Question

\log _(2)(x^(2)+x-2)<=\log _(2)(2x+10)

Answer 1

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Answer 2

To solve the inequality $\log_2(x^2 + x - 2) \leq \log_2(2x + 10)$, we can use logarithmic properties. In this case, since the base of the logarithm is the same (base 2), we can eliminate the logarithm and rewrite the inequality as an exponential equation:

$x^2 + x - 2 \leq 2x + 10$

Now we can solve this quadratic inequality:

1. Move all terms to one side of the inequality:

$x^2 + x - 2 - (2x + 10) \leq 0$

Simplifying:

$x^2 - x - 12 \leq 0$

2. Factorize the quadratic equation:

$(x - 4)(x + 3) \leq 0$

3. Determine the sign of the expression within each factor:

For $x - 4 \leq 0$, the values of $x$ that satisfy this condition are $x \leq 4$.

For $x + 3 \leq 0$, the values of $x$ that satisfy this condition are $x \geq -3$.

4. Create an interval where the expression is less than or equal to zero:

To satisfy $(x - 4)(x + 3) \leq 0$, we need either:

\begin{align*} x - 4 & \leq 0 \quad \text{and} \quad x + 3 > 0 \ \text{or} \ x - 4 & > 0 \quad \text{and} \quad x + 3 \leq 0 \end{align*}

The first case ($x - 4 \leq 0$ and $x + 3 > 0$) gives us $x \in [-3, 4]$.

The second case ($x - 4 > 0$ and $x + 3 \leq 0$) does not give us any valid solution.

5. Combine the solutions:

The solution to the original inequality is $x \in [-3, 4]$.