| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | Given that points \( A \) and \( B \) lie on the graph of \( y = \log_2 x \) and the midpoint of \(\overline{AB}\) is \((6, 2)\), we need to find the positive difference between the \( x \)-coordinates of \( A \) and \( B \). First, let's denote the coordinates of points \( A \) and \( B \) as \( (x_1, \log_2 x_1) \) and \( (x_2, \log_2 x_2) \), respectively. The midpoint formula gives us: \[ \left( \frac{x_1 + x_2}{2}, \frac{\log_2 x_1 + \log_2 x_2}{2} \right) = (6, 2) \] From the \( y \)-coordinate of the midpoint, we have: \[ \frac{\log_2 x_1 + \log_2 x_2}{2} = 2 \] Multiplying both sides by 2, we get: \[ \log_2 x_1 + \log_2 x_2 = 4 \] Using the properties of logarithms, this can be rewritten as: \[ \log_2 (x_1 x_2) = 4 \] Exponentiating both sides with base 2, we obtain: \[ x_1 x_2 = 2^4 = 16 \] Next, from the \( x \)-coordinate of the midpoint, we have: \[ \frac{x_1 + x_2}{2} = 6 \] Multiplying both sides by 2, we get: \[ x_1 + x_2 = 12 \] We now have the system of equations: \[ x_1 + x_2 = 12 \] \[ x_1 x_2 = 16 \] To find the positive difference between the \( x \)-coordinates of \( A \) and \( B \), we solve for \( x_1 \) and \( x_2 \). These are the roots of the quadratic equation: \[ t^2 - (x_1 + x_2)t + x_1 x_2 = 0 \] Substituting the known values, we get: \[ t^2 - 12t + 16 = 0 \] We solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} = \frac{12 \pm 4\sqrt{5}}{2} = 6 \pm 2\sqrt{5} \] Thus, the \( x \)-coordinates of \( A \) and \( B \) are \( 6 + 2\sqrt{5} \) and \( 6 - 2\sqrt{5} \). The positive difference between these coordinates is: \[ (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \] Therefore, the positive difference between the \( x \)-coordinates of \( A \) and \( B \) is: \[ \boxed{D} \] |
D |
| deepseek-r1 | Points \( A \) and \( B \) lie on the graph of \( y = \log_{2}x \). The midpoint of \( \overline{AB} \) is \( (6, 2) \). We need to find the positive difference between the \( x \)-coordinates of \( A \) and \( B \). 1. Let \( A = (a, \log_{2}a) \) and \( B = (b, \log_{2}b) \). The midpoint coordinates give us: \[ \left( \frac{a + b}{2}, \frac{\log_{2}a + \log_{2}b}{2} \right) = (6, 2) \] 2. From the midpoint coordinates, we derive two equations: \[ \frac{a + b}{2} = 6 \implies a + b = 12 \] \[ \frac{\log_{2}a + \log_{2}b}{2} = 2 \implies \log_{2}a + \log_{2}b = 4 \implies \log_{2}(ab) = 4 \implies ab = 2^4 = 16 \] 3. Solving the system of equations \( a + b = 12 \) and \( ab = 16 \): - The quadratic equation \( x^2 - 12x + 16 = 0 \) has roots \( a \) and \( b \). - Using the quadratic formula: \[ x = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} = \frac{12 \pm 4\sqrt{5}}{2} = 6 \pm 2\sqrt{5} \] 4. The \( x \)-coordinates are \( 6 + 2\sqrt{5} \) and \( 6 - 2\sqrt{5} \). The positive difference is: \[ (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \] Thus, the positive difference between the \( x \)-coordinates of \( A \) and \( B \) is \(\boxed{D}\). \[ \boxed{DDDDD} \] |
D |
| gpto1 | DDDDD | D |
