💡First, determine the total adult population by using the number of employed adults and the percentage of employed adults.Next, calculate the total population of the village based on the proportion of adults in the population.Then, find the new adult population after a 5% growth.Finally, calculate the new total population by maintaining the same proportion of adults and children.

💡Okay, so I have this problem where I'm given that (1 < x < 10), and I need to figure out the relationship between (a), (b), and (c). The definitions are:- (a = log x^{2})- (b = log (log x))- (c = (log x)^{2})And the options are:A: (c > a > b)B: (c > b > a)C: (a > c > b)D: (a > b > c)Alright, let me break this down step by step.First, I need to understand what each of these expressions represents.Starting with (a = log x^{2}). I know that (log x^{2}) can be simplified using logarithm properties. Specifically, (log x^{n} = n log x), so (a = 2 log x). That seems straightforward.Next, (b = log (log x)). Hmm, this is a logarithm of a logarithm, which might be a bit tricky. Since (1 < x < 10), (log x) will be between 0 and 1 because (log 1 = 0) and (log 10 = 1). So, (log x) is between 0 and 1, which means (log (log x)) will be negative because the logarithm of a number between 0 and 1 is negative. So, (b) is negative.Then, (c = (log x)^{2}). This is just the square of (log x). Since (log x) is between 0 and 1, squaring it will make it smaller, but it will still be positive. So, (c) is positive but less than (log x).Now, let's see if I can compare (a), (b), and (c).First, I know (a = 2 log x). Since (log x) is between 0 and 1, (a) will be between 0 and 2. Specifically, if (x) is between 1 and 10, (log x) is between 0 and 1, so (a) is between 0 and 2.Next, (c = (log x)^{2}). As I mentioned, since (log x) is between 0 and 1, squaring it will make it smaller. So, (c) is less than (log x), which is less than 1. Therefore, (c) is between 0 and 1.Now, comparing (a) and (c): (a = 2 log x), and (c = (log x)^{2}). Let's see when (2 log x) is greater than ((log x)^{2}).Let me set (y = log x), so (0 < y < 1). Then, the comparison becomes (2y) vs. (y^{2}).So, (2y > y^{2}) when (2y - y^{2} > 0), which simplifies to (y(2 - y) > 0). Since (y > 0) and (2 - y > 0) (because (y < 2), which is certainly true since (y < 1)), the product is positive. Therefore, (2y > y^{2}), which means (a > c).So, (a > c).Now, what about (b)? As I noted earlier, (b = log (log x)), and since (log x) is between 0 and 1, (log (log x)) is negative. So, (b) is negative, while both (a) and (c) are positive. Therefore, (b) is the smallest.So, putting it all together: (a > c > b).Let me double-check to make sure I didn't make any mistakes.1. (a = 2 log x), which is positive and between 0 and 2.2. (c = (log x)^{2}), which is positive and less than (log x), so between 0 and 1.3. Comparing (a) and (c), since (2 log x > (log x)^{2}) for (0 < log x < 1), (a > c).4. (b = log (log x)) is negative because (log x) is between 0 and 1, so (b) is less than both (a) and (c).Therefore, the order is (a > c > b), which corresponds to option C.I think that makes sense. I don't see any errors in my reasoning.**Final Answer**boxed{C}

💡First, I need to determine how much Alicia earns per hour in cents. Since she earns 25 per hour and there are 100 cents in a dollar, her hourly wage in cents is:25 dollars/hour × 100 cents/dollar = 2500 cents/hourNext, I need to calculate the local tax deduction, which is 2% of her hourly wage. To find 2% of 2500 cents:2% of 2500 cents = 0.02 × 2500 = 50 centsTherefore, 50 cents per hour of Alicia's wages are used to pay local taxes.

