Question

Log0. 1(x2-3x)=-1 решите пожалуйста

Answer 1

$log_{0.1} (x^{2} -3x)=-1$

ОДЗ:

x²-3x>0

x(x-3)>0

x<0

x>3

$log_{0.1} (x^{2} -3x)=-1*log_{0.1} 0.1$

$log_{0.1} (x^{2} -3x)=log_{0.1} (0.1)^{-1} =log_{0.1} 10$

$x^{2} -3x=10$

$x^{2} -3x-10=0$

$(x+2)(x-5)=0$

$x_{1} =-2$

$x_{2} =5$

Ответ: х₁=-2; х₂=5

Answer 2

To solve the equation log₀.₁(x² - 3x) = -1, we need to rewrite the equation in exponential form. In general, for any logarithmic equation, we can convert it to exponential form using the property: logᵦ(y) = x is equivalent to ᵦˣ = y.

In this case, the base is ₀.₁ (0.1), the exponent is -1, and the result of the logarithm is x² - 3x. So, the exponential form is:

₀.₁ˢ = x² - 3x

Now, we can solve for x by converting the equation to a quadratic equation:

₀.₁ˢ = x² - 3x

Since ₀.₁ is equivalent to 1/10, we can rewrite the equation as:

(1/10)ˢ = x² - 3x

To eliminate the fractional exponent, we can rewrite the equation using fractional exponents:

10⁻ˢ = x² - 3x

Now, to remove the negative exponent, we can take the reciprocal of both sides:

1/10ˢ = 1/(x² - 3x)

Now, we can equate the exponents and solve the quadratic equation:

10ˢ = x² - 3x

x² - 3x - 10ˢ = 0

This is a quadratic equation in standard form. To solve for x, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

In our equation, a = 1, b = -3, and c = -10ˢ. Substituting these values into the formula:

x = (3 ± √((-3)² - 4 * 1 * (-10ˢ))) / 2 * 1

x = (3 ± √(9 + 40ˢ)) / 2

Now we have two possible solutions for x:

1. x = (3 + √(9 + 40ˢ)) / 2
2. x = (3 - √(9 + 40ˢ)) / 2

Without a specific value for s or the logarithm, we cannot provide numerical solutions. However, the above expressions represent the solutions to the equation log₀.₁(x² - 3x) = -1.