Question

Log (4-x)=log2 два снизу (6-2x)

Answer 1

Логарифмы сокращаются если у них одинаковые основы
Только ОДЗ 4-х >0 х<4 и 6-2х >0 х<3 значит ОДЗ = от -бесконечности к 3 не включивши
4-х=6-2х
-2=-х
Х=2

Answer 2

To solve the equation log(4-x) = log₂(2 / (6-2x)), we can use the properties of logarithms. The equation states that the logarithm of (4-x) to an unspecified base is equal to the logarithm base 2 of (2 / (6-2x)). To find the value of x, we can equate the expressions inside the logarithms and solve for x.

Step 1: Set the expressions inside the logarithms equal to each other:

4 - x = 2 / (6 - 2x)

Step 2: Eliminate the fraction by multiplying both sides by (6 - 2x):

(6 - 2x)(4 - x) = 2

Step 3: Expand and simplify:

24 - 8x - 2x + x^2 = 2

Step 4: Rearrange to form a quadratic equation:

x^2 - 10x + 22 = 0

Step 5: Solve the quadratic equation using factoring or the quadratic formula:

The quadratic equation doesn't factor easily, so let's use the quadratic formula:

x = [ -(-10) ± √((-10)^2 - 4(1)(22)) ] / (2*1)

x = [ 10 ± √(100 - 88) ] / 2

x = [ 10 ± √12 ] / 2

Step 6: Simplify the square root:

x = [ 10 ± 2√3 ] / 2

Step 7: Reduce the fraction:

x = 5 ± √3

So the two possible solutions for x are:

1. x ≈ 5 + √3 (approximately 6.73)
2. x ≈ 5 - √3 (approximately 3.27)

Please note that when dealing with logarithmic equations, we must check for extraneous solutions, which are solutions that don't satisfy the original equation because they lead to undefined results in the domain of the logarithm. We need to ensure that both solutions are valid in the context of the original equation.