| gpt-4o-2024-08-06 | To solve this problem, we need to find the positive difference between the \(x\)-coordinates of points \(A\) and \(B\) on the graph of \(y = \log_2 x\), given that the midpoint of \(\overline{AB}\) is \((6, 2)\). Since \(A\) and \(B\) lie on the graph of \(y = \log_2 x\), their coordinates can be expressed as \(A = (x_1, \log_2 x_1)\) and \(B = (x_2, \log_2 x_2)\). The midpoint of \(\overline{AB}\) is given by: \[ \left( \frac{x_1 + x_2}{2}, \frac{\log_2 x_1 + \log_2 x_2}{2} \right) = (6, 2) \] From the \(x\)-coordinate of the midpoint, we have: \[ \frac{x_1 + x_2}{2} = 6 \implies x_1 + x_2 = 12 \] From the \(y\)-coordinate of the midpoint, we have: \[ \frac{\log_2 x_1 + \log_2 x_2}{2} = 2 \implies \log_2 x_1 + \log_2 x_2 = 4 \] Using the property of logarithms, \(\log_2 x_1 + \log_2 x_2 = \log_2 (x_1 x_2)\), we have: \[ \log_2 (x_1 x_2) = 4 \implies x_1 x_2 = 2^4 = 16 \] Now, we have the system of equations: \[ x_1 + x_2 = 12 \] \[ x_1 x_2 = 16 \] These are the sum and product of the roots of the quadratic equation: \[ t^2 - (x_1 + x_2)t + x_1 x_2 = 0 \implies t^2 - 12t + 16 = 0 \] To find the roots, we use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 16}}{2} \] \[ t = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} \] \[ t = \frac{12 \pm \sqrt{16 \cdot 5}}{2} = \frac{12 \pm 4\sqrt{5}}{2} \] \[ t = 6 \pm 2\sqrt{5} \] Thus, the \(x\)-coordinates of \(A\) and \(B\) are \(6 + 2\sqrt{5}\) and \(6 - 2\sqrt{5}\). The positive difference between the \(x\)-coordinates is: \[ (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \] Therefore, the positive difference between the \(x\)-coordinates of \(A\) and \(B\) is \(\boxed{4\sqrt{5}}\). The answer is \(\textbf{(D)}\), so the final answer is DDDDD. |
D |
| DeepSeek-Coder-V2-Lite-Instruct | To solve the problem, we need to determine the positive difference between the \( x \)-coordinates of points \( A \) and \( B \) on the graph of \( y = \log_2 x \), given that the midpoint of \(\overline{AB}\) is \((6, 2)\). 1. **Identify the coordinates of points \( A \) and \( B \):** Since the midpoint of \(\overline{AB}\) is \((6, 2)\), let the coordinates of \( A \) be \((x_1, \log_2 x_1)\) and the coordinates of \( B \) be \((x_2, \log_2 x_2)\). 2. **Use the midpoint formula:** The midpoint formula states that the coordinates of the midpoint are the averages of the coordinates of the endpoints. Therefore, we have: \[ \left( \frac{x_1 + x_2}{2}, \frac{\log_2 x_1 + \log_2 x_2}{2} \right) = (6, 2) \] This gives us two equations: \[ \frac{x_1 + x_2}{2} = 6 \quad \text{and} \quad \frac{\log_2 x_1 + \log_2 x_2}{2} = 2 \] 3. **Solve for \( x_1 \) and \( x_2 \):** From the first equation: \[ \frac{x_1 + x_2}{2} = 6 \implies x_1 + x_2 = 12 \] From the second equation: \[ \frac{\log_2 x_1 + \log_2 x_2}{2} = 2 \implies \log_2 x_1 + \log_2 x_2 = 4 \] Using the property of logarithms, \(\log_2 x_1 + \log_2 x_2 = \log_2 (x_1 x_2)\), we get: \[ \log_2 (x_1 x_2) = 4 \implies x_1 x_2 = 2^4 = 16 \] 4. **Form a system of equations:** We now have two equations: \[ x_1 + x_2 = 12 \quad \text{and} \quad x_1 x_2 = 16 \] 5. **Solve the system of equations:** These are the sum and product of the roots of a quadratic equation. Let \( x_1 \) and \( x_2 \) be the roots of the quadratic equation: \[ t^2 - (x_1 + x_2)t + x_1 x_2 = 0 \implies t^2 - 12t + 16 = 0 \] 6. **Solve the quadratic equation:** Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} = \frac{12 \pm 4\sqrt{5}}{2} = 6 \pm 2\sqrt{5} \] Therefore, the \( x \)-coordinates of \( A \) and \( B \) are \( 6 + 2\sqrt{5} \) and \( 6 - 2\sqrt{5} \). 7. **Calculate the positive difference:** The positive difference between the \( x \)-coordinates is: \[ |(6 + 2\sqrt{5}) - (6 - 2\sqrt{5})| = |6 + 2\sqrt{5} - 6 + 2\sqrt{5}| = |4\sqrt{5}| = 4\sqrt{5} \] Thus, the positive difference between the \( x \)-coordinates of \( A \) and \( B \) is \( 4\sqrt{5} \). Therefore, the correct answer is: \[ \boxed{D} \] |
D |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Understand the problem We are given that points A and B lie on the graph of y = log2x, and the midpoint of segment AB is (6, 2). We need to find the positive difference between the x-coordinates of A and B. ## Step 2: Use the midpoint formula The midpoint formula states that the midpoint of a segment with endpoints (x1, y1) and (x2, y2) is ((x1 + x2)/2, (y1 + y2)/2). In this case, the midpoint is (6, 2), so we can set up the equation: (x1 + x2)/2 = 6 and (y1 + y2)/2 = 2. ## Step 3: Simplify the equation Simplifying the equation, we get x1 + x2 = 12 and y1 + y2 = 4. ## Step 4: Use the equation of the graph Since points A and B lie on the graph of y = log2x, we can substitute y1 = log2x1 and y2 = log2x2 into the equation y1 + y2 = 4. We get log2x1 + log2x2 = 4. ## Step 5: Use the properties of logarithms Using the property of logarithms that loga(x) + loga(y) = loga(xy), we can rewrite the equation as log2(x1x2) = 4. ## Step 6: Solve for x1x2 Raising 2 to the power of both sides of the equation, we get x1x2 = 2^4 = 16. ## Step 7: Solve for x1 and x2 Now we have two equations: x1 + x2 = 12 and x1x2 = 16. We can solve this system of equations to find the values of x1 and x2. ## Step 8: Solve the quadratic equation Rearranging the first equation, we get x2 = 12 - x1. Substituting this into the second equation, we get x1(12 - x1) = 16. Expanding and rearranging, we get x1^2 - 12x1 + 16 = 0. ## Step 9: Factor the quadratic equation Factoring the quadratic equation, we get (x1 - 8)(x1 - 2) = 0. ## Step 10: Solve for x1 Solving for x1, we get x1 = 8 or x1 = 2. ## Step 11: Solve for x2 If x1 = 8, then x2 = 12 - 8 = 4. If x1 = 2, then x2 = 12 - 2 = 10. ## Step 12: Calculate the positive difference The positive difference between the x-coordinates of A and B is |x1 - x2| = |8 - 4| = 4 or |2 - 10| = 8. ## Step 13: Choose the correct answer Since the answer choices do not include 4, the correct answer must be 8. The final answer is: $\boxed{8}$ |
