Question

Log2(x-2)+log2(x-3)=1 ОДЗ x-3>0,=> x>3 log_2((x-2)(x-3)) = 1 log_2(x^2-5x+6) =1 x^2-5x+6 =2 x^2-5x+4 = 0 D = 25-16 = 9 x = (5-3)/2 = 1 (не входит в ОДЗ) x = (5+3)/2 = 4 Ответ:4 почему одз только (х-3) ? и откуда взяли 5 и 6 распишие

Answer 1

Solving the Equation

To solve the equation log2(x-2) + log2(x-3) = 1, we can use the properties of logarithms.

First, let's combine the logarithms using the property log(a) + log(b) = log(a * b). Applying this property to the equation, we get:

log2((x-2)(x-3)) = 1

Next, we can rewrite the equation in exponential form. The logarithmic equation log(base a)(b) = c is equivalent to a^c = b. Applying this to our equation, we have:

2^1 = (x-2)(x-3)

Simplifying further, we have:

2 = (x-2)(x-3)

Now, let's solve this quadratic equation.

Solving the Quadratic Equation

To solve the quadratic equation 2 = (x-2)(x-3), we can expand the equation and bring it to standard form.

Expanding the equation, we get:

2 = x^2 - 5x + 6

Rearranging the terms, we have:

x^2 - 5x + 4 = 0

Now, we can factorize the quadratic equation:

x^2 - 4x - x + 4 = 0

x(x - 4) - 1(x - 4) = 0

(x - 1)(x - 4) = 0

From the factored form, we can see that the equation will be satisfied when either (x - 1) = 0 or (x - 4) = 0.

Solving for x, we have two possible solutions:

x - 1 = 0 => x = 1

x - 4 = 0 => x = 4

Domain of the Equation

Now let's discuss the domain of the equation. The domain refers to the set of values that x can take in order for the equation to be valid.

In this case, we have the equation log2(x-2) + log2(x-3) = 1. For logarithmic functions, the argument (the value inside the logarithm) must be greater than zero. Therefore, we need to find the values of x that satisfy the condition x - 2 > 0 and x - 3 > 0.

Solving these inequalities, we get:

x - 2 > 0 => x > 2

x - 3 > 0 => x > 3

Combining the two inequalities, we find that x must be greater than 3.

Therefore, the domain of the equation is x > 3.

Summary

In summary, the equation log2(x-2) + log2(x-3) = 1 has two solutions: x = 1 and x = 4. However, the domain of the equation is x > 3. Therefore, the only solution that satisfies both the equation and the domain is x = 4.

Please note that the values 5 and 6 were mentioned in the explanation as part of the expansion and factorization of the quadratic equation, but they are not directly related to the solution or the domain of the equation.