answer：Okay, so I need to evaluate log base 16 of 2, which is written as log₁₆(2). Hmm, I remember that logarithms are the inverse of exponentials, so log₁₆(2) is asking the question: "16 raised to what power gives me 2?" In mathematical terms, if log₁₆(2) = x, then 16^x = 2.Alright, so I need to find x such that 16^x equals 2. I know that 16 is a power of 2, specifically 2^4 because 2 multiplied by itself four times is 16 (2*2=4, 4*2=8, 8*2=16). So, 16 can be written as 2^4. Let me substitute that into the equation.So, replacing 16 with 2^4, the equation becomes (2^4)^x = 2. Now, when you raise a power to another power, you multiply the exponents. So, (2^4)^x is the same as 2^(4x). Therefore, the equation simplifies to 2^(4x) = 2.Now, since both sides of the equation have the same base, which is 2, I can set the exponents equal to each other. That means 4x must equal 1 because 2^1 is 2. So, 4x = 1. To solve for x, I divide both sides by 4, which gives me x = 1/4.Wait, let me double-check that. If I plug x = 1/4 back into the original equation, does it hold true? Let's see: 16^(1/4). The fourth root of 16 is 2 because 2^4 is 16. So, 16^(1/4) is indeed 2. That checks out. Therefore, x = 1/4 is the correct solution.Alternatively, I could use the change of base formula for logarithms, which states that log_b(a) = log_c(a) / log_c(b), where c is any positive number. If I choose c to be 2, then log₁₆(2) becomes log₂(2) / log₂(16). I know that log₂(2) is 1 because any number to the power of 1 is itself. And log₂(16) is 4 because 2^4 is 16. So, log₁₆(2) = 1/4, which confirms my earlier result.Is there another way to think about this? Maybe using exponent rules or properties of logarithms. Let's see. I recall that log_b(a) is the exponent you need to raise b to get a. Since 16 is 2^4, and 2 is 2^1, I can express both in terms of base 2. So, log₁₆(2) is the exponent x such that (2^4)^x = 2^1. As before, this simplifies to 2^(4x) = 2^1, leading to 4x = 1 and x = 1/4.I think I've approached this problem from a couple of different angles now, and each time I arrive at the same answer. That gives me confidence that x = 1/4 is indeed the correct value for log₁₆(2). It's always good to verify the solution by plugging it back into the original equation to ensure it works, and in this case, it does.Another thought: sometimes, when dealing with logarithms, it's helpful to express both the base and the argument in terms of a common base. Here, both 16 and 2 are powers of 2, which made the problem straightforward. If they weren't, I might have to use the change of base formula or natural logarithms to find the value. But in this scenario, recognizing that 16 is 2^4 was key.I also remember that logarithms can be interpreted as exponents, so log₁₆(2) is essentially asking how many times I need to multiply 16 by itself to get 2. Since 16 is larger than 2, the exponent must be a fraction between 0 and 1. Specifically, since 16 is 2^4, taking the fourth root (which is the same as raising to the 1/4 power) brings it down to 2. That makes sense because roots are fractional exponents.Just to recap, the steps I took were:1. Recognize that 16 is 2^4.2. Rewrite the logarithmic equation in terms of base 2.3. Use the power rule for exponents to simplify.4. Set the exponents equal to each other since the bases are the same.5. Solve for x.6. Verify the solution by plugging it back into the original equation.This systematic approach helps ensure that I don't miss any steps and that my solution is accurate. It's especially useful when dealing with logarithms and exponents, which can sometimes be tricky due to their inverse relationship.I also think about the graphical representation of logarithmic functions. The function log₁₆(x) would pass through (1,0) because 16^0 = 1, and it would increase slowly as x increases. Since 2 is less than 16, the logarithm of 2 should be a positive number less than 1, which aligns with our solution of 1/4.In summary, by expressing both 16 and 2 as powers of 2 and applying the properties of exponents and logarithms, I was able to determine that log₁₆(2) equals 1/4. This problem reinforced my understanding of how logarithms and exponents are inverses and how recognizing common bases can simplify the process.

