Question

Log6-x((x-6)^2/x-2) >= 2

Answer 1

Ответ:

решение на фотографии.

Answer 2

To solve the inequality $\log_{6-x}\left(\frac{(x-6)^2}{x-2}\right) \geq 2$, we'll follow these steps:

1. Ensure that the argument of the logarithm is positive and the base of the logarithm is greater than 1.
2. Express everything in terms of a single logarithm.
3. Eliminate the logarithm.
4. Solve the resulting inequality.

Let's proceed step by step:

1. Ensure the Argument and Base are Valid:

   • The argument of the logarithm must be positive:

     $\frac{(x-6)^2}{x-2} > 0$

     This means that both $(x-6)^2$ and $x-2$ must have the same sign, and neither can be zero.

   • The base of the logarithm, $6-x$, must be greater than 1:

     $6-x > 1$

     Solving this inequality gives $x < 5$.

2. Express Everything in Terms of a Single Logarithm:

   Rewrite the inequality using a common base for both sides:

   $\log_{6-x}\left(\frac{(x-6)^2}{x-2}\right) \geq 2$

   $\frac{\ln\left(\frac{(x-6)^2}{x-2}\right)}{\ln(6-x)} \geq 2$

3. Eliminate the Logarithm:

   Raise both sides to the power of $\ln(6-x)$ (since $\ln(6-x) > 0$):

   $\ln\left(\frac{(x-6)^2}{x-2}\right) \geq 2\ln(6-x)$

4. Solve the Resulting Inequality:

   Now, we need to solve this inequality. Be careful with the domain of the logarithms.

   $\frac{(x-6)^2}{x-2} \geq e^{2\ln(6-x)}$

   $\frac{(x-6)^2}{x-2} \geq (6-x)^2$

   $(x-6)^2 \geq (x-2)(x-10)$

   This is a quadratic inequality. Let's simplify:

   $x^2 - 8x + 36 \geq x^2 - 12x + 20$

   $4x \leq -16$

   $x \leq -4$

However, this solution is not valid in the original inequality because it would make the argument of the logarithm negative ($6-x$ would be negative), which is not allowed.

Thus, the original inequality has no valid solutions.

Answer 3

To solve the inequality $\log_6(x)((x-6)^2/(x-2)) \geq 2$, we can start by getting rid of the logarithm and simplifying the expression. The base of the logarithm is 6, so we can rewrite the equation as an exponential equation:

$6^2 \leq (x-6)^2/(x-2)$

Now, let's solve for $x$:

$6^2 \leq (x-6)^2/(x-2)$

$36 \leq (x-6)^2/(x-2)$

Now, multiply both sides by $x-2$ to eliminate the fraction:

$36(x-2) \leq (x-6)^2$

Expand the right side:

$36x - 72 \leq x^2 - 12x + 36$

Rearrange the terms and set the inequality to zero:

$x^2 - 12x + 36 - 36x + 72 \geq 0$

$x^2 - 48x + 108 \geq 0$

Now, we have a quadratic inequality. Let's solve it. We can first simplify the inequality by dividing all terms by 3:

$(1/3)x^2 - 16x + 36 \geq 0$

Now, we can factor the quadratic:

$(1/3)(x^2 - 48x + 108) \geq 0$

$(1/3)(x-6)(x-18) \geq 0$

Now, we have factored the inequality. We can analyze the sign of each factor and determine the sign of the expression for different intervals on the number line.

1. $x-6$ changes sign at $x = 6$.
2. $x-18$ changes sign at $x = 18$.

Now, we can create a sign chart:

	$x-6$	$x-18$	$(1/3)(x-6)(x-18)$
$x < 6$	$-$	$-$	$-$
$6 < x < 18$	$+$	$-$	$-$
$x > 18$	$+$	$+$	$+$

The expression $(1/3)(x-6)(x-18)$ is greater than or equal to zero in the intervals $[6, 18]$ and $x = 6$ (including the endpoints). So, the solution to the inequality is:

$x \in [6, 18]$

In interval notation, the solution is $[6, 18]$.