💡First, I need to calculate the value of (2^4 + 2^4 + 2^4).I notice that all the terms are the same, so I can factor out (2^4):[2^4 + 2^4 + 2^4 = 3 times 2^4]Next, I recognize that (3) can be written as (2 + 1), so:[3 times 2^4 = (2 + 1) times 2^4 = 2 times 2^4 + 1 times 2^4]Using the exponent rule (a^m times a^n = a^{m+n}), I can simplify each term:[2 times 2^4 = 2^{1+4} = 2^5][1 times 2^4 = 2^4]Adding these together:[2^5 + 2^4 = 32 + 16 = 48]However, this result doesn't match any of the provided answer choices. I must have made a mistake in my approach.Let me try a different method by directly calculating each term:[2^4 = 16][2^4 + 2^4 + 2^4 = 16 + 16 + 16 = 48]Again, the result is (48), which still doesn't match the answer choices. It seems there's an inconsistency between my calculations and the provided options. Perhaps I need to re-examine the problem or consider alternative interpretations.

💡Okay, so I have this function f(x) = a(e^x + a) - x, and I need to do two things: first, discuss its monotonicity, and second, prove that when a is greater than 0, f(x) is greater than 2 ln a + 3/2. Alright, let's start with the first part.**Part 1: Monotonicity of f(x)**Monotonicity refers to whether the function is increasing or decreasing over certain intervals. To determine this, I remember that I need to look at the derivative of the function. If the derivative is positive, the function is increasing; if it's negative, the function is decreasing.So, let's find the derivative of f(x). The function is f(x) = a(e^x + a) - x. Let's differentiate term by term.The derivative of a(e^x) is a*e^x because the derivative of e^x is e^x. The derivative of a*a is 0 because it's a constant. Then, the derivative of -x is -1. So putting it all together, f'(x) = a*e^x - 1.Now, to discuss the monotonicity, I need to analyze where f'(x) is positive, negative, or zero.Case 1: a ≤ 0If a is less than or equal to zero, then a*e^x is going to be less than or equal to zero because e^x is always positive. So, a*e^x ≤ 0. Then, f'(x) = a*e^x - 1 ≤ -1, which is always negative. So, if a ≤ 0, the derivative is always negative, meaning the function is monotonically decreasing everywhere.Case 2: a > 0If a is positive, then a*e^x is positive. Now, we can set the derivative equal to zero to find critical points.Set f'(x) = 0:a*e^x - 1 = 0 a*e^x = 1 e^x = 1/a x = ln(1/a) x = -ln(a)So, the critical point is at x = -ln(a). Now, we can analyze the intervals around this point.For x < -ln(a):Since e^x is increasing, when x is less than -ln(a), e^x < 1/a. Therefore, a*e^x < 1, so f'(x) = a*e^x - 1 < 0. Thus, the function is decreasing on (-∞, -ln(a)).For x > -ln(a):Similarly, e^x > 1/a, so a*e^x > 1, which means f'(x) = a*e^x - 1 > 0. Therefore, the function is increasing on (-ln(a), ∞).So, summarizing part 1:- If a ≤ 0, f(x) is monotonically decreasing on ℝ.- If a > 0, f(x) is decreasing on (-∞, -ln(a)) and increasing on (-ln(a), ∞).