Question

Log2 (x-3)+ log 2 (2x+1)=2

Answer 1

Log2 (x-3)+ log 2 (2x+1)=2
2x²+x-6x-3 = 4
2x²-5x-7=0
D=25+4*2*7= 25 + 56=81
x1= $\frac{5+9}{4} = \frac{14}{4} = \frac{7}{2} =3.5$
x2= $\frac{5-9}{4}=-1$  ⇒  -1∉(3;+∞),значит
ответ:3.5

Answer 2

$log_2(x-3)+log_2(2x+1)=2\\ log_2(x-3)(2x+1)=log_24\\ (x-3)(2x+1)=4\\ 2x^2+x-6x-3-4=0\\ 2x^2-5x-7=0\\ D=25+4*2*7=25+56=81=9^2\\ x_{1,2}= \frac{5{\pm}9}{4} = \left \{ {{x_1=3,5} \atop {x_2=-1}} \right.$

корни нашли 3,5 и -1, однако, накладывая ограничение (под логарифмом не может быть отрицательного числа) получаем, что корень -1 не подходит нашему уравнению

Ответ: х=3,5

Answer 3

To solve the equation log2(x-3) + log2(2x+1) = 2, we can use the properties of logarithms to simplify and solve for x.

First, we can apply the property of logarithms that states log(a) + log(b) = log(ab):

log2(x-3) + log2(2x+1) = 2

log2[(x-3)(2x+1)] = 2

Next, we can rewrite the equation using the exponentiation property of logarithms:

2^2 = (x-3)(2x+1)

4 = (x-3)(2x+1)

Expanding the right side of the equation:

4 = 2x^2 - 5x - 3

Rearranging the equation and setting it equal to zero:

2x^2 - 5x - 7 = 0

Now, we can solve this quadratic equation. There are multiple methods to solve quadratics, such as factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula here:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = -5, and c = -7:

x = (-(-5) ± √((-5)^2 - 4 * 2 * (-7))) / (2 * 2)

x = (5 ± √(25 + 56)) / 4

x = (5 ± √81) / 4

x = (5 ± 9) / 4

There are two possible solutions:

1. x = (5 + 9) / 4 = 14 / 4 = 3.5

2. x = (5 - 9) / 4 = -4 / 4 = -1

So, the equation log2(x-3) + log2(2x+1) = 2 has two solutions: x = 3.5 and x = -1.