answer：First, I need to determine how many puppies are left after the pet store sold 7 out of 13 puppies.Subtracting the sold puppies from the total gives me 13 - 7 = 6 puppies remaining.Next, these 6 puppies are distributed equally into 3 cages. To find out how many puppies are in each cage, I divide the total number of remaining puppies by the number of cages.So, 6 puppies ÷ 3 cages = 2 puppies per cage.Therefore, each cage contains 2 puppies.

answer：Alright, so I've got this problem here that says: For any positive real numbers x and y, the expression (2 - x/(e y))(ln x - ln y) - 1/m is less than or equal to zero. I need to find the range of the real number m. The options are A: (0,1], B: (0,e], C: (-∞,0) ∪ [1,∞), and D: (-∞,0) ∪ [e,∞).Hmm, okay. Let me try to unpack this step by step. First, I need to understand what the inequality is saying. It involves logarithms and a fraction with m in the denominator. The variables x and y are positive real numbers, so ln x and ln y are defined.Let me rewrite the inequality to make it a bit clearer:(2 - x/(e y))(ln x - ln y) - 1/m ≤ 0Maybe I can simplify this a bit. The term (ln x - ln y) is the same as ln(x/y), right? So, I can rewrite the inequality as:(2 - x/(e y)) * ln(x/y) - 1/m ≤ 0Hmm, that looks a bit cleaner. Now, let's think about what this expression represents. It seems like a product of two terms: (2 - x/(e y)) and ln(x/y), minus 1/m. The whole thing is supposed to be less than or equal to zero for any positive x and y.I wonder if I can make a substitution to simplify this further. Let me set t = x/y. Since x and y are positive, t will also be positive. So, substituting t in, we get:(2 - t/e) * ln t - 1/m ≤ 0Okay, so now the inequality depends only on t, which is a positive real number. So, for all t > 0, (2 - t/e) * ln t - 1/m ≤ 0.Let me rearrange this inequality to solve for 1/m:(2 - t/e) * ln t ≤ 1/mSo, 1/m is greater than or equal to (2 - t/e) * ln t for all t > 0.Hmm, so 1/m has to be at least as big as the maximum value of (2 - t/e) * ln t over all positive t. That means that 1/m must be greater than or equal to the maximum of the function f(t) = (2 - t/e) * ln t.Therefore, to find the range of m, I need to find the maximum value of f(t) and then set 1/m to be greater than or equal to that maximum.Alright, let's define f(t) = (2 - t/e) * ln t. I need to find its maximum for t > 0.To find the maximum, I can take the derivative of f(t) with respect to t and set it equal to zero.So, f(t) = (2 - t/e) * ln tLet's compute f'(t):First, use the product rule: f'(t) = derivative of the first term times the second term plus the first term times the derivative of the second term.Derivative of (2 - t/e) with respect to t is -1/e.Derivative of ln t with respect to t is 1/t.So, f'(t) = (-1/e) * ln t + (2 - t/e) * (1/t)Simplify this:f'(t) = (-ln t)/e + (2 - t/e)/tLet me write that as:f'(t) = (-ln t)/e + 2/t - 1/eHmm, okay. So, f'(t) = (-ln t)/e + 2/t - 1/eTo find critical points, set f'(t) = 0:(-ln t)/e + 2/t - 1/e = 0Multiply both sides by e to eliminate denominators:-ln t + 2e/t - 1 = 0So, -ln t + 2e/t - 1 = 0Let me rearrange this:2e/t - ln t - 1 = 0Hmm, this seems a bit tricky to solve algebraically. Maybe I can make a substitution or see if there's a value of t that satisfies this equation.Let me try t = e, since e is a common base in logarithms and might simplify things.Plugging t = e:2e/e - ln e - 1 = 2 - 1 - 1 = 0Oh, that works! So, t = e is a critical point.Now, I should check if this is a maximum or a minimum. To do that, I can look at the second derivative or analyze the behavior around t = e.Alternatively, since we're looking for the maximum of f(t), and we have only one critical point, it's likely that this is the maximum.Let me compute f(t) at t = e:f(e) = (2 - e/e) * ln e = (2 - 1) * 1 = 1 * 1 = 1So, f(e) = 1Now, let's check the behavior of f(t) as t approaches 0 and as t approaches infinity.As t approaches 0 from the right:ln t approaches -infinity, but (2 - t/e) approaches 2. So, f(t) approaches 2 * (-infinity) = -infinity.As t approaches infinity:(2 - t/e) approaches -infinity, and ln t approaches infinity. So, f(t) approaches (-infinity) * infinity, which is -infinity.Therefore, the function f(t) tends to -infinity as t approaches both 0 and infinity, and it has a critical point at t = e where f(t) = 1. Since this is the only critical point and the function tends to -infinity on both ends, this must be the global maximum.So, the maximum value of f(t) is 1, achieved at t = e.Therefore, going back to our inequality:(2 - t/e) * ln t ≤ 1/mSince the maximum of the left side is 1, we have:1 ≤ 1/mWhich implies that:1/m ≥ 1So, 1/m is greater than or equal to 1.Now, solving for m:If 1/m ≥ 1, then m ≤ 1.But we also need to consider the sign of m. Since 1/m must be positive (because the left side can be negative or positive, but in our case, we found that the maximum is 1, which is positive, so 1/m must be positive). Therefore, m must be positive.So, m is positive and m ≤ 1.Therefore, the range of m is (0, 1].Looking back at the options, that's option A.Wait, but let me double-check. The original inequality is:(2 - x/(e y))(ln x - ln y) - 1/m ≤ 0We transformed it to:(2 - t/e) * ln t ≤ 1/mWe found that the maximum of the left side is 1, so 1 ≤ 1/m, which gives m ≤ 1.But m must be positive because 1/m is positive (since the left side can be positive or negative, but in this case, the maximum is positive, so 1/m must be positive). So, m > 0.Hence, m ∈ (0, 1].So, the answer should be A: (0,1].But wait, let me think again. The problem says "for any positive real numbers x and y". So, the inequality must hold for all x and y > 0. We found that the maximum of the left side is 1, so 1/m must be at least 1. So, m must be at most 1, but m must also be positive because if m were negative, 1/m would be negative, and the left side can be positive, which would violate the inequality.Therefore, m must be in (0,1].Yes, that seems correct. So, the answer is A.**Final Answer**boxed{A}

answer：First, I need to determine the total number of ways to choose three different numbers from the set ({-3, -2, -1, 1, 2, 3, 4}). Since there are 7 elements in the set, the number of ways to choose 3 different numbers is calculated using the combination formula:[binom{7}{3} = frac{7 times 6 times 5}{3 times 2 times 1} = 35]Next, I need to count the number of favorable outcomes where the product of the three selected numbers is negative. A product is negative if there is an odd number of negative numbers in the selection. In the given set, there are 3 negative numbers ({-3, -2, -1}) and 4 positive numbers ({1, 2, 3, 4}).There are two scenarios to consider for a negative product:1. **Selecting 1 negative and 2 positive numbers:** [ binom{3}{1} times binom{4}{2} = 3 times 6 = 18 text{ ways} ]2. **Selecting 3 negative numbers:** [ binom{3}{3} = 1 text{ way} ]Adding these favorable outcomes together gives:[18 + 1 = 19]Finally, the probability that the product of the three chosen numbers is negative is:[text{Probability} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}} = frac{19}{35}]Therefore, the probability that the product is negative is (boxed{textbf{(A) }dfrac{19}{35}}).