**Part 2: Proving f(x) > 2 ln a + 3/2 when a > 0**Alright, now I need to show that for a > 0, f(x) is always greater than 2 ln a + 3/2. Hmm, okay.From part 1, we know that when a > 0, f(x) has a minimum at x = -ln(a). So, if I can find the minimum value of f(x) and show that it's greater than 2 ln a + 3/2, then the inequality will hold for all x.Let's compute f(-ln(a)):f(-ln(a)) = a(e^{-ln(a)} + a) - (-ln(a))First, simplify e^{-ln(a)}. Remember that e^{ln(b)} = b, so e^{-ln(a)} = 1/e^{ln(a)} = 1/a.So, substituting back:f(-ln(a)) = a(1/a + a) + ln(a)Simplify inside the parentheses:1/a + a = (1 + a^2)/aSo, f(-ln(a)) = a*(1 + a^2)/a + ln(a) = (1 + a^2) + ln(a)Therefore, the minimum value of f(x) is 1 + a^2 + ln(a).Now, we need to show that 1 + a^2 + ln(a) > 2 ln a + 3/2.Let's rearrange the inequality:1 + a^2 + ln(a) > 2 ln a + 3/2 Subtract 2 ln a from both sides: 1 + a^2 - ln(a) > 3/2 Subtract 3/2 from both sides: a^2 - ln(a) - 1/2 > 0So, define g(a) = a^2 - ln(a) - 1/2. We need to show that g(a) > 0 for all a > 0.To analyze g(a), let's find its derivative to see its behavior.g'(a) = 2a - (1/a)Set g'(a) = 0 to find critical points:2a - 1/a = 0 Multiply both sides by a: 2a^2 - 1 = 0 2a^2 = 1 a^2 = 1/2 a = sqrt(1/2) = √2 / 2 ≈ 0.707So, the critical point is at a = √2 / 2.Now, let's analyze the intervals around this critical point.For a < √2 / 2:Pick a value less than √2 / 2, say a = 0.5.g'(0.5) = 2*(0.5) - 1/0.5 = 1 - 2 = -1 < 0So, g(a) is decreasing on (0, √2 / 2).For a > √2 / 2:Pick a value greater than √2 / 2, say a = 1.g'(1) = 2*1 - 1/1 = 2 - 1 = 1 > 0So, g(a) is increasing on (√2 / 2, ∞).Therefore, the minimum of g(a) occurs at a = √2 / 2.Let's compute g(√2 / 2):g(√2 / 2) = ( (√2 / 2)^2 ) - ln(√2 / 2) - 1/2 = ( (2 / 4) ) - ln(√2 / 2) - 1/2 = (1/2) - ln(√2 / 2) - 1/2 = - ln(√2 / 2)Simplify ln(√2 / 2):ln(√2 / 2) = ln(√2) - ln(2) = (1/2) ln(2) - ln(2) = (-1/2) ln(2)So, g(√2 / 2) = - (-1/2 ln(2)) = (1/2) ln(2) ≈ 0.3466 > 0Since the minimum value of g(a) is positive, and g(a) tends to infinity as a approaches 0 or infinity, we can conclude that g(a) > 0 for all a > 0.Therefore, 1 + a^2 + ln(a) > 2 ln a + 3/2 for all a > 0, which implies that f(x) > 2 ln a + 3/2 for all x when a > 0.**Final Answer**The function ( f(x) ) is monotonically decreasing for ( a leq 0 ) and has a minimum at ( x = -ln(a) ) for ( a > 0 ). When ( a > 0 ), ( f(x) ) is always greater than ( 2ln a + frac{3}{2} ). Thus, the inequality is proven.boxed{f(x) > 2ln a + frac{3}{2} text{ for } a > 0}

💡Okay, so I have this problem about two circles, and I need to figure out their positional relationship. The options are disjoint, intersecting, externally tangent, or internally tangent. Hmm, let me try to remember how to approach this.First, I think I need to find the centers and radii of both circles. The equations given are in the general form, so I should convert them into the standard form of a circle's equation, which is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius.Starting with the first circle: (x^2 + y^2 - 6x = 0). I need to complete the square for the x-terms. Let me rewrite it:(x^2 - 6x + y^2 = 0).To complete the square for (x^2 - 6x), I take half of -6, which is -3, square it to get 9, and add it to both sides:((x^2 - 6x + 9) + y^2 = 9).This simplifies to:((x - 3)^2 + y^2 = 9).So, the center of the first circle is at (3, 0) and the radius is (sqrt{9} = 3).Now, moving on to the second circle: (x^2 + y^2 + 8y + 12 = 0). Again, I need to complete the square, this time for the y-terms.Rewriting the equation:(x^2 + y^2 + 8y = -12).Completing the square for (y^2 + 8y), I take half of 8, which is 4, square it to get 16, and add it to both sides:(x^2 + (y^2 + 8y + 16) = -12 + 16).This simplifies to:(x^2 + (y + 4)^2 = 4).So, the center of the second circle is at (0, -4) and the radius is (sqrt{4} = 2).Alright, now I have both centers and radii. The first circle has center (3, 0) and radius 3, and the second circle has center (0, -4) and radius 2.Next, I need to find the distance between the two centers to determine their positional relationship. The distance formula between two points ((x_1, y_1)) and ((x_2, y_2)) is:(d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}).Plugging in the centers:(d = sqrt{(0 - 3)^2 + (-4 - 0)^2}).Calculating the differences:(d = sqrt{(-3)^2 + (-4)^2}).Squaring the values:(d = sqrt{9 + 16}).Adding them up:(d = sqrt{25}).Taking the square root:(d = 5).So, the distance between the centers is 5 units.Now, I need to compare this distance with the sum and difference of the radii to determine the positional relationship.The radii are 3 and 2, so their sum is (3 + 2 = 5) and their difference is (3 - 2 = 1).Since the distance between the centers is equal to the sum of the radii, that means the circles are touching each other at exactly one point from the outside. Therefore, they are externally tangent.Wait, let me just make sure I didn't mix up internal and external tangent. If the distance between centers is equal to the sum, it's external tangent because they touch from outside. If it were equal to the difference, it would be internal tangent, meaning one is inside the other and they touch at one point. Since 5 is the sum, it's external.So, the answer should be C: Externally Tangent.**Final Answer**boxed{C}

💡Okay, so I have this problem where I need to find how many points with positive x-coordinates lie on two or more of the graphs of the functions ( y = log_2 x ), ( y = log_x 2 ), ( y = log_4 x ), and ( y = x - 2 ). The options are from A to E, with E being 7. Hmm, let me try to figure this out step by step.First, I know that to find intersection points between two functions, I need to set them equal to each other and solve for x. Since all these functions are defined for positive x, I don't have to worry about negative x-values. Let me list out all possible pairs of functions and see where they intersect.1. **Intersection between ( y = log_2 x ) and ( y = log_x 2 ):** - Let me set them equal: ( log_2 x = log_x 2 ). - I remember that ( log_b a = frac{1}{log_a b} ), so ( log_x 2 = frac{1}{log_2 x} ). - Let me let ( t = log_2 x ), so the equation becomes ( t = frac{1}{t} ). - Multiplying both sides by t gives ( t^2 = 1 ), so ( t = 1 ) or ( t = -1 ). - If ( t = 1 ), then ( log_2 x = 1 ) which means ( x = 2^1 = 2 ). - If ( t = -1 ), then ( log_2 x = -1 ) which means ( x = 2^{-1} = frac{1}{2} ). - So, the intersection points are at ( x = 2 ) and ( x = frac{1}{2} ).2. **Intersection between ( y = log_2 x ) and ( y = log_4 x ):** - Set them equal: ( log_2 x = log_4 x ). - I know that ( log_4 x = frac{log_2 x}{log_2 4} = frac{log_2 x}{2} ). - So, ( log_2 x = frac{log_2 x}{2} ). - Subtracting ( frac{log_2 x}{2} ) from both sides: ( frac{log_2 x}{2} = 0 ). - This implies ( log_2 x = 0 ), so ( x = 2^0 = 1 ). - So, the intersection point is at ( x = 1 ).3. **Intersection between ( y = log_2 x ) and ( y = x - 2 ):** - Set them equal: ( log_2 x = x - 2 ). - This seems trickier. Maybe I can try plugging in some values. - Let me try ( x = 2 ): ( log_2 2 = 1 ), and ( 2 - 2 = 0 ). Not equal. - ( x = 4 ): ( log_2 4 = 2 ), and ( 4 - 2 = 2 ). Equal! So, ( x = 4 ) is a solution. - Let me check