Question

1)Log(x^2-x-12)по основанию 2 <3 2)Log(x-4) по основанию 2 >или равно 1 3)(1/3) в степени (x+3/4) +1 > или равно 1

Answer 1

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Answer 2

To solve these logarithmic inequalities, we need to apply certain properties of logarithms and algebraic manipulation. Let's work through each inequality step-by-step:

1. $\log_2(x^2-x-12) < 3$

Step 1: Solve for the logarithmic expression:

$x^2 - x - 12 > 2^3$

Step 2: Simplify the right-hand side:

$x^2 - x - 12 > 8$

Step 3: Move all terms to one side of the inequality:

$x^2 - x - 12 - 8 > 0$

$x^2 - x - 20 > 0$

Step 4: Factor the quadratic expression:

$(x - 5)(x + 4) > 0$

Step 5: Find the intervals where the expression is greater than zero:

For $x - 5 > 0$ and $x + 4 > 0$:

$x > 5$ and $x > -4$

The solution is $x > 5$.

2. $\log_2(x - 4) \geq 1$

Step 1: Solve for the logarithmic expression:

$x - 4 \geq 2^1$

Step 2: Simplify the right-hand side:

$x - 4 \geq 2$

Step 3: Move all terms to one side of the inequality:

$x - 4 - 2 \geq 0$

$x - 6 \geq 0$

Step 4: Find the interval where the expression is greater than or equal to zero:

$x \geq 6$

The solution is $x \geq 6$.

3. $\left(\frac{1}{3}\right)^{(x+3/4)} + 1 \geq 1$

Step 1: Simplify the left-hand side:

$\left(\frac{1}{3}\right)^{(x+3/4)} + 1 \geq 1$

Step 2: Subtract 1 from both sides:

$\left(\frac{1}{3}\right)^{(x+3/4)} \geq 0$

Step 3: For any positive base, the exponential function is positive for any real exponent. Thus, the inequality is satisfied for any value of $x$.

The solution is $x \in \mathbb{R}$ (all real numbers).

In summary:

1. $x > 5$
2. $x \geq 6$
3. $x \in \mathbb{R}$ (all real numbers)