answer：Okay, so I have this problem where I need to find the 2008th term in a sequence of positive integers that are relatively prime to 75. The sequence is given in increasing order, and the first few terms are 1, 2, 4, 7, 8, and so on. I need to figure out what the 2008th term is. Hmm, let me try to break this down step by step.First, I remember that two numbers are relatively prime if their greatest common divisor (GCD) is 1. So, I need to list all positive integers that don't share any common factors with 75 except for 1. To do this, I should probably factorize 75 to understand its prime components. 75 can be factored into prime numbers as 3 multiplied by 5 squared, so 75 = 3 × 5². This means that any number that is not divisible by 3 or 5 will be relatively prime to 75. Therefore, the sequence we're looking at includes all positive integers except those divisible by 3 or 5.Now, to find the 2008th term, I think I need a systematic way to count how many numbers are relatively prime to 75 up to a certain point. Maybe I can use the concept of the totient function, which Euler's totient function φ(n) gives the number of integers less than or equal to n that are relatively prime to n. But in this case, since we're dealing with all positive integers, not just up to 75, I might need a different approach.Wait, actually, since the sequence is periodic in a way, every 75 numbers, the pattern of numbers relatively prime to 75 repeats. So, if I can figure out how many numbers in each block of 75 are relatively prime to 75, I can determine how many complete blocks fit into 2008 terms and then find the position within the next block.Let me calculate how many numbers between 1 and 75 are relatively prime to 75. Using the inclusion-exclusion principle, I can find this. The total numbers from 1 to 75 is 75. Now, subtract the numbers divisible by 3 or 5.Numbers divisible by 3: floor(75/3) = 25Numbers divisible by 5: floor(75/5) = 15But wait, some numbers are divisible by both 3 and 5, which is 15. So, numbers divisible by 15: floor(75/15) = 5Using inclusion-exclusion, the count of numbers divisible by 3 or 5 is 25 + 15 - 5 = 35. Therefore, the numbers relatively prime to 75 are 75 - 35 = 40. So, in each block of 75 numbers, there are 40 numbers that are relatively prime to 75.That means the sequence of numbers relatively prime to 75 repeats every 75 numbers, with 40 terms in each cycle. So, to find the 2008th term, I can figure out how many complete cycles of 40 fit into 2008 and then find the position within the next cycle.Let me calculate how many complete cycles there are in 2008 terms. So, 2008 divided by 40 gives me the number of complete cycles, and the remainder will tell me the position in the next cycle.2008 ÷ 40 = 50.2Wait, 40 × 50 = 2000, so 2008 - 2000 = 8. So, there are 50 complete cycles, and the 2008th term is the 8th term in the next cycle.But each cycle corresponds to 75 numbers, right? So, each cycle is 75 numbers, and within each cycle, there are 40 numbers that are relatively prime to 75. So, the first cycle is numbers 1 to 75, the second cycle is 76 to 150, and so on.Therefore, the 50th cycle ends at 75 × 50 = 3750. The next cycle starts at 3751, and the 2008th term is the 8th term in this 51st cycle.Now, I need to list the numbers from 1 to 75 that are relatively prime to 75 and find the 8th term. Then, add 3750 to it to get the actual value in the 51st cycle.Let me list the numbers from 1 to 75 that are relatively prime to 75. Since 75 is 3 × 5², any number not divisible by 3 or 5 is relatively prime to 75.Starting from 1:1 (relatively prime)2 (relatively prime)3 (divisible by 3)4 (relatively prime)5 (divisible by 5)6 (divisible by 3)7 (relatively prime)8 (relatively