Question
upstudy study bank question image url

\( 6 ^ { - | x - 3 | } \cdot \log _ { 3 } ( 6 x - x ^ { 2 } - 6 ) \geq 1 \)

Ask by Ramirez Gough. in Russia
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\( x = 3 \)

Solution

Рассмотрим неравенство: \[ 6^{-\,|x - 3|} \cdot \log_{3}(6x - x^{2} - 6) \geq 1 \] Для решения данного неравенства выполним следующие шаги: ### Шаг 1: Определение области определения Логарифм определен тогда и только тогда, когда его аргумент положителен: \[ 6x - x^{2} - 6 > 0 \] Решим данное неравенство: \[ -x^{2} + 6x - 6 > 0 \\ x^{2} - 6x + 6 < 0 \] Найдем корни квадратного уравнения \(x^{2} - 6x + 6 = 0\): \[ D = (-6)^{2} - 4 \cdot 1 \cdot 6 = 36 - 24 = 12 \\ x = \frac{6 \pm \sqrt{12}}{2} = 3 \pm \sqrt{3} \] Неравенство \(x^{2} - 6x + 6 < 0\) выполняется для: \[ 3 - \sqrt{3} < x < 3 + \sqrt{3} \] Таким образом, область определения: \[ x \in (3 - \sqrt{3},\ 3 + \sqrt{3}) \] ### Шаг 2: Исследование исходного неравенства Исходное неравенство можно записать как: \[ 6^{-\,|x - 3|} \cdot \log_{3}(6x - x^{2} - 6) \geq 1 \] Поскольку \(6^{-\,|x - 3|} > 0\) для всех \(x\), можно умножить обе части неравенства на \(6^{\,|x - 3|}\), не меняя знака неравенства: \[ \log_{3}(6x - x^{2} - 6) \geq 6^{\,|x - 3|} \] Теперь рассмотрим оба случая для \(|x - 3|\): 1. **Когда \(x = 3\):** \[ \log_{3}(6 \cdot 3 - 3^{2} - 6) = \log_{3}(18 - 9 - 6) = \log_{3}(3) = 1 \] \[ 6^{\,|3 - 3|} = 6^{0} = 1 \] Неравенство становится: \[ 1 \geq 1 \quad \text{— выполняется.} \] 2. **Когда \(x \neq 3\):** Для \(x \neq 3\), \(|x - 3| > 0\), следовательно, \(6^{\,|x - 3|} > 1\). Однако, из области определения \(3 - \sqrt{3} < x < 3 + \sqrt{3}\), находим: \[ \log_{3}(6x - x^{2} - 6) \leq \log_{3}(3) = 1 \] Следовательно, для \(x \neq 3\): \[ \log_{3}(6x - x^{2} - 6) < 6^{\,|x - 3|} \] Таким образом, неравенство не выполняется для \(x \neq 3\). ### Шаг 3: Заключение Единственным решением неравенства является: \[ x = 3 \] **Ответ:** Единственным решением неравенства является \( x = 3 \).

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Beyond the Answer

To solve the inequality \( 6^{- |x - 3|} \cdot \log_{3}(6x - x^{2} - 6) \geq 1 \), we can break it down step by step. First, rewrite the expression inside the logarithm: \[ 6x - x^{2} - 6 = -x^{2} + 6x - 6 = - (x^{2} - 6x + 6) \] To ensure that the logarithm is defined and non-negative, we need: 1. \(6x - x^2 - 6 > 0\) 2. \(6^{- |x - 3|} > 0\) (which is always true since it's a positive exponential function). Next, factor or analyze \(x^2 - 6x + 6 = 0\) to find where the quadratic is positive. The roots can be found using the quadratic formula: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = 3 \pm \sqrt{3} \] The quadratic opens upwards, so it is positive outside the interval \((3 - \sqrt{3}, 3 + \sqrt{3})\). Now, we can determine the constraints on \(x\): - The expression \(6x - x^2 - 6\) must be positive, leading to \(x < 3 - \sqrt{3}\) or \(x > 3 + \sqrt{3}\). Next, we substitute into the main inequality and analyze the two parts: \[ 6^{- |x - 3|} \cdot \log_{3}(6x - x^{2} - 6) \geq 1 \] Since \(6^{- |x - 3|}\) decreases as \(|x - 3|\) increases, we need \( \log_{3}(6x - x^{2} - 6) \) to be sufficiently large in those ranges. This fraction \( \frac{\log_{3}(6x - x^{2} - 6)}{6^{- |x - 3|}} \) shows that as \(x\) moves away from 3, both terms may counteract. To have a clearer view, let’s evaluate specific points in the ranges \((-\infty, 3 - \sqrt{3})\) and \( (3 + \sqrt{3}, \infty)\). Testing \(x = 4\) gives negative results on \(6x - x^2 - 6\), and similar tests show the behavior near boundaries will be critical. Thus, after careful analysis and numeric testing in those intervals, the exact values satisfying the inequality can be obtained through detailed plotting or further numerical checks, leading you to approximate answers where needed. Analyzing the roots and testing values will help you find the right bounds for the real solution. Overall, maintain focus on the logarithmic condition and the behavior of the exponential function to reach a conclusion.

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy