Малко логаритмични задачки

[zlobil]

Мъча се тука над една курсова със задачи от средното,които доста съм ги позабравил.Ако някой има време и желание,ще му бъда много благодарен
п.с. стойностите след логаритъма са основи

log0,25 (x^2 + 2x - ^2 - log0,5 (10+3x-x^2)=1;


log3 1/koren ot (log3 *x) = log9*log9*x/3;

2*koren(log2^3) = 3*koren(log9*4x-0,75);

log1/2*(koren(x+1) -x)<2;

log3*(1-2x)&ge;log3*(5x-2);


log2*log3* (x-1)/(x+1) < log1/2*log1/3 * (x+1)/(x-1)

[Реклама]

[Tony_89]

log0,25 (x^2 + 2x - 8 )^2 - log0,5 (10+3x-x^2)=1;

log_0,25(x² + 2*x -8 )² - log_0,5(10 + 3*x - x²) = 1

log_1/4(x² + 2*x -8 )² - log_1/2(10 + 3*x - x²) = 1

2*log_(1/2)²(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

2*(1/2)*log_1/2(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

log_1/2(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

ДС:

|x² + 2*x -8 > 0
|10 + 3*x - x² > 0

x² + 2*x -8 > 0
(x-2)*(x+4) > 0
x E (-oo;-4) U (2;+oo)

10 + 3*x - x² > 0 /*(-1)
x² - 3*x - 10 < 0
(x+2)*(x-5) < 0
|x E (-2;5)
|x E (-oo;-4) U (2;+oo) =>
=> ДС E (2;5)

log_1/2(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

log_1/2((x² + 2*x -8 )/(10 + 3*x - x²)) = 1

(x² + 2*x -8 )/(10 + 3*x - x²) = 1/2

2*x² + 4*x - 16 = 10 + 3*x - x²

3*x² + x - 26 = 0

x₁ = (-1-SQRT(313))/6
x₂ = (SQRT(313)-1)/6

313 E (17²;18²) =>
=> SQRT(313) E (17;1 =>
=> x₁ E (-19/6;-3), x₂ E (8/3;17/6)
ДС E (2;5) =>
=> само x₂ = (SQRT(313)-1)/6 е решение

Ако нещо не е ясно питай

[Tony_89]

log0,25 (x^2 + 2x - 8 )^2 - log0,5 (10+3x-x^2)=1;

log_0,25(x² + 2*x -8 )² - log_0,5(10 + 3*x - x²) = 1

log_1/4(x² + 2*x -8 )² - log_1/2(10 + 3*x - x²) = 1

2*log_(1/2)²(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

2*(1/2)*log_1/2(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

log_1/2(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

ДС:

|x² + 2*x -8 > 0
|10 + 3*x - x² > 0

x² + 2*x -8 > 0
(x-2)*(x+4) > 0
x E (-oo;-4) U (2;+oo)

10 + 3*x - x² > 0 /*(-1)
x² - 3*x - 10 < 0
(x+2)*(x-5) < 0
|x E (-2;5)
|x E (-oo;-4) U (2;+oo) =>
=> ДС: x E (2;5)

log_1/2(x² + 2*x -8 ) - log_1/2(10 + 3*x - x²) = 1

log_1/2((x² + 2*x -8 )/(10 + 3*x - x²)) = 1

(x² + 2*x -8 )/(10 + 3*x - x²) = 1/2

2*x² + 4*x - 16 = 10 + 3*x - x²

3*x² + x - 26 = 0

x₁ = (-1-SQRT(313))/6
x₂ = (SQRT(313)-1)/6

313 E (17²;18²) =>
=> SQRT(313) E ( 17;18 ) =>
=> x₁ E (-19/6;-3), x₂ E (8/3;17/6)
ДС: x E (2;5) =>
=> само x₂ = (SQRT(313)-1)/6 е решение

Ако нещо не е ясно питай

[Tony_89]

log3 1/koren ot (log3 *x) = log9*log9*x/3;

2*koren(log2^3) = 3*koren(log9*4x-0,75);

log1/2*(koren(x+1) -x)<2;

log3*(1-2x)&ge;log3*(5x-2);


log2*log3* (x-1)/(x+1) < log1/2*log1/3 * (x+1)/(x-1)

2.

log₃(1/SQRT(log₃x)) = log₉(log₉(x/3))

ДС:

|log₃x >= 0
|SQRT(log₃x) <> 0
|1/log₃x > 0
|log₉(x/3) > 0
|x/3 > 0

log₃x >= 0
x >= 1

SQRT(log₃x) <> 0
log₃x <> 0
x <> 1

1/log₃x > 0
log₃x > 0
x > 1

log₉(x/3) > 0
x/3 > 1
x > 3

x/3 > 0
x > 0

|x >= 1
|x <> 1
|x > 1
|x > 3
|x > 0 =>
=> ДС: x E (3;+oo)

log₃(1/SQRT(log₃x)) = log₉(log₉(x/3))

log₃(1/(log₃x)^1/2) = log_3²(log_3²(x/3))

log₃(1/(log₃x))^1/2 = (1/2)*log₃((1/2)*log₃(x/3))

