Question

Log2(x-3)+log2(x+2)≤1  

Answer 1

ОДЗ х-3>0 и х+2>0⇒x>3
log(2)(x-3)(x+2)≤1
(x-3)(x+2)≤2
x²+2x-3x-6-2≤0
x²-x-8≤0
D=1+32=33
x1=(1-√33)/2    x2=(1+√33/2)
         +          _          +
___________________________
        (1-√33)/2     (1+√33)/2 
x∈[(1-√33)/2  ;(1+√33)/2  ]

Answer 2

To solve the inequality $\log_2(x-3) + \log_2(x+2) \leq 1$, we can use logarithmic properties and algebraic techniques. Here's the step-by-step solution:

Step 1: Combine the logarithms using the logarithmic property $\log_a(x) + \log_a(y) = \log_a(xy)$. Apply this property to the given inequality: $\log_2((x-3)(x+2)) \leq 1$.

Step 2: Convert the inequality to exponential form. Rewrite the inequality using the definition of logarithms: $\log_a(b) \leq c$ is equivalent to $a^c \leq b$. Apply this to our inequality: $2^1 \leq (x-3)(x+2)$.

Step 3: Simplify the inequality: $2 \leq (x-3)(x+2)$.

Step 4: Expand the right side of the inequality: $2 \leq x^2 - x - 6$.

Step 5: Move all terms to one side to obtain a quadratic inequality: $x^2 - x - 8 \geq 0$.

Step 6: Factorize the quadratic: $(x-4)(x+2) \geq 0$.

Step 7: Determine the sign of the quadratic expression for different intervals on the number line. We can do this by considering the signs of the factors $(x-4)$ and $(x+2)$ separately.

For $(x-4)$: When $x < 4$, $(x-4)$ is negative. When $x > 4$, $(x-4)$ is positive.

For $(x+2)$: When $x < -2$, $(x+2)$ is negative. When $x > -2$, $(x+2)$ is positive.

Step 8: Combine the signs:

• $(x-4)$ is negative for $x < 4$.
• $(x+2)$ is negative for $x < -2$.

From this analysis, we find the intervals where the inequality holds true:

Interval 1: $x < -2$ Interval 2: $4 < x$

Step 9: Write the solution in interval notation: $(- \infty, -2] \cup (4, \infty)$

Therefore, the solution to the inequality $\log_2(x-3) + \log_2(x+2) \leq 1$ is $(- \infty, -2] \cup (4, \infty)$.