💡Okay, so I have this problem about a sequence ( {a_n} ) where the sum of the first ( n ) terms is given by ( S_n = 2a_n - 3n ) for all positive integers ( n ). The problem has two parts: first, I need to find the values of ( a_1 ), ( a_2 ), and ( a_3 ). Second, I have to determine if there's a constant ( lambda ) such that ( {a_n + lambda} ) forms a geometric sequence. If it does, I need to find ( lambda ) and the general term ( a_n ). If not, I have to explain why.Starting with part (1), finding ( a_1 ), ( a_2 ), and ( a_3 ). I know that ( S_n ) is the sum of the first ( n ) terms, so ( S_n = a_1 + a_2 + dots + a_n ). Given that ( S_n = 2a_n - 3n ), I can use this formula for different values of ( n ) to find the required terms.For ( n = 1 ):( S_1 = a_1 ). According to the given formula, ( S_1 = 2a_1 - 3(1) ). So, substituting ( S_1 ) with ( a_1 ), we have:( a_1 = 2a_1 - 3 ).Let me solve this equation for ( a_1 ):Subtract ( 2a_1 ) from both sides:( a_1 - 2a_1 = -3 )( -a_1 = -3 )Multiply both sides by -1:( a_1 = 3 )Okay, so ( a_1 = 3 ).Moving on to ( n = 2 ):( S_2 = a_1 + a_2 ). According to the formula, ( S_2 = 2a_2 - 3(2) = 2a_2 - 6 ).We already know ( a_1 = 3 ), so substituting into ( S_2 ):( 3 + a_2 = 2a_2 - 6 )Let me solve for ( a_2 ):Subtract ( a_2 ) from both sides:( 3 = a_2 - 6 )Add 6 to both sides:( 9 = a_2 )So, ( a_2 = 9 ).Now, for ( n = 3 ):( S_3 = a_1 + a_2 + a_3 ). Using the formula, ( S_3 = 2a_3 - 3(3) = 2a_3 - 9 ).We know ( a_1 = 3 ) and ( a_2 = 9 ), so substituting into ( S_3 ):( 3 + 9 + a_3 = 2a_3 - 9 )Simplify the left side:( 12 + a_3 = 2a_3 - 9 )Subtract ( a_3 ) from both sides:( 12 = a_3 - 9 )Add 9 to both sides:( 21 = a_3 )So, ( a_3 = 21 ).Alright, part (1) seems done. I have ( a_1 = 3 ), ( a_2 = 9 ), and ( a_3 = 21 ).Moving on to part (2). I need to check if there's a constant ( lambda ) such that ( {a_n + lambda} ) is a geometric sequence. A geometric sequence has the property that the ratio between consecutive terms is constant. So, if ( {a_n + lambda} ) is geometric, then ( frac{a_{n+1} + lambda}{a_n + lambda} ) should be the same for all ( n ).Alternatively, another approach is to use the property that in a geometric sequence, the square of the middle term is equal to the product of the adjacent terms. So, for ( n = 1, 2, 3 ), we can set up an equation using ( a_1 + lambda ), ( a_2 + lambda ), and ( a_3 + lambda ).Let me denote ( b_n = a_n + lambda ). If ( {b_n} ) is geometric, then ( b_2^2 = b_1 b_3 ).Substituting the known values:( b_1 = a_1 + lambda = 3 + lambda )( b_2 = a_2 + lambda = 9 + lambda )( b_3 = a_3 + lambda = 21 + lambda )So, the equation becomes:( (9 + lambda)^2 = (3 + lambda)(21 + lambda) )Let me expand both sides:Left side: ( (9 + lambda)^2 = 81 + 18lambda + lambda^2 )Right side: ( (3 + lambda)(21 + lambda) = 3*21 + 3lambda + 21lambda + lambda^2 = 63 + 24lambda + lambda^2 )Set them equal:( 81 + 18lambda + lambda^2 = 63 + 24lambda + lambda^2 )Subtract ( lambda^2 ) from both sides:( 81 + 18lambda = 63 + 24lambda )Subtract ( 63 ) and ( 18lambda ) from both sides:( 81 - 63 = 24lambda - 18lambda )( 18 = 6lambda )Divide both sides by 6:( lambda = 3 )So, ( lambda = 3 ). Therefore, ( b_n = a_n + 3 ) should be a geometric sequence.Let me verify this. Compute ( b_1 = 3 + 3 = 6 ), ( b_2 = 9 + 3 = 12 ), ( b_3 = 21 + 3 = 24 ).Check if this is a geometric sequence:The ratio between ( b_2 ) and ( b_1 ) is ( 12 / 6 = 2 ).The ratio between ( b_3 ) and ( b_2 ) is ( 24 / 12 = 2 ).So, yes, the common ratio is 2. Therefore, ( {b_n} ) is a geometric sequence with first term 6 and common ratio 2.Now, to find the general term ( a_n ), since ( b_n = a_n + 3 ) is geometric, we can write:( b_n = b_1 times r^{n-1} )Where ( b_1 = 6 ) and ( r = 2 ). So,( b_n = 6 times 