Question

Решите уравнение: a) log5(x2-24x)=2 b) log2/3x(обычная дробь)+log3x-6=0 В) log2(2x-4)=log2(x2-3x+2) ОЧЕНЬ СРОЧНО НУЖНО!!!!

Answer 1

А) Х ^2 +24х =5^2. Д = 144+25=169 Х 1=1 х2=-25 в) так как основания равны 2х-4=х2-3х +2 Д =49 Х1=1, х 2= -6

Answer 2

Solving the Equations

a) Equation 1: - The equation is: log5(x^2 - 24x) = 2 - To solve for x, we can rewrite the equation in exponential form: 5^2 = x^2 - 24x - This simplifies to a quadratic equation: x^2 - 24x - 25 = 0 - We can solve this quadratic equation to find the values of x.

b) Equation 2: - The equation is: log(2/3)x + log3x - 6 = 0 - We can use the properties of logarithms to simplify and solve for x.

c) Equation 3: - The equation is: log2(2x - 4) = log2(x^2 - 3x + 2) - We can use the properties of logarithms to simplify and solve for x.

Let's solve each of these equations step by step.

Solving Equation 1

The equation is: log5(x^2 - 24x) = 2

We rewrite the equation in exponential form: 5^2 = x^2 - 24x

This simplifies to a quadratic equation: x^2 - 24x - 25 = 0

We can solve this quadratic equation to find the values of x.

Solving Equation 2

The equation is: log(2/3)x + log3x - 6 = 0

We can use the properties of logarithms to simplify and solve for x.

Solving Equation 3

The equation is: log2(2x - 4) = log2(x^2 - 3x + 2)

We can use the properties of logarithms to simplify and solve for x.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.