D |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Understand the problem We are given that points $A$ and $B$ lie on the graph of $y=\log_{2}x$, and the midpoint of $\overline{AB}$ is $(6, 2)$. We need to find the positive difference between the $x$-coordinates of $A$ and $B$. ## Step 2: Use the midpoint formula The midpoint formula states that the midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. In this case, the midpoint is $(6, 2)$, so we can set up the equation $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = (6, 2)$. ## Step 3: Substitute the values of $y_1$ and $y_2$ Since $A$ and $B$ lie on the graph of $y=\log_{2}x$, we can substitute $y_1 = \log_{2}x_1$ and $y_2 = \log_{2}x_2$ into the equation. This gives us $\left(\frac{x_1+x_2}{2}, \frac{\log_{2}x_1 + \log_{2}x_2}{2}\right) = (6, 2)$. ## Step 4: Simplify the equation Using the properties of logarithms, we can simplify the equation to $\left(\frac{x_1+x_2}{2}, \log_{2}\sqrt{x_1x_2}\right) = (6, 2)$. ## Step 5: Equate the $x$ and $y$ coordinates Equating the $x$ and $y$ coordinates, we get two equations: $\frac{x_1+x_2}{2} = 6$ and $\log_{2}\sqrt{x_1x_2} = 2$. ## Step 6: Solve the first equation Solving the first equation, we get $x_1+x_2 = 12$. ## Step 7: Solve the second equation Solving the second equation, we get $\sqrt{x_1x_2} = 2^2 = 4$, so $x_1x_2 = 16$. ## Step 8: Use the quadratic formula We can use the quadratic formula to solve for $x_1$ and $x_2$. Let $x_1 = x$ and $x_2 = 12-x$. Then, we have $x(12-x) = 16$. ## Step 9: Expand and simplify Expanding and simplifying, we get $12x - x^2 = 16$, which can be rewritten as $x^2 - 12x + 16 = 0$. ## Step 10: Solve the quadratic equation Solving the quadratic equation, we get $x = \frac{12 \pm \sqrt{144-64}}{2} = \frac{12 \pm \sqrt{80}}{2} = \frac{12 \pm 4\sqrt{5}}{2} = 6 \pm 2\sqrt{5}$. ## Step 11: Find the positive difference The positive difference between the $x$-coordinates of $A$ and $B$ is $|x_1 - x_2| = |(6 + 2\sqrt{5}) - (6 - 2\sqrt{5})| = 4\sqrt{5}$. The final answer is: $\boxed{4\sqrt{5}}$ |
D |
| Qwen2-72B-Instruct | Given the midpoint of $\overline{AB}$ as $(6, 2)$, we can use the properties of logarithms to find the x-coordinates of points $A$ and $B$. Since the midpoint's y-coordinate is $2$, this means that the average of the y-coordinates of $A$ and $B$ (which are both of the form $\log_2{x}$) is $2$. Let's denote the x-coordinates of $A$ and $B$ as $x_A$ and $x_B$ respectively. The y-coordinates of $A$ and $B$ are $\log_2{x_A}$ and $\log_2{x_B}$ respectively. The midpoint formula for the y-coordinates gives us: \[ \frac{\log_2{x_A} + \log_2{x_B}}{2} = 2 \] \[ \log_2{x_A} + \log_2{x_B} = 4 \] Using the property of logarithms that $\log_a{b} + \log_a{c} = \log_a{bc}$, we get: \[ \log_2{x_Ax_B} = 4 \] \[ x_Ax_B = 2^4 = 16 \] Now, let's consider the x-coordinates. The midpoint formula for the x-coordinates is: \[ \frac{x_A + x_B}{2} = 6 \] \[ x_A + x_B = 12 \] We now have a system of equations: \[ x_Ax_B = 16 \] \[ x_A + x_B = 12 \] Let's solve this system. We can express $x_B$ from the second equation as $x_B = 12 - x_A$ and substitute it into the first equation: \[ x_A(12 - x_A) = 16 \] \[ 12x_A - x_A^2 = 16 \] \[ x_A^2 - 12x_A + 16 = 0 \] Solving