2^{n-1} )Therefore,( a_n = b_n - 3 = 6 times 2^{n-1} - 3 )Let me check if this formula works for the first few terms:For ( n = 1 ):( a_1 = 6 times 2^{0} - 3 = 6 - 3 = 3 ) ✔️For ( n = 2 ):( a_2 = 6 times 2^{1} - 3 = 12 - 3 = 9 ) ✔️For ( n = 3 ):( a_3 = 6 times 2^{2} - 3 = 24 - 3 = 21 ) ✔️Looks good. So, the general term is ( a_n = 6 times 2^{n-1} - 3 ).Wait, just to be thorough, let me check if this formula satisfies the original condition ( S_n = 2a_n - 3n ).Compute ( S_n ) using the formula for ( a_n ):( S_n = a_1 + a_2 + dots + a_n )Since ( a_n = 6 times 2^{n-1} - 3 ), let's write ( S_n ) as:( S_n = sum_{k=1}^{n} (6 times 2^{k-1} - 3) )This can be split into two sums:( S_n = 6 sum_{k=1}^{n} 2^{k-1} - 3 sum_{k=1}^{n} 1 )The first sum is a geometric series with first term 1 and ratio 2:( sum_{k=1}^{n} 2^{k-1} = 2^{n} - 1 )The second sum is just ( 3n ).So, substituting back:( S_n = 6(2^{n} - 1) - 3n = 6 times 2^{n} - 6 - 3n )Now, compute ( 2a_n - 3n ):( 2a_n - 3n = 2(6 times 2^{n-1} - 3) - 3n = 12 times 2^{n-1} - 6 - 3n = 6 times 2^{n} - 6 - 3n )Which matches ( S_n ). So, the formula is consistent with the given condition.Therefore, everything checks out. There exists a constant ( lambda = 3 ), and the general term is ( a_n = 6 times 2^{n-1} - 3 ).**Final Answer**(1) The values are ( a_1 = boxed{3} ), ( a_2 = boxed{9} ), and ( a_3 = boxed{21} ).(2) Yes, such a constant ( lambda ) exists. The value is ( lambda = boxed{3} ) and the general term is ( a_n = boxed{6 times 2^{n-1} - 3} ).

💡Alright, so I've got this problem here that I need to prove. It says that for any numbers ( a_1, a_2, ldots, a_n ) in the interval [0, 2], where ( n ) is at least 2, the sum of the absolute differences between every pair of these numbers is less than or equal to ( n^2 ). And not only that, but I also need to figure out for which specific values of ( a_1, a_2, ldots, a_n ) this sum actually equals ( n^2 ).Hmm, okay. Let me try to unpack this step by step. First, the problem is about absolute differences between pairs of numbers. So, for each pair ( (i, j) ), we calculate ( |a_i - a_j| ) and then sum all of these up. The claim is that this total sum can't exceed ( n^2 ), and we need to find when it does reach that maximum.I think it might help to start by considering a small value of ( n ) to get an intuition. Let's take ( n = 2 ). Then, we have two numbers, say ( a_1 ) and ( a_2 ), both in [0, 2]. The sum of absolute differences would be ( |a_1 - a_2| + |a_2 - a_1| ), which simplifies to ( 2|a_1 - a_2| ). The maximum value of ( |a_1 - a_2| ) occurs when one is 0 and the other is 2, giving ( 2 times 2 = 4 ). Since ( n^2 = 4 ) when ( n = 2 ), this checks out. So, for ( n = 2 ), the maximum is achieved when one number is 0 and the other is 2.Let me try ( n = 3 ). Now, we have three numbers ( a_1, a_2, a_3 ) in [0, 2]. The sum of absolute differences would be:( |a_1 - a_2| + |a_1 - a_3| + |a_2 - a_1| + |a_2 - a_3| + |a_3 - a_1| + |a_3 - a_2| )Which simplifies to ( 2(|a_1 - a_2| + |a_1 - a_3| + |a_2 - a_3|) ). To maximize this, we want to maximize each of these differences. The maximum difference between any two numbers is still 2, achieved when one is 0 and the other is 2. But with three numbers, how do we arrange them to maximize the sum?If we set one number to 0, one to 2, and the third somewhere in between, say 0 or 2. Wait, actually, if we set two numbers to 0 and one to 2, then the differences would be:Between 0 and 0: 0Between 0 and 2: 2Between 0 and 2: 2So, the total sum would be ( 0 + 2 + 2 + 2 + 0 + 0 = 6 ). But ( n^2 = 9 ) here, so 6 is less than 9. Hmm, that's not achieving the maximum.Wait, maybe if we spread the numbers out more? Let's set one number to 0, one to 2, and the third somewhere else. Let's say 0, 1, and 