Question

Log2 2x+log2 3x=log2 3 + log2 2

Answer 1

Ответ: х=1.

Пошаговое объяснение:

log₂(2x)+log₂(3x)=log₂3+log₂2=log(2x*3x)=log₂(3*2)

ОДЗ: x>0

log(6x²)=log₂6

6x²=6  |÷6

x²=1

x²-1=0

(x-1)(x+1)=0

x₁=1  ∈ОДЗ       x₂=-1 ∉ОДЗ.

Answer 2

To solve the equation log2(2x) + log2(3x) = log2(3) + log2(2), we can use logarithm properties to simplify it. The logarithm properties we'll be using are:

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

Let's start by applying these properties to the given equation:

log2(2x) + log2(3x) = log2(3) + log2(2)

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

log2(2x * 3x) = log2(3 * 2)

Simplify further:

log2(6x^2) = log2(6)

Now, since both sides have the same base (logarithm base 2), the arguments inside the logarithms must be equal:

6x^2 = 6

Now, to solve for x, divide both sides by 6:

x^2 = 1

Now, take the square root of both sides:

x = ±1

So, the equation has two solutions: x = 1 and x = -1. You can plug each of these values back into the original equation to verify their validity.