this quadratic equation using the quadratic formula, where $a=1$, $b=-12$, and $c=16$, we get: \[ x_A = \frac{-(-12) \pm \sqrt{(-12)^2 - 4*1*16}}{2*1} \] \[ x_A = \frac{12 \pm \sqrt{144 - 64}}{2} \] \[ x_A = \frac{12 \pm \sqrt{80}}{2} \] \[ x_A = \frac{12 \pm 4\sqrt{5}}{2} \] \[ x_A = 6 \pm 2\sqrt{5} \] Thus, the two possible values for $x_A$ are $6 + 2\sqrt{5}$ and $6 - 2\sqrt{5}$. Since $x_A$ and $x_B$ are interchangeable in this context, we can say that the x-coordinates of $A$ and $B$ are $6 + 2\sqrt{5}$ and $6 - 2\sqrt{5}$ respectively. The positive difference between these coordinates is: \[ (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \] Therefore, the answer is $\boxed{\textbf{(D)}~4\sqrt{5}}$, which when duplicated five times becomes DDDDD. |
D |
| baichuan4 | To solve this problem, we first need to understand the properties of the logarithmic function $y = \log_{2}x$ and the midpoint formula for a line segment. The midpoint formula states that if two points $A(x_1, y_1)$ and $B(x_2, y_2)$ lie on a line segment, then the midpoint $M$ of that line segment has coordinates given by: $$ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) $$ In this problem, we are given that the midpoint $M$ of the line segment $\overline{AB}$ is $(6, 2)$. We can use this information to derive an equation that relates the $x$-coordinates of points $A$ and $B$. Let's denote the $x$-coordinates of $A$ and $B$ as $x_A$ and $x_B$, respectively. Since $A$ and $B$ lie on the graph of $y = \log_{2}x$, their $y$-coordinates are $\log_{2}x_A$ and $\log_{2}x_B$, respectively. Using the midpoint formula, we have: $$ 6 = \frac{x_A + x_B}{2} $$ $$ 2 = \frac{\log_{2}x_A + \log_{2}x_B}{2} $$ From the first equation, we can derive: $$ x_A + x_B = 12 $$ From the second equation, we can derive: $$ \log_{2}x_A + \log_{2}x_B = 4 $$ Using the logarithm property $\log_{a}(bc) = \log_{a}b + \log_{a}c$, we can rewrite the second equation as: $$ \log_{2}(x_Ax_B) = 4 $$ Exponentiating both sides with base 2, we get: $$ x_Ax_B = 2^4 = 16 $$ Now, we have two equations: 1. $x_A + x_B = 12$ 2. $x_Ax_B = 16$ We can solve this system of equations by substituting $x_B = 12 - x_A$ into the second equation: $$ x_A(12 - x_A) = 16 $$ $$ 12x_A - x_A^2 = 16 $$ $$ x_A^2 - 12x_A + 16 = 0 $$ This is a quadratic equation that we can solve using the quadratic formula: $$ x_A = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} $$ $$ x_A = \frac{12 \pm \sqrt{144 - 64}}{2} $$ $$ x_A = \frac{12 \pm \sqrt{80}}{2} $$ $$ x_A = \frac{12 \pm 4\sqrt{5}}{2} $$ $$ x_A = 6 \pm 2\sqrt{5} $$ Since $x_A$ and $x_B$ are the $x$-coordinates of points $A$ and $B$, respectively, and $x_B = 12 - x_A$, we have: $$ x_B = 12 - (6 \pm 2\sqrt{5}) $$ $$ x_B = 6 \mp 2\sqrt{5} $$ The positive difference between the $x$-coordinates of $A$ and $B$ is: $$ |x_A - x_B| = |(6 + 2\sqrt{5}) - (6 - 2\sqrt{5})| $$ $$ |x_A - x_B| = |4\sqrt{5}| $$ $$ |x_A - x_B| = 4\sqrt{5} $$ Therefore, the correct answer is $\textbf{(D)}~4\sqrt{5}$. |
D |