if there's another solution. Maybe ( x = 1 ): ( log_2 1 = 0 ), and ( 1 - 2 = -1 ). Not equal. - Maybe ( x = frac{1}{2} ): ( log_2 frac{1}{2} = -1 ), and ( frac{1}{2} - 2 = -frac{3}{2} ). Not equal. - It seems like ( x = 4 ) is the only solution here.4. **Intersection between ( y = log_x 2 ) and ( y = log_4 x ):** - Set them equal: ( log_x 2 = log_4 x ). - Let me express both in terms of base 2. - ( log_x 2 = frac{1}{log_2 x} ) and ( log_4 x = frac{log_2 x}{2} ). - So, ( frac{1}{log_2 x} = frac{log_2 x}{2} ). - Let ( t = log_2 x ), so ( frac{1}{t} = frac{t}{2} ). - Multiply both sides by 2t: ( 2 = t^2 ). - So, ( t = sqrt{2} ) or ( t = -sqrt{2} ). - If ( t = sqrt{2} ), then ( log_2 x = sqrt{2} ), so ( x = 2^{sqrt{2}} ). - If ( t = -sqrt{2} ), then ( log_2 x = -sqrt{2} ), so ( x = 2^{-sqrt{2}} ). - These are valid positive x-values, so we have two more intersection points.5. **Intersection between ( y = log_x 2 ) and ( y = x - 2 ):** - Set them equal: ( log_x 2 = x - 2 ). - Again, this seems tricky. Let me try some values. - ( x = 2 ): ( log_2 2 = 1 ), and ( 2 - 2 = 0 ). Not equal. - ( x = 4 ): ( log_4 2 = frac{1}{2} ), and ( 4 - 2 = 2 ). Not equal. - ( x = 1 ): ( log_1 2 ) is undefined because base 1 is invalid. - ( x = frac{1}{2} ): ( log_{1/2} 2 = -1 ), and ( frac{1}{2} - 2 = -frac{3}{2} ). Not equal. - Maybe ( x = 3 ): ( log_3 2 ) is approximately 0.631, and ( 3 - 2 = 1 ). Not equal. - It seems like there might not be an intersection here, or at least not an easy one to find. Maybe I can check if ( x = 2^{sqrt{2}} ) or ( x = 2^{-sqrt{2}} ) satisfy this equation, but that might be complicated. I'll come back to this.6. **Intersection between ( y = log_4 x ) and ( y = x - 2 ):** - Set them equal: ( log_4 x = x - 2 ). - Let me try some values. - ( x = 4 ): ( log_4 4 = 1 ), and ( 4 - 2 = 2 ). Not equal. - ( x = 2 ): ( log_4 2 = frac{1}{2} ), and ( 2 - 2 = 0 ). Not equal. - ( x = 1 ): ( log_4 1 = 0 ), and ( 1 - 2 = -1 ). Not equal. - ( x = 3 ): ( log_4 3 ) is approximately 0.792, and ( 3 - 2 = 1 ). Not equal. - Maybe ( x = 16 ): ( log_4 16 = 2 ), and ( 16 - 2 = 14 ). Not equal. - It seems like there might not be an intersection here either, or at least not an easy one to find.7. **Intersection between ( y = log_x 2 ) and ( y = x - 2 ):** - Wait, I think I already did this one above. It seems like there isn't an easy solution, so maybe there are no intersection points here.8. **Intersection between ( y = log_2 x ) and ( y = x - 2 ):** - I already found that ( x = 4 ) is a solution.9. **Intersection between ( y = log_x 2 ) and ( y = log_4 x ):** - I found two solutions: ( x = 2^{sqrt{2}} ) and ( x = 2^{-sqrt{2}} ).10. **Intersection between ( y = log_4 x ) and ( y = x - 2 ):** - I didn't find any easy solutions, so maybe there are none.11. **Intersection between ( y = log_2 x ) and ( y = log_x 2 ):** - I found two solutions: ( x = 2 ) and ( x = frac{1}{2} ).12. **Intersection between ( y = log_2 x ) and ( y = log_4 x ):** - I found one solution: ( x = 1 ).13. **Intersection between ( y = log_x 2 ) and ( y = x - 2 ):** - I didn't find any solutions.14. **Intersection between ( y = log_4 x ) and ( y = x - 2 ):** - I didn't find any solutions.15. **Intersection between ( y = log_2 x ) and ( y = x - 2 ):** - I found one solution: ( x = 4 ).16. **Intersection between ( y = log_x 2 ) and ( y = log_4 x ):** - I found two solutions: ( x = 2^{sqrt{2}} ) and ( x = 2^{-sqrt{2}} ).17. **Intersection between ( y = log_x 2 ) and ( y = x - 2 ):** - I didn't find any solutions.18. **Intersection between ( y = log_4 x ) and ( y = x - 2 ):** - I didn't find any solutions.Wait, I think I might be repeating some pairs here. Let me make sure I cover all unique pairs without repetition.The unique pairs are:- ( y = log_2 x ) and ( y = log_x 2 )- ( y = log_2 x ) and ( y = log_4 x )- ( y = log_2 x ) and ( y = x - 2 )- ( y = log_x 2 ) and ( y = log_4 x )- ( y = log_x 2 ) and ( y = x - 2 )- ( y = log_4 x ) and ( y = x - 2 )I think I covered all these pairs. So, from the above, the intersection points I found are:- ( x = 2 ) and ( x = frac{1}{2} ) from the first pair.- ( x = 1 ) from the second pair.- ( x = 4 ) from the third pair.- ( x = 2^{sqrt{2}} ) and ( x = 2^{-sqrt{2}} ) from the fourth pair.- No solutions from the fifth and sixth pairs.So, that gives me a total of 5 intersection points: ( x = frac{1}{2} ), ( x = 1 ), ( x = 2 ), ( x = 4 ), ( x = 2^{sqrt{2}} ), and ( x = 2^{-sqrt{2}} ). Wait, that's 6 points. But let me double-check if ( x = 2^{sqrt{2}} ) and ( x = 2^{-sqrt{2}} ) are distinct from the others.Yes, ( 2^{sqrt{2}} ) is approximately 2.665, and ( 2^{-sqrt{2}} ) is approximately 0.375, which are different from ( frac{1}{2} ), 1, 2, and 4. So, that's 6 distinct points.But wait, the options only go up to E.7, and the highest is 7. But I think I might have made a mistake because the answer choices are A.3, B.4, C.5, D.6, E.7.Wait, in the initial problem, the user mentioned the final answer was B.4, but in my detailed thought process, I arrived at 6 points. Hmm, maybe I overcounted.Let me go back and check each intersection:1. ( y = log_2 x ) and ( y = log_x 2 ): 2 points.2. ( y = log_2 x ) and ( y = log_4 x ): 1 point.3. ( y = log_2 x ) and ( y = x - 2 ): 1 point.4. ( y = log_x 2 ) and ( y = log_4 x ): 2 points.5. ( y = log_x 2 ) and ( y = x - 2 ): 0 points.6. ( y = log_4 x ) and ( y = x - 2 ): 0 points.So, total points: 2 + 1 + 1 + 2 = 6. But the answer choices only go up to 7, and the initial answer was B.4. Maybe I'm missing something.Wait, perhaps some of these intersection points are the same. For example, does ( x = 2 ) satisfy more than two functions? Let me check:- At ( x = 2 ): - ( y = log_2 2 = 1 ) - ( y = log_2 2 = 1 ) - ( y = log_4 2 = 0.5 ) - ( y = 2 - 2 = 0 ) - So, only ( y = log_2 x ) and ( y = log_x 2 ) intersect here.At ( x = frac{1}{2} ): - ( y = log_2 frac{1}{2} = -1 ) - ( y = log_{1/2} 2 = -1 ) - ( y = log_4 frac{1}{2} = -0.5 ) - ( y = frac{1}{2} - 2 = -1.5 ) - So, only ( y = log_2 x ) and ( y = log_x 2 ) intersect here.At ( x = 1 ): - ( y = log_2 1 = 0 ) - ( y = log_1 2 ) is undefined - ( y = log_4 1 = 0 ) - ( y = 1 - 2 = -1 ) - So, ( y = log_2 x ) and ( y = log_4 x ) intersect here.At ( x = 4 ): - ( y = log_2 4 = 2 ) - ( y = log_4 2 = 0.5 ) - ( y = log_4 4 = 1 ) - ( y = 4 - 2 = 2 ) - So, ( y = log_2 x ) and ( y = x - 2 ) intersect here.At ( x = 2^{sqrt{2}} ) and ( x = 2^{-sqrt{2}} ): - These are unique points where ( y = log_x 2 ) and ( y = log_4 x ) intersect.So, all these points are distinct. Therefore, there are 6 intersection points. But the answer choices only go up to 7, and the initial answer was B.4. Maybe I'm misunderstanding the question.Wait, the question says "points, with positive x-coordinates, lie on two or more of these