Question

Log2(1-x) +log2(3-x)=3

Answer 1

Ответ:

лЛчьа чюжвжвдадпбебжвдч чи дт

Answer 2

To solve the equation $\log_2(1-x) + \log_2(3-x) = 3$, we can use the properties of logarithms to simplify and solve for $x$.

First, let's apply the logarithmic property that states $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$:

$\log_2((1-x) \cdot (3-x)) = 3$

Now we can rewrite the equation in exponential form:

$2^3 = (1-x) \cdot (3-x)$

Simplify:

$8 = 3 - 4x + x^2$

Rearrange the terms:

$x^2 - 4x - 5 = 0$

Now we have a quadratic equation. To solve for $x$, we can either factor the quadratic or use the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula states that for an equation $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation $x^2 - 4x - 5 = 0$, $a = 1$, $b = -4$, and $c = -5$. Plugging these values into the quadratic formula:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1}$

Simplify under the square root:

$x = \frac{4 \pm \sqrt{16 + 20}}{2}$

$x = \frac{4 \pm \sqrt{36}}{2}$

$x = \frac{4 \pm 6}{2}$

So the solutions for $x$ are:

$x_1 = \frac{4 + 6}{2} = 5$ $x_2 = \frac{4 - 6}{2} = -1$

However, we should check the validity of these solutions by plugging them back into the original equation, as the logarithms involved might introduce restrictions on the domain.

Plugging $x = 5$ into the original equation:

$\log_2(1-5) + \log_2(3-5) = \log_2(-4) + \log_2(-2)$

Logarithms of negative numbers are not defined in the real number system, so $x = 5$ is not a valid solution.

Plugging $x = -1$ into the original equation:

$\log_2(1-(-1)) + \log_2(3-(-1)) = \log_2(2) + \log_2(4)$

Now, both arguments of the logarithms are positive, so the solution $x = -1$ is valid.

Therefore, the solution to the equation $\log_2(1-x) + \log_2(3-x) = 3$ is $x = -1$.