| doubao | Let the coordinates of $A$ be $(x_1, \log_2 x_1)$ and the coordinates of $B$ be $(x_2, \log_2 x_2)$. The midpoint formula gives $\frac{x_1+x_2}{2}=6$ and $\frac{\log_2 x_1+\log_2 x_2}{2}=2$. The second equation can be simplified to $\log_2 (x_1x_2)=4$, so $x_1x_2=2^4=16$. From the first equation, $x_1+x_2=12$. Now $(x_2-x_1)^2=(x_1+x_2)^2-4x_1x_2=12^2-4\times16=144-64=80$, so $|x_2-x_1|=\sqrt{80}=4\sqrt{5}$. The positive difference is $4\sqrt{5}$. So the answer is D. DDDDD. | D |
| gpto1 | DDDDD | D |
| kimi | To solve the problem, we start by letting the coordinates of points \( A \) and \( B \) be \( (a, \log_2 a) \) and \( (b, \log_2 b) \) respectively. Given that the midpoint of \( \overline{AB} \) is \( (6, 2) \), we can use the midpoint formula to set up the following equations: \[ \left( \frac{a + b}{2}, \frac{\log_2 a + \log_2 b}{2} \right) = (6, 2) \] From this, we get two separate equations: 1. \(\frac{a + b}{2} = 6\) 2. \(\frac{\log_2 a + \log_2 b}{2} = 2\) First, solve the first equation for \( a + b \): \[ a + b = 12 \] Next, solve the second equation for \( \log_2 a + \log_2 b \): \[ \frac{\log_2 a + \log_2 b}{2} = 2 \implies \log_2 a + \log_2 b = 4 \] Using the properties of logarithms, we can combine the logs: \[ \log_2 (a \cdot b) = 4 \implies a \cdot b = 2^4 = 16 \] Now we have a system of equations: 1. \( a + b = 12 \) 2. \( ab = 16 \) These are the sum and product of the roots of a quadratic equation. We can form the quadratic equation whose roots are \( a \) and \( b \): \[ x^2 - (a + b)x + ab = 0 \implies x^2 - 12x + 16 = 0 \] Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} = \frac{12 \pm 4\sqrt{5}}{2} = 6 \pm 2\sqrt{5} \] Thus, the \( x \)-coordinates of points \( A \) and \( B \) are \( 6 + 2\sqrt{5} \) and \( 6 - 2\sqrt{5} \). The positive difference between the \( x \)-coordinates of \( A \) and \( B \) is: \[ (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \] Therefore, the positive difference between the \( x \)-coordinates of \( A \) and \( B \) is \(\boxed{4\sqrt{5}}\). |
D |
| minimax | Let's denote the coordinates of points $A$ and $B$ as $(x_1, y_1)$ and $(x_2, y_2)$ respectively, where $y_1 = \log_2(x_1)$ and $y_2 = \log_2(x_2)$ because they lie on the graph of $y = \log_2(x)$. The midpoint $M$ of $\overline{AB}$ has coordinates $(6, 2)$. The midpoint formula gives us: $$ M_x = \frac{x_1 + x_2}{2} = 6 $$ $$ M_y = \frac{y_1 + y_2}{2} = 2 $$ From $M_y$, we have: $$ \frac{\log_2(x_1) + \log_2(x_2)}{2} = 2 $$ Using the properties of logarithms, we can combine the logs: $$ \log_2(\sqrt{x_1x_2}) = 2 $$ This implies that: $$ \sqrt{x_1x_2} = 2^2 = 4 $$ Squaring |
D |
| wenxinyiyan | Let the points $A$ and $B$ be $(x_1, \log_2{x_1})$ and $(x_2, \log_2{x_2})$ respectively. The midpoint of $\overline{AB}$ is given to be $(6, 2)$. Therefore, by the midpoint formula, we have: $$\frac{x_1 + x_2}{2} = 6 \quad \text{and} \quad \frac{\log_2{x_1} + \log_2{x_2}}{2} = 2$$From the first equation, we get: $$x_1 + x_2 = 12 \quad \text{(1)}$$From the second equation, we get: $$\log_2{x_1} + \log_2{x_2} = 4$$Using the properties of logarithms, we can