/16; 2) хlgх-2 = 1000; 3) х2-(log4х)/2 = 8;
4) хlg3х-5lgх = 0,0001; 5) (√х)log5х-1 = 5; 6) 10lg2х + 9хlgх = 100.
Умова:
Відповідь:
1) хlog0,5х = 1/16
ОДЗ: х > 0
log0,5хlog0,5х = log0,51/16
log0,5х • log0,5х = log0,5(0,5)4
(log0,5х)2 = 4
log0,5х = 2 log0,5х = -2
х = 0,52 х = 0,5-2
х = 0,25 х = 4;
2) хlgх-2 = 1000
ОДЗ: х > 0
lg(хlgх-2) = lg1000
(lgх - 2) • lgх = 3
заміна: lgх = t
(t - 2) • t = 3
t2 - 2t - 3 = 0
Д = (-2)2 - 4 • (-3) = 16
t1;2 = (2±4)/2 = 3; -1
lgх = -1 lgх = 3
х = 10-1 х = 103
х = 0,1 х = 1000;
3) х2-(log4х)/2 = 8
ОДЗ: х > 0
пролагарифмуємо обидві частини:
(2 - (log4х)/2))log4х = log48
2log4х - (log42х)/2 - log48 = 0
4log4х - log42х - 2log48 = 0
2log48 = log482 = log464 = log443 = 3
4log4х - log42х - 3 = 0
-log42х + 4log4х - 3 = 0
заміна: log4х = t
-t2 + 4t - 3 = 0
Д = 42 - 4 • (-1) • (-3) = 4
t1;2 = (-4±2)/2 = -3; -1
log4х = t1 log4х = t2
log4х = -3 log4х = -1
х = 4-3 х = 4-1
х = 1/64 х = 1/4;
4) хlg3х-5lgх = 0,0001
ОДЗ: х > 0
lgх(lg3х - 5lgх) = lg0,0001
нехай: lgх = у
у(у3 - 5у) = -4
у4 - 5у2 + 4 = 0
заміна: у2 = t
t2 - 5t + 4 = 0
Д = (-5)2 + 4 • 4 = 9
t1;2 = (5±3)/2 = 4; 1
у2 = t1 у2 = t2
у2 = 4 у2 = 1
у1,2 = ±2 у3,4 = ±1
lgх = у1 lgх = у2 lgх = у3 lgх = у1
lgх = -2 lgх = 2 lgх = -1 lgх = 1
х1 = 0,01 х2 = 100 х3 = 0,1 х4 = 10;
5) (√х)log5х-1 = 5
ОДЗ: х > 0
х1/2log5х-1 = 5
хlog5х1/2-1 = 5
log5хlog5х1/2-1 = log55
log5х(log5х1/2 - 1) = log55
log5х(log5х1/2 - log55) = log55
log5х • log5х1/2 - log5х = log55
log5х3/2 - log5х = log55
log5х3/2/х = log55
log5х0,5 = log55
х0,5 = 5
√х = 5
х = 25;
6) 10lg2х + 9хlgх = 100
(10lgх)lgх + 9хlgх = 10
хlgх + 9 • хlgх = 10
10 • хlgх = 10
хlgх = 1
lgх • хlgх = lg1
lg2х = 0
lgх = 0
х = 100
х = 1.
ОДЗ: х > 0
log0,5хlog0,5х = log0,51/16
log0,5х • log0,5х = log0,5(0,5)4
(log0,5х)2 = 4
log0,5х = 2 log0,5х = -2
х = 0,52 х = 0,5-2
х = 0,25 х = 4;
2) хlgх-2 = 1000
ОДЗ: х > 0
lg(хlgх-2) = lg1000
(lgх - 2) • lgх = 3
заміна: lgх = t
(t - 2) • t = 3
t2 - 2t - 3 = 0
Д = (-2)2 - 4 • (-3) = 16
t1;2 = (2±4)/2 = 3; -1
lgх = -1 lgх = 3
х = 10-1 х = 103
х = 0,1 х = 1000;
3) х2-(log4х)/2 = 8
ОДЗ: х > 0
пролагарифмуємо обидві частини:
(2 - (log4х)/2))log4х = log48
2log4х - (log42х)/2 - log48 = 0
4log4х - log42х - 2log48 = 0
2log48 = log482 = log464 = log443 = 3
4log4х - log42х - 3 = 0
-log42х + 4log4х - 3 = 0
заміна: log4х = t
-t2 + 4t - 3 = 0
Д = 42 - 4 • (-1) • (-3) = 4
t1;2 = (-4±2)/2 = -3; -1
log4х = t1 log4х = t2
log4х = -3 log4х = -1
х = 4-3 х = 4-1
х = 1/64 х = 1/4;
4) хlg3х-5lgх = 0,0001
ОДЗ: х > 0
lgх(lg3х - 5lgх) = lg0,0001
нехай: lgх = у
у(у3 - 5у) = -4
у4 - 5у2 + 4 = 0
заміна: у2 = t
t2 - 5t + 4 = 0
Д = (-5)2 + 4 • 4 = 9
t1;2 = (5±3)/2 = 4; 1
у2 = t1 у2 = t2
у2 = 4 у2 = 1
у1,2 = ±2 у3,4 = ±1
lgх = у1 lgх = у2 lgх = у3 lgх = у1
lgх = -2 lgх = 2 lgх = -1 lgх = 1
х1 = 0,01 х2 = 100 х3 = 0,1 х4 = 10;
5) (√х)log5х-1 = 5
ОДЗ: х > 0
х1/2log5х-1 = 5
хlog5х1/2-1 = 5
log5хlog5х1/2-1 = log55
log5х(log5х1/2 - 1) = log55
log5х(log5х1/2 - log55) = log55
log5х • log5х1/2 - log5х = log55
log5х3/2 - log5х = log55
log5х3/2/х = log55
log5х0,5 = log55
х0,5 = 5
√х = 5
х = 25;
6) 10lg2х + 9хlgх = 100
(10lgх)lgх + 9хlgх = 10
хlgх + 9 • хlgх = 10
10 • хlgх = 10
хlgх = 1
lgх • хlgх = lg1
lg2х = 0
lgх = 0
х = 100
х = 1.