2. Then the differences would be:|0 - 1| = 1|0 - 2| = 2|1 - 0| = 1|1 - 2| = 1|2 - 0| = 2|2 - 1| = 1Adding these up: 1 + 2 + 1 + 1 + 2 + 1 = 8. That's closer to 9, but still not quite there.Wait, maybe if all three numbers are either 0 or 2? Let's try two 0s and one 2. Then, as before, the sum is 6. Alternatively, two 2s and one 0: same thing, sum is 6. So, that's not enough.Alternatively, maybe setting all three numbers to 0 or 2 in some other way? But with three numbers, you can't have more than two distinct values if you're only using 0 and 2. Hmm.Wait, maybe the maximum occurs when two numbers are at 0 and one is at 2, but that only gives a sum of 6, which is less than 9. So, perhaps for ( n = 3 ), the maximum isn't achieved? Or maybe I'm missing something.Wait, maybe I need to consider more numbers. Let's try ( n = 4 ). If I set two numbers to 0 and two numbers to 2, then the sum of absolute differences would be:Each 0 will have differences with the other 0 (0), and with the two 2s (2 each). Similarly, each 2 will have differences with the other 2 (0), and with the two 0s (2 each).So, for each 0: 0 + 2 + 2 = 4For each 2: 2 + 2 + 0 = 4But since there are two 0s and two 2s, the total sum would be 4 + 4 + 4 + 4 = 16. But ( n^2 = 16 ) here, so that works. So, for ( n = 4 ), setting half the numbers to 0 and half to 2 achieves the maximum.Wait, so for even ( n ), setting half to 0 and half to 2 gives the maximum sum of ( n^2 ). But for odd ( n ), like ( n = 3 ), we can't split them evenly, so maybe the maximum is less than ( n^2 )?Let me check ( n = 3 ) again. If I set one number to 0, one to 2, and the third to something else, say 1, the sum was 8. If I set all three numbers to 0 and 2, but since we can't split them evenly, maybe the maximum is 8, which is less than 9. So, perhaps for odd ( n ), the maximum is less than ( n^2 ), and for even ( n ), it's exactly ( n^2 ).Okay, that seems to make sense. So, the strategy is to set as many numbers as possible to 0 and 2, with half in each group if ( n ) is even, and as close as possible if ( n ) is odd.But let's try to generalize this. Suppose we have ( k ) numbers set to 2 and ( n - k ) numbers set to 0. Then, the sum of absolute differences would be:For each pair where one is 2 and the other is 0, the difference is 2. There are ( k(n - k) ) such pairs. Since each pair is counted twice in the sum (once as ( |a_i - a_j| ) and once as ( |a_j - a_i| )), the total contribution is ( 2 times 2k(n - k) ).Wait, no. Actually, in the sum ( sum_{i,j=1}^{n} |a_i - a_j| ), each pair ( (i, j) ) is counted once, regardless of order. So, if ( a_i = 2 ) and ( a_j = 0 ), the difference is 2, and this occurs ( k(n - k) ) times. Similarly, if ( a_i = 0 ) and ( a_j = 2 ), it's also 2, but since we're summing over all ( i, j ), each unordered pair is counted twice. Wait, no, in the sum, ( i ) and ( j ) are both from 1 to ( n ), so for each unordered pair ( {i, j} ), we have both ( |a_i - a_j| ) and ( |a_j - a_i| ), which are the same. So, the total number of terms where ( a_i = 2 ) and ( a_j = 0 ) is ( 2k(n - k) ), each contributing 2. So, the total sum would be ( 2 times 2k(n - k) = 4k(n - k) ).Wait, but in the case of ( n = 4 ), ( k = 2 ), so ( 4 times 2 times 2 = 16 ), which matches ( n^2 = 16 ). For ( n = 2 ), ( k = 1 ), so ( 4 times 1 times 1 = 4 ), which matches ( n^2 = 4 ). For ( n = 3 ), if we set ( k = 1 ), then ( 4 times 1 times 2 = 8 ), which is less than ( 9 ). If we set ( k = 2 ), then ( 4 times 2 times 1 = 8 ), same result. So, indeed, for odd ( n ), the maximum is ( 4k(n - k) ), which is less than ( n^2 ).So, to maximize the sum, we need to maximize ( 4k(n - k) ). This is a quadratic in ( k ), which is maximized when ( k = n/2 ). But since ( k ) must be an