Question

Log2(x+1)+log2(x+3)>3

Answer 1

Ответ:

$log_2(x+1)+log_2(x+3) > 3\ \ ,\ \ \ \ ODZ:\ \displaystyle \left \{ {{x+1 > 0} \atop {x+3 > 0}} \right.\ ,\ \ x > -1\ \ ,\\\\log_2(x+1)(x+3) > log_22^3$

Логарифмическая функция с основанием a=2>1 возрастающая, значит

$(x+1)(x+3) > 8\\\\x^2+4x+3 > 8\\\\x^2+4x-5 > 0\ \ ,\ \ x_1=-5\ ,\ x_2=1\ \ (teorema\ Vieta)\\\\(x+5)(x-1) > 0$

Метод интервалов решения неравенств.

Знаки функции:   $+++(-5)---(1)+++$

Выбираем интервалы со знаком плюс :   $x\in (-\infty ;-5)\cup (\, 1\, ;+\infty )$  .

Учтём ОДЗ, получим ответ.

Ответ:  $x\in (\, 1\, ;+\infty )\ .$  

Answer 2

Ответ:

х>1

Пошаговое объяснение:

log2(x+1)(x+3)>log2(8)

2>1

(x+1)(x+3)>8

x²+3+4x>8

x²+4x-5>0

(x-1)(x+5)>0

x>-1 ОДЗ

х<-5 U x>1

с учетом ОДЗ

х>1

Answer 3

To solve the inequality log2(x+1) + log2(x+3) > 3, we can start by using logarithmic properties to simplify the equation. Recall that the sum of logarithms is equivalent to the logarithm of the product:

log2(x+1) + log2(x+3) = log2[(x+1)(x+3)]

Now, the inequality becomes:

log2[(x+1)(x+3)] > 3

Next, we can rewrite the inequality in exponential form using the definition of logarithms:

2^3 < (x+1)(x+3)

8 < (x+1)(x+3)

Expanding the right side of the inequality:

8 < x^2 + 4x + 3

Rearranging the terms:

x^2 + 4x + 3 - 8 > 0

x^2 + 4x - 5 > 0

Now, we need to find the values of x that satisfy this inequality. We can solve it by factoring or using the quadratic formula. Let's use factoring:

(x + 5)(x - 1) > 0

Now we have two factors, (x + 5) and (x - 1), and we need to determine the sign of the inequality for different intervals of x.

Case 1: (x + 5) > 0 and (x - 1) > 0 x > -5 and x > 1 (Take the intersection) x > 1

Case 2: (x + 5) < 0 and (x - 1) < 0 x < -5 and x < 1 (Take the union) x < -5

Therefore, the solution to the inequality is x < -5 or x > 1.