Question

2log^2x с основанием 5+5logxпо основанию 5 +2=0

Answer 1

$2\log^2_5x+5\log_5x+2=0\\O.D.3.:\;x>0\\\log_5x=t,\;\log^2_5x=t^2\\2t^2+5t+2=0\\D=25-4\cdot2\cdot2=9\\t_{1,2}=\frac{-5\pm3}4\\t_1=-2,\;t_2=-\frac12\\\log_5x=-2\Rightarrow x_1=5^{-2}=\frac1{25}\\\log_5x=-\frac12\Rightarrow x_2=5^{-\frac12}=\frac1{\sqrt2}$

Answer 2

Ответ:

x(1)=1/25

x(2)=1/sqrt(5)

Пошаговое объяснение:

Обозначим y=log(x) по основанию 5.

2×y^2+5×y+2=0.

Решаем квадратное ур-е

y(1)=-2

y(2)=-(1/2)

Обратная подстановка

x(1)=5^(-2)=1/25

x(2)=5^(-(1/2))=1/sqrt(5)

Answer 3

To solve the equation 2log^2(x) base 5 + 5log(x) base 5 + 2 = 0, we can use the substitution method.

Let's make a substitution:

Let y = log(x) base 5.

Now the equation becomes:

2y^2 + 5y + 2 = 0

To solve this quadratic equation, we can factor it:

(2y + 1)(y + 2) = 0

Now set each factor to zero and solve for y:

1. 2y + 1 = 0 2y = -1 y = -1/2

2. y + 2 = 0 y = -2

Now, recall that y = log(x) base 5.

1. log(x) base 5 = -1/2

To get rid of the logarithm, rewrite the equation in exponential form:

x = 5^(-1/2) or x = 1/√5

2. log(x) base 5 = -2

Again, rewrite the equation in exponential form:

x = 5^(-2) or x = 1/25

So the solutions for the original equation are:

1. x = 1/√5 (approximately 0.447)
2. x = 1/25 (0.04)

Please note that we need to check if these solutions are valid since logarithms are only defined for positive real numbers. Both solutions satisfy this condition, so they are valid.