Question

Lod2(2x+1)>log2(x-1)

Answer 1

ОДЗ :
$\left \{ {{2x+1\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} \right. \\\\ \left \{ {{x\ \textgreater \ -0,5} \atop {x\ \textgreater \ 1}} \right. \\\\x\ \textgreater \ 1\\\\\\log _{2}(2x+1)\ \textgreater \ log _{2}(x-1)\\\\2x+1\ \textgreater \ x-1\\\\x\ \textgreater \ -2$
x ∈ (- 2 ; + ∞)
С учётом ОДЗ окончательный ответ: 
x ∈ (1 ; + ∞)

Answer 2

To solve the inequality $\log_2(2x+1) > \log_2(x-1)$, we need to be careful with the domain of the logarithmic functions. Since the base of the logarithm is 2, the argument of the logarithm must be greater than 0.

Step 1: Find the domain restrictions For $\log_2(2x+1)$ to be defined, we need $2x+1 > 0$ since we cannot take the logarithm of a non-positive number. Solving this inequality:

$2x + 1 > 0$ $2x > -1$ $x > -\frac{1}{2}$

For $\log_2(x-1)$ to be defined, we need $x - 1 > 0$. Solving this inequality:

$x - 1 > 0$ $x > 1$

Step 2: Identify the critical points

The critical points are the values of $x$ where $\log_2(2x+1)$ and $\log_2(x-1)$ are equal. Therefore:

$2x+1 = x-1$ $2x = x - 2$ $x = -2$

Step 3: Test intervals

We now have three intervals: $x < -\frac{1}{2}$, $-\frac{1}{2} < x < 1$, and $x > 1$. We need to test the inequality within each interval to find the solution.

Interval 1: $x < -\frac{1}{2}$ Choose $x = -1$ (since it's a simple value less than $-\frac{1}{2}$):

$\log_2(2(-1)+1) = \log_2(-1)$ -> not defined (logarithm of a non-positive number)

Interval 2: $-\frac{1}{2} < x < 1$ Choose $x = 0$ (since it's a simple value within the interval):

$\log_2(2(0)+1) = \log_2(1) = 0$

$\log_2(0-1) = \log_2(-1)$ -> not defined (logarithm of a non-positive number)

Since $0$ is not greater than $-1$, this interval does not satisfy the inequality.

Interval 3: $x > 1$ Choose $x = 2$ (since it's a simple value greater than $1$):

$\log_2(2(2)+1) = \log_2(5)$

$\log_2(2-1) = \log_2(1) = 0$

Since $\log_2(5)$ is greater than $0$, this interval satisfies the inequality.

Step 4: Final solution

The solution to the inequality $\log_2(2x+1) > \log_2(x-1)$ is $x > 1$ (for $x$ greater than 1).

However, we need to keep in mind that $x > -\frac{1}{2}$ is also part of the domain, but for that interval, the inequality does not hold because $\log_2(2x+1)$ is not defined. So the final solution is $x > 1$.