Question

Log2 ( 2x + 1 ) < log2 5

Answer 1

2x+1>0

2x> - 1  /:2

x> - 1/2

D=(- 1/2;+∞)

$log_2(2x+1)$

Ответ

x∈(- 1/2;2)

Answer 2

To solve the inequality $\log_2(2x + 1) < \log_2(5)$, you can use the properties of logarithms.

First, remember that if you have $\log_a(b) < \log_a(c)$, you can simplify it to $b < c$. In this case, you have $\log_2(2x + 1) < \log_2(5)$, so you can simplify it to $2x + 1 < 5$.

Now, solve for $x$:

$2x + 1 < 5$

Subtract 1 from both sides:

$2x < 5 - 1$

$2x < 4$

Now, divide by 2 to isolate $x$:

$x < \frac{4}{2}$

$x < 2$

So, the solution to the inequality is $x < 2$.