Решите пожалуйста самостоятельную работу по алгебре 2 вариант

1
log(3)5=x вариант 3

2
x>-1
log(1/2)(x+1)=2
x+1=0,25
x=-3/4

x>0
x^6=64
x=2

Ага, жди))) Решат тебе сейчас.

1) 3^(x) = 5 -----> x = log3 5
2)
log2 16 = log(0,5) (x+1) + 2 ----> ОДЗ: x > -1
log2 (2^4) = log(2^(-1)) (x+1) + 2
4 = log2 (x+1) + 2
log2 (x+1) = 2
(x+1) = 2^2 = 4 ------> x = 3

log(x) 64 = 6
x^6 = 64 -----> x = 2
3)
3*log2 x - log16 x + log2 x^2 = 38 ---> ОДЗ: x > 0
3*log2 x - log(2^4) x + 2*log2 x = 38
5*log2 x - (1/4)*log2 x = 38
19*log2 x = (19*2)*4
log2 x = 8 ------> x = 2^8 = 256

log2 (6-x) + log2 (8-x) - log2 (3-x) = 4
ОДЗ: x не = числам 6; 8 и 3
log2 {(6-x)(8-x) / (3-x)} = 4
{(6-x)(8-x) / (3-x)} = 2^4
(6-x)(8-x) = 16*(3-x)
48 - 8x - 6x + x^2 = 48 - 16x
x^2 - 2x = 0
x*(x - 2) = 0 -----> x1 = 0; x2 = 2
4)
log2 (0,5x^2)* log2 x = 1 ---> ОДЗ: x > 0
(log2 (0,5) + log2 (x^2)) * log2 x = 1
(log2 (2^(-1)) + 2*log2 x) * log2 x = 1
(- 1 + 2*log2 x) * log2 x = 1
2*(log2 x)^2 - log2 x - 1 = 0 ---> log2 x = t
2t^2 - t - 1 = 0 ----> t1 = -1/2[ t2= 1
log2 x = t1 = -1/2 -----> x1 = 2^(-1/2) = V2/2
log2 x = t1 = 1 -----> x2 = 2^1 = 2

{ 2*log5 x + log5 y = 1 ---> (*) на 2
{ log5 x - 2*log5 y = 8
=> ОДЗ: x > 0; y > 0
{ 4*log5 x + 2*log5 y = 2
{ log5 x - 2*log5 y = 8
=> сложить ур-ния:
5*log5 x = 10
log5 x = 2 --------> x = 5^2 = 25
=>
{ log5 x - 2*log5 y = 8
2 - 2*log5 y = 8
2*log5 y = -6
log5 y = -3 --------> y = 5^(-3) = 1/125
5)
{log5 (25x^2)}^2= log2 16 ---> ОДЗ: x > 0
{log5 25 + log5 x^2}^2= log2 (2^4)
{2 + 2*log5 x}^2= 4
4 + 8*log5 x + 4*{log5 x}^2 = 4
4*log5 x * (2 + log5 x) = 0
log5 x = 0 ------> x1 = 5^0 = 1
(2 + log5 x) = 0 ----> log5 x = -2 ----> x2 = 5^(-2) = 1/25

Ответ 5

Если играешь в доту, то получишь в попу ??