Question

Log2(x+2)+log2(x-3)=log2(3+1)

Answer 1

ОДЗ :

$\left \{ {{x+2>0} \atop {x-3>0}} \right.\\\\\left \{ {{x>-2} \atop {x>3}} \right. \Rightarrow x\in(3;+\infty)$

$log_{2}(x+2)+log_{2}(x-3)=log_{2}(3+1)\\\\log_{2}(x+2)(x-3)=log_{2}4\\\\(x+2)(x-3)=4\\\\x^{2}-3x+2x-6-4=0\\\\x^{2}-x-10=0\\\\D=(-1)^{2}-4*(-10)=1+40=41\\\\x_{1}=\frac{1+\sqrt{41}}{2}\\\\x_{2}=\frac{1-\sqrt{41}}{2}-neyd\\\\Otvet:\frac{1+\sqrt{41}}{2}$

Answer 2

To solve the equation log2(x+2) + log2(x-3) = log2(3+1), we can use logarithmic properties to simplify it.

First, we can combine the logarithms on the left side of the equation using the product rule of logarithms. According to the product rule, the sum of logarithms is equal to the logarithm of the product. So, we can rewrite the equation as:

log2((x+2)(x-3)) = log2(4)

Since the logarithms on both sides of the equation have the same base (base 2), we can drop the logarithm notation and focus on the arguments:

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

Now, we have a quadratic equation. Expanding the left side, we get:

x^2 - 3x + 2x - 6 = 4

Simplifying further:

x^2 - x - 10 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Let's factor it:

(x - 5)(x + 2) = 0

Now, we have two possible solutions:

1. x - 5 = 0 x = 5

2. x + 2 = 0 x = -2

Therefore, the solutions to the equation are x = 5 and x = -2.