Question

Log2(x+1)=1+log2(x-3)​

Answer 1

Смотри.,..................

Answer 2

Ответ: 7.

Ограничения: х>-1 и х>3

Т.о. х>3

Решение:

Log2(x+1)=1+log2(x-3)​

Log2(x+1)=log2(2)+log2(x-3)​

Log2(x+1)=log2(2*(x-3)​)

x+1=2(x-3)

х+1=2х-6

2х-х=1+6

х=7

Answer 3

To solve the equation $\log_2(x+1) = 1 + \log_2(x-3)$, we'll start by using properties of logarithms to simplify and isolate $x$.

First, we'll use the property $\log_a(b) = c$ can be written as $b = a^c$. Applying this to our equation, we get:

$x+1 = 2^{1+\log_2(x-3)}$.

Now, we can use the properties of logarithms to further simplify:

$x+1 = 2 \times 2^{\log_2(x-3)}$ (using $a^{b+c} = a^b \times a^c$).

Since $2^{\log_2(x-3)} = x-3$, we can substitute and continue:

$x+1 = 2 \times (x-3)$.

Now, distribute the 2 on the right side:

$x+1 = 2x - 6$.

Now, isolate $x$ on one side of the equation:

$x - 2x = -6 - 1$.

$-x = -7$.

Now, multiply both sides by -1 to solve for $x$:

$x = 7$.

However, we need to check if this value satisfies the original equation by substituting $x = 7$ into $\log_2(x+1) = 1 + \log_2(x-3)$:

$\log_2(7+1) = 1 + \log_2(7-3)$.

$\log_2(8) = 1 + \log_2(4)$.

$3 = 1 + 2$.

The left-hand side does not equal the right-hand side, so $x = 7$ is not a valid solution to the original equation.

Let's reevaluate our steps and see if there's a mistake or if the equation has no valid solutions.