Logarithmic Equations

Question 1: Inequality with Logarithms

The Question: If the value of $(\log_{\frac{1}{2}} \pi + \log_{\pi} \frac{1}{2})$ is $m$, then which of the following options are true for m. (Note: The question in the image is extremely blurry and appears to contain typos. The version above is a plausible interpretation of a common problem type. The literal text in the image does not lead to any of the given answers.)

Core Concept: We can analyze this expression using the change of base formula and the property that for any positive number $k \neq 1$, the sum $k + \frac{1}{k} \ge 2$.

Detailed Solution (based on the interpretation $m = \log_2 \pi + \log_{\pi} 2$):

1. Let’s assume the expression was intended to be $m = \log_2 \pi + \log_{\pi} 2$.
2. Use the Change of Base formula to relate the terms. Let $k = \log_2 \pi$. Then: $$\log_{\pi} 2 = \frac{\log_2 2}{\log_2 \pi} = \frac{1}{k}$$
3. Rewrite m: The expression becomes $m = k + \frac{1}{k}$.
4. Analyze the value of k: We know that $2^1 = 2$ and $2^2 = 4$. Since $\pi \approx 3.14$, we know that $1 < \log_2 \pi < 2$. So, $k$ is a positive number not equal to 1.
5. Apply the inequality: For any positive number $k \neq 1$, it is a known property that $k + \frac{1}{k} > 2$.
6. Therefore, we can conclude that $m > 2$.
7. Conclusion: Even with this common interpretation, none of the options provided in the image (e.g., $m \ge 3.178$ or $2 < m < 2.2$) are correct, as a more precise calculation shows $m \approx 2.257$. The question is flawed as presented.

Final Answer: Based on the visual information and options, the question is ill-defined.

Question 2: Function with the Floor Function

The Question: Consider the function $f(x) = \log_{10}(x - \lfloor x \rfloor)$ (where $\lfloor x \rfloor$ is the greatest integer less or equal to $x$), and $D \subset \mathbb{R}$ is the set of points at which $f$ is defined. Which of the following options are correct?

Detailed Solution:

Let’s first understand the argument of the logarithm: $x - \lfloor x \rfloor$. This is the definition of the fractional part of $x$, often written as ${x}$.

• Domain (D): The argument must be strictly positive.
  • The range of the fractional part function ${x}$ is $[0, 1)$.
  • For the log to be defined, we need ${x} > 0$.
  • This means we must exclude all values of $x$ where the fractional part is 0. This happens precisely when $x$ is an integer.
  • Therefore, the domain $D$ is the set of all real numbers that are not integers ($D = \mathbb{R} \setminus \mathbb{Z}$).

Now let’s evaluate the statements:

• “The set D is infinite.”: TRUE. There are infinitely many non-integer real numbers.
• “The cardinality of the set $\mathbb{R} \setminus D$ is infinite.”: The set $\mathbb{R} \setminus D$ means “all real numbers that are not in D”. This is the set of integers, $\mathbb{Z}$. The set of integers is infinite. TRUE.
• “The graph of $f(x)$ have an infinite number of vertical asymptotes.”: A vertical asymptote occurs when the log’s argument approaches 0. This happens whenever $x$ approaches any integer from the right side (e.g., as $x \to 1^+$, $x - \lfloor x \rfloor \to 0$). Since there are infinitely many integers, there are infinitely many vertical asymptotes. TRUE.
• “f is an invertible function on D.”: FALSE. The function is periodic. For example, $f(1.5) = \log_{10}(1.5 - 1) = \log_{10}(0.5)$. Also, $f(2.5) = \log_{10}(2.5 - 2) = \log_{10}(0.5)$. Since different inputs give the same output, it fails the Horizontal Line Test and is not invertible.
• “Range of the function f is $\mathbb{R}$”: FALSE. The argument of the log, $x - \lfloor x \rfloor$, is always in the interval $(0, 1)$ for our domain. The logarithm of a number in this interval is always negative. The range is $(-\infty, 0)$.

Final Answer: The correct options are:

• The set D is infinite.
• The cardinality of the set $\mathbb{R} \setminus D$ is infinite.
• The graph of $f(x)$ have an infinite number of vertical asymptotes.

Question 3: Solving a Logarithmic Equation

The Question: Suppose $x$ and $y$ are positive real number. If $\log_x(2) + \log_y(8) = 0$, then which of the following options are true for $x$ and $y$?

• $x=2, y=\frac{1}{8}$
• $xy^3 = 1$
• $x=27, y=\frac{1}{3}$
• $x^3y = 1$

Detailed Solution:

1. Start with the equation: $\log_x(2) + \log_y(8) = 0$

2. Use the Change of Base formula to convert both logarithms to a common base, like the natural log ($\ln$).

   $$\frac{\ln 2}{\ln x} + \frac{\ln 8}{\ln y} = 0$$

3. Simplify the terms: We know that $8 = 2^3$, so $\ln(8) = \ln(2^3) = 3\ln(2)$.

   $$\frac{\ln 2}{\ln x} + \frac{3\ln 2}{\ln y} = 0$$

4. Isolate the variables: Since $\ln(2) \neq 0$, we can divide the entire equation by $\ln(2)$.

   $$\frac{1}{\ln x} + \frac{3}{\ln y} = 0$$
   $$\frac{1}{\ln x} = -\frac{3}{\ln y}$$

5. Solve for a relationship between y and x:

   • Cross-multiply: $\ln y = -3 \ln x$
   • Use the Power Rule of logarithms in reverse to move the -3 into the exponent:
   • $\ln y = \ln(x^{-3})$
   • Now that we have $\ln(\text{something}) = \ln(\text{something else})$, we can equate the arguments:
   • $y = x^{-3} \implies y = \frac{1}{x^3}$

6. Rewrite the relationship:

   • Multiply both sides by $x^3$:
   • $x^3y = 1$

7. Evaluate the options:

   • "$x=2, y=\frac{1}{8}$": Let’s check if this satisfies our relationship $x^3y=1$. $(2)^3(\frac{1}{8}) = 8(\frac{1}{8}) = 1$. This is TRUE.
   • "$xy^3 = 1$": FALSE. Our derived relationship is $x^3y=1$.
   • "$x=27, y=\frac{1}{3}$": Let’s check: $(27)^3(\frac{1}{3}) = (19683)(\frac{1}{3}) \neq 1$. FALSE.
   • "$x^3y = 1$": TRUE. This is the relationship we derived.

Final Answer: The correct options are:

• $x=2, y=\frac{1}{8}$
• $x^3y = 1$