graphs." So, each intersection point is counted once, regardless of how many functions intersect there. So, if three functions intersect at a point, it's still just one point.But in my case, each intersection is between two functions, so each point is unique. So, if I have 6 intersection points, each involving two functions, then the total number of points is 6.But the answer choices are A.3, B.4, C.5, D.6, E.7. So, D.6 would be the answer. But the initial answer was B.4. Maybe I made a mistake in counting.Wait, let me check the intersections again:1. ( y = log_2 x ) and ( y = log_x 2 ): 2 points.2. ( y = log_2 x ) and ( y = log_4 x ): 1 point.3. ( y = log_2 x ) and ( y = x - 2 ): 1 point.4. ( y = log_x 2 ) and ( y = log_4 x ): 2 points.5. ( y = log_x 2 ) and ( y = x - 2 ): 0 points.6. ( y = log_4 x ) and ( y = x - 2 ): 0 points.Total: 2 + 1 + 1 + 2 = 6 points.But maybe some of these points are the same. For example, does ( x = 2^{sqrt{2}} ) coincide with any other point? Let me calculate ( 2^{sqrt{2}} ) is approximately 2.665, which is different from 2, 1, 4, and ( 2^{-sqrt{2}} ) is approximately 0.375, which is different from ( frac{1}{2} ), 1, 2, 4.So, all 6 points are distinct. Therefore, the answer should be D.6.But the initial answer was B.4. Maybe the user made a mistake in their initial thought process. Let me check their thought process again.In the initial thought process, they found 4 points: ( x = 1 ), ( x = 2 ), ( x = frac{1}{2} ), and ( x = 4 ). They didn't consider the intersections between ( y = log_x 2 ) and ( y = log_4 x ), which gives two more points. So, maybe they missed those.Therefore, the correct answer should be D.6.

💡First, I recognize that the problem involves a square with a side length of 10 units. When a square is cut diagonally from one corner to the opposite corner, it results in two congruent right-angled triangles.Each of these triangles will have two sides that are the original sides of the square, so each of these sides will be 10 units long. The third side of the triangle is the diagonal of the square, which can be calculated using the Pythagorean theorem.Using the Pythagorean theorem for a right-angled triangle with legs of 10 units each:[text{Diagonal} = sqrt{10^2 + 10^2} = sqrt{100 + 100} = sqrt{200} = 10sqrt{2} text{ units}]Therefore, each triangle has sides of 10 units, 10 units, and (10sqrt{2}) units.

💡First, I need to understand the movements of Car A. It starts by traveling 25 km along the main road, then takes a right turn and goes 15 km. After that, it takes a left turn and travels another 25 km before returning to the main road. This means Car A has effectively moved 25 km away from its starting point and then 25 km back towards the main road, resulting in a net movement of 25 km along the main road.Next, I consider the total initial distance between the two cars, which is 150 km. After Car A has moved 25 km, the remaining distance between Car A and Car B is 150 km minus 25 km, which equals 125 km.However, the problem states that the distance between the two cars at this point is 38 km. This indicates that Car B has also moved towards Car A. To find out how far Car B has traveled, I subtract the remaining distance (38 km) from the distance after Car A's movement (125 km). This gives me 125 km minus 38 km, which equals 87 km.Therefore, Car B has traveled 87 km along the main road.