combine the two logarithms on the left side: $$\log_2{x_1x_2} = 4$$This gives us: $$x_1x_2 = 2^4 = 16 \quad \text{(2)}$$Now, we have two equations (1) and (2) involving $x_1$ and $x_2$. We can solve these equations to find the values of $x_1$ and $x_2$. From equation (1), we can express $x_2$ in terms of $x_1$: $$x_2 = 12 - x_1$$Substituting this into equation (2), we get: $$x_1(12 - x_1) = 16$$Expanding and rearranging, we get a quadratic equation: $$x_1^2 - 12x_1 + 16 = 0$$Solving this quadratic equation, we get two possible values for $x_1$: $$x_1 = 6 + 2\sqrt{5} \quad \text{or} \quad x_1 = 6 - 2\sqrt{5}$$Since $x_2 = 12 - x_1$, the corresponding values for $x_2$ are: $$x_2 = 6 - 2\sqrt{5} \quad \text{or} \quad x_2 = 6 + 2\sqrt{5}$$Without loss of generality, let's assume $x_1 = 6 + 2\sqrt{5}$ and $x_2 = 6 - 2\sqrt{5}$. The positive difference between the $x$-coordinates of $A$ and $B$ is then: $$|x_1 - x_2| = |(6 + 2\sqrt{5}) - (6 - 2\sqrt{5})| = 4\sqrt{5}$$Therefore, the answer is $\textbf{(D)}~4\sqrt{5}$. |
D |
| zhipu | 1. **Identify the coordinates of points $A$ and $B$:** Let the coordinates of point $A$ be $(x_1, y_1)$ and the coordinates of point $B$ be $(x_2, y_2)$. Since both points lie on the graph of $y = \log_2 x$, we have: \[ y_1 = \log_2 x_1 \quad \text{and} \quad y_2 = \log_2 x_2. \] 2. **Use the midpoint formula:** The midpoint of $\overline{AB}$ is given as $(6, 2)$. By the midpoint formula, we know: \[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (6, 2). \] This gives us two equations: \[ \frac{x_1 + x_2}{2} = 6 \quad \text{and} \quad \frac{y_1 + y_2}{2} = 2. \] 3. **Solve for $x_1 + x_2$ and $y_1 + y_2$:** From the equations above, we find: \[ x_1 + x_2 = 12 \quad \text{and} \quad y_1 + y_2 = 4. \] 4. **Relate $y_1$ and $y_2$ to $x_1$ and $x_2$:** Since $y_1 = \log_2 x_1$ and $y_2 = \log_2 x_2$, we can write: \[ \log_2 x_1 + \log_2 x_2 = 4. \] Using the logarithmic property $\log_b a + \log_b c = \log_b (ac)$, we have: \[ \log_2 (x_1 x_2) = 4. \] Therefore, \[ x_1 x_2 = 2^4 = 16. \] 5. **Find the difference between $x_1$ and $x_2$:** Let $d = x_2 - x_1$. We know: \[ x_2 = x_1 + d. \] Substituting into $x_1 + x_2 = 12$, we get: \[ x_1 + (x_1 + d) = 12 \implies 2x_1 + d = 12. \] Also, substituting into $x_1 x_2 = 16$, we get: \[ x_1(x_1 + d) = 16. \] 6. **Solve the system of equations:** From $2x_1 + d = 12$, solve for $d$: \[ d = 12 - 2x_1. \] Substitute into $x_1(x_1 + d) = 16$: \[ x_1(x_1 + 12 - 2x_1) = 16 \implies x_1(12 - x_1) = 16. \] Rearrange and solve the quadratic equation: \[ x_1^2 - 12x_1 + 16 = 0. \] Using the quadratic formula, $x_1 = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} = \frac{12 \pm 4\sqrt{5}}{2} = 6 \pm 2\sqrt{5}$. 7. **Calculate $d$:** If $x_1 = 6 - 2\sqrt{5}$, then $x_2 = 6 + 2\sqrt{5}$ (and vice versa), so: \[ d = x_2 - x_1 = (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5}. \] Thus, the positive difference between the $x$-coordinates of $A$ and $B$ is $\boxed{\textbf{(D)}~4\sqrt{5}}$. DDDDD |
D |
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