Question

Log3(x - 1)+ log3(x -2) =log3(2)

Answer 1

Ответ:

х=3

Объяснение:

Log3(x-1)+Log3(x-2)=Log3(2)

ОДЗ:

х-1>0           x-2>0

x>1               x>2

x∈(2;+∞)

Log3((x-1)(x-2))=Log3(2)

(x-1)(x-2)=2

x²-3x+2=2

x²-3x=0

x(x-3)=0

x1=0 (не подходит под ОДЗ)

х2=3

Answer 2

Ответ:

x = 3.

Объяснение:

log₃(x - 1) + log₃(x - 2) = log₃2

ОДЗ:

x - 1 > 1;   x > 2;  x ≠ 0;  ⇒ x > 2.

$log_{3} \left(\dfrac{(x-1) \cdot (x-2)}{2} \right)=0\\\\ \dfrac{(x-1) \cdot (x-2)}{2} =3^{0} \\\\\dfrac{(x-1) \cdot (x-2)}{2} =1\\\\(x-1) \cdot (x-2)=2\\x^{2} -3x+2=2\\x^{2} -3x+2-2=0\\x^{2} -3x=0\\x(x-3)=0\\x_{1} =0; \: \: x_{2} =3$

х₁ = 0 - не удовлетворяет области определения.

Answer 3

To solve the equation:

log₃(x - 1) + log₃(x - 2) = log₃(2)

You can use the properties of logarithms to simplify and solve for x. First, recall the properties of logarithms:

1. logₐ(b) + logₐ(c) = logₐ(b * c)
2. logₐ(b) - logₐ(c) = logₐ(b / c)

Using property 1, you can combine the two logarithms on the left side:

log₃((x - 1) * (x - 2)) = log₃(2)

Now, since both sides of the equation have the same base (log₃), you can equate the arguments:

(x - 1) * (x - 2) = 2

Expand the left side of the equation:

x² - 2x - x + 2 = 2

Combine like terms:

x² - 3x + 2 = 2

Now, subtract 2 from both sides to isolate the x² term:

x² - 3x = 0

Factor the left side:

x(x - 3) = 0

Now, set each factor equal to zero and solve for x:

1. x = 0
2. x - 3 = 0 x = 3

So, the solutions to the equation are x = 0 and x = 3. However, you need to check if these solutions are valid by ensuring that the arguments of the logarithms are positive.

For x = 0: log₃(0 - 1) + log₃(0 - 2) = log₃(-1) + log₃(-2)

Logarithms of negative numbers are not defined in the real number system, so x = 0 is not a valid solution.

For x = 3: log₃(3 - 1) + log₃(3 - 2) = log₃(2) + log₃(1)

Log₃(1) is always equal to 0, so the equation becomes:

log₃(2) + 0 = log₃(2)

This is true, so x = 3 is a valid solution.

Therefore, the only valid solution to the equation is x = 3.