(1/2)log₃(1/(log₃x)) = (1/2)*log₃((1/2)*(log₃x - log₃3))

log₃(1/(log₃x)) = log₃((1/2)*(log₃x - 1))

Нека log₃x = u
ДС: x E (3;+oo) =>
=> u E (1;+oo)

log₃(1/u) = log₃((1/2)*(u - 1))
1/u = (u - 1)/2
u² - u = 2
u² - u - 2 = 0

u₁ = -1 < 1 => не е решение
u₂ = 2

u = 2

log₃x = 2

x = 9

4.

log_1/2(SQRT(x+1) - x) < 2

ДС:

|x+1 >= 0
|SQRT(x+1) - x > 0

x+1 >= 0
x >= -1

SQRT(x+1) - x > 0
SQRT(x+1) > x

I случай:
|x < 0
|x >= -1
x E [-1;0)
При x E [-1;0), x < 0 и SQRT(x+1) >= 0 =>
=> x E [-1;0) е решение

II случай:
|x >= 0
|x >= -1
x E [0;+oo)
SQRT(x+1) > x
SQRT(x+1)² > x²
x + 1 > x²
x² - x - 1 < 0
x₁ = (1-SQRT(5))/2
x₂ = (1+SQRT(5))/2
|x E ((1-SQRT(5))/2;(1+SQRT(5))/2)
|x E [0;+oo)
x E [0;(1+SQRT(5))/2)

x E I случай U II случай =>
=>ДС: x E [-1;(1+SQRT(5))/2)

log_1/2(SQRT(x+1) - x) < 2
SQRT(x+1) - x > 1/4 /*4
4*SQRT(x+1) > 4*x + 1

I случай:
|4*x + 1 < 0
|ДС

|x<-1/4
|x E [-1;(1+SQRT(5))/2)

x E [-1;-1/4)

При x E [-1;-1/4), 4*x + 1 < 0 и 4*SQRT(x+1) >= 0 =>
=> x E [-1;-1/4) е решение

II случай:
|4*x + 1 >= 0
|ДС

x E [-1/4;(1+SQRT(5))/2)

4*SQRT(x+1) > 4*x + 1
(4*SQRT(x+1))² > (4*x + 1)²
16*x² - 8*x - 15 < 0
x₁ = -3/4
x₂ = 5/4

|x E (-3/4;5/4)
|x E [-1/4;(1+SQRT(5))/2)

x E [-1/4;5/4)

x E I случай U II случай =>

=> x E [-1;5/4)

[Tony_89]

5.

log₃(1 - 2*x) >= log₃(5*x - 2)

ДС:

|1 - 2*x > 0
|5*x - 2 > 0

1 - 2*x > 0
x < 1/2

5*x - 2 > 0
x > 2/5 =>
=> ДС: x E (2/5;1/2)

log₃(1 - 2*x) >= log₃(5*x - 2)
1 - 2*x >= 5*x - 2
|x <= 3/7
|ДС

|x <= 3/7
|x E (2/5;1/2) =>

=> x E (2/5;3/7]

6.

log₂(log₃((x-1)/(x+1))) < log_1/2(log_1/3((x+1)/(x-1)))

log₂(log₃((x-1)/(x+1))) < log_2^-1(log_3^-1(((x-1)/(x+1))^-1))

log₂(log₃((x-1)/(x+1))) < (1/(-1))*log₂((1/(-1))*(-1)*log₃((x-1)/(x+1)))

log₂(log₃((x-1)/(x+1))) < -log₂(log₃((x-1)/(x+1)))

2*log₂(log₃((x-1)/(x+1))) < 0 /*(1/2)

log₂(log₃((x-1)/(x+1))) < 0

ДС:

|x+1 <>0
|x-1 <> 0
|(x-1)/(x+1) > 0
|log₃((x-1)/(x+1)) > 0

x <> +-1

(x-1)/(x+1) > 0
x E (-oo;-1) U (1;+oo)

log₃((x-1)/(x+1)) > 0
(x-1)/(x+1) > 1
(x-1)/(x+1) - 1 > 0
(x-1-(x+1))/(x+1) > 0
-2/(x+1) > 0 /*(-1)
2/(x+1) < 0
x+1 < 0
x < -1 =>
=> ДС: x E (-oo;-1)

log₂(log₃((x-1)/(x+1))) < 0

log₃((x-1)/(x+1)) < 1

(x-1)/(x+1) < 3

(x-1)/(x+1) - 3 < 0

(x-1-3*(x+1))(x+1) < 0

(-2*x - 4)/(x+1) < 0 /*(-1/2)

(x+2)/(x+1) > 0

|x E (-oo;-2) U (-1;+oo)
|ДС

|x E (-oo;-2) U (-1;+oo)
|x E (-oo;-1) =>

=> x E (-oo;-2)

[Tony_89]

На места без да искам съм paste-вал твоите готови условия на примерите

[Tony_89]

Третото условие - 1/log3x > 0 от ДС на 2 задача е излишно.

[zlobil]

Човек,не знаеш колко съм ти благодарен.Точно на време Мерси!!!!!