prime)9 (divisible by 3)10 (divisible by 5)11 (relatively prime)12 (divisible by 3)13 (relatively prime)14 (relatively prime)15 (divisible by 3 and 5)16 (relatively prime)17 (relatively prime)18 (divisible by 3)19 (relatively prime)20 (divisible by 5)21 (divisible by 3)22 (relatively prime)23 (relatively prime)24 (divisible by 3)25 (divisible by 5)26 (relatively prime)27 (divisible by 3)28 (relatively prime)29 (relatively prime)30 (divisible by 3 and 5)31 (relatively prime)32 (relatively prime)33 (divisible by 3)34 (relatively prime)35 (divisible by 5)36 (divisible by 3)37 (relatively prime)38 (relatively prime)39 (divisible by 3)40 (relatively prime)41 (relatively prime)42 (divisible by 3)43 (relatively prime)44 (relatively prime)45 (divisible by 3 and 5)46 (relatively prime)47 (relatively prime)48 (divisible by 3)49 (relatively prime)50 (divisible by 5)51 (divisible by 3)52 (relatively prime)53 (relatively prime)54 (divisible by 3)55 (divisible by 5)56 (relatively prime)57 (divisible by 3)58 (relatively prime)59 (relatively prime)60 (divisible by 3 and 5)61 (relatively prime)62 (relatively prime)63 (divisible by 3)64 (relatively prime)65 (divisible by 5)66 (divisible by 3)67 (relatively prime)68 (relatively prime)69 (divisible by 3)70 (divisible by 5)71 (relatively prime)72 (divisible by 3)73 (relatively prime)74 (relatively prime)75 (divisible by 3 and 5)Wait, that was a bit tedious, but I think I can list the numbers that are relatively prime to 75 in order:1, 2, 4, 7, 8, 11, 13, 14, 16, 17, 19, 22, 23, 26, 28, 29, 31, 32, 34, 37, 38, 41, 43, 44, 46, 47, 49, 52, 53, 56, 58, 59, 61, 62, 64, 67, 68, 71, 73, 74.Let me count these to make sure there are 40 numbers. Hmm, 1, 2, 4, 7, 8, 11, 13, 14, 16, 17, 19, 22, 23, 26, 28, 29, 31, 32, 34, 37, 38, 41, 43, 44, 46, 47, 49, 52, 53, 56, 58, 59, 61, 62, 64, 67, 68, 71, 73, 74. That's 40 numbers, so that checks out.Now, the 8th term in this list is 14. So, in each cycle of 75 numbers, the 8th number that's relatively prime to 75 is 14. Therefore, in the 51st cycle, which starts at 3751, the 8th term would be 3750 + 14 = 3764.Wait, let me verify that. The first cycle ends at 75, the second at 150, and so on. So, the 50th cycle ends at 75 × 50 = 3750. The next cycle starts at 3751, so the first term in the 51st cycle is 3751, which corresponds to 1 in the original list. Therefore, the 8th term in the 51st cycle would be 3750 + 14 = 3764.But just to make sure I didn't make a mistake in the list, let me recount the 8th term. Starting from 1:1 (1st), 2 (2nd), 4 (3rd), 7 (4th), 8 (5th), 11 (6th), 13 (7th), 14 (8th). Yes, the 8th term is 14. So, adding that to 3750 gives 3764.Therefore, the 2008th term in the sequence is 3764.**Final Answer**The value of (a_{2008}) is boxed{3764}.

answer：First, I need to determine the total number of student-days worked by all three schools. For Allen school: 7 students × 3 days = 21 student-daysFor Balboa school: 4 students × 5 days = 20 student-daysFor Carver school: 5 students × 9 days = 45 student-daysAdding these together gives the total student-days:21 + 20 + 45 = 86 student-daysNext, I'll calculate the daily wage per student by dividing the total amount paid by the total number of student-days:744 ÷ 86 = 8.65116279 per student-dayFinally, to find the total earnings for Balboa school, I'll multiply the number of student-days for Balboa by the daily wage:20 student-days × 8.65116279 ≈ 173.02However, since the answer choices don't include 173.02, I'll consider rounding the daily wage to the nearest dollar:8.65116279 ≈ 9 per student-dayNow, calculating the total earnings for Balboa school with the rounded daily wage:20 student-days × 9 = 180Therefore, the students from Balboa school earned a total of 180.