integer, for even ( n ), ( k = n/2 ) gives the maximum ( n^2 ). For odd ( n ), the maximum is ( 4 times lfloor n/2 rfloor times lceil n/2 rceil ), which is less than ( n^2 ).Therefore, the maximum sum ( sum_{i,j=1}^{n} |a_i - a_j| ) is ( n^2 ) when ( n ) is even, achieved by setting exactly half of the ( a_i )s to 0 and the other half to 2. For odd ( n ), the maximum is less than ( n^2 ), specifically ( 4 times lfloor n/2 rfloor times lceil n/2 rceil ).Wait, but the problem statement says "for any ( a_1, a_2, ldots, a_n in [0,2] )", so it's not restricted to just 0 and 2. Maybe there's a way to get a higher sum by choosing some numbers in between 0 and 2? Let me think.Suppose I have three numbers: 0, 1, and 2. The sum of absolute differences is 8, as calculated earlier. If I instead set all three numbers to 0 and 2, but since I can't split them evenly, the maximum is 8, which is less than 9. So, in this case, setting some numbers to intermediate values doesn't help in increasing the sum beyond what's possible with just 0 and 2.Wait, actually, maybe it does? Let me try with ( n = 3 ). Suppose I set one number to 0, one to 2, and the third to some value ( x ) between 0 and 2. Then, the sum of absolute differences would be:|0 - 2| + |0 - x| + |2 - 0| + |2 - x| + |x - 0| + |x - 2|Which simplifies to:2 + x + 2 + (2 - x) + x + (2 - x) = 2 + x + 2 + 2 - x + x + 2 - xSimplifying:2 + x + 2 + 2 - x + x + 2 - x = 2 + 2 + 2 + 2 + (x - x + x - x) = 8So, regardless of the value of ( x ), the sum remains 8. Interesting. So, for ( n = 3 ), even if we set one number to an intermediate value, the sum doesn't increase beyond 8. So, setting numbers to intermediate values doesn't help in increasing the sum beyond what's possible with just 0 and 2.Therefore, it seems that the maximum sum is achieved when as many numbers as possible are set to 0 and 2, with the rest set to either 0 or 2 as well, but for odd ( n ), we can't split them evenly, so the maximum is slightly less.Wait, but in the case of ( n = 3 ), setting one number to 0, one to 2, and the third to 0 or 2 doesn't increase the sum beyond 8. So, it seems that the maximum is indeed achieved when we have as many numbers as possible set to 0 and 2, with the rest set to either 0 or 2.So, generalizing this, for any ( n ), the maximum sum ( sum_{i,j=1}^{n} |a_i - a_j| ) is achieved when we set as many numbers as possible to 0 and 2, with the rest set to either 0 or 2. For even ( n ), this means exactly half are 0 and half are 2, giving the sum ( n^2 ). For odd ( n ), we can't split them evenly, so the maximum sum is ( 4 times lfloor n/2 rfloor times lceil n/2 rceil ), which is less than ( n^2 ).Therefore, the inequality ( sum_{i,j=1}^{n} |a_i - a_j| leq n^2 ) holds for any ( a_1, a_2, ldots, a_n in [0,2] ), and equality holds if and only if ( n ) is even and exactly half of the ( a_i )s are 0 and the other half are 2.Wait, but let me double-check this with another example. Let's take ( n = 4 ). If I set two numbers to 0 and two to 2, the sum is 16, which is ( 4^2 ). If I set three numbers to 0 and one to 2, the sum would be:Each 0 has differences with the other 0s (0) and with the 2 (2). There are three 0s, so each contributes 2 once, totaling 6. The 2 has differences with the three 0s, each contributing 2, totaling 6. So, the total sum is 6 + 6 = 12, which is less than 16. So, indeed, setting half to 0 and half to 2 gives the maximum.Similarly, for ( n = 5 ), setting two to 0 and three to 2, the sum would be:Each 0 has differences with the three 2s, contributing 2 each, so 3 * 2 = 6 per 0, but there are two 0s, so 12. Each 2 has differences with the two 0s, contributing 2 each, so 2 * 2 = 4 per 2, and there are three 2s, so 12. So, total sum is 12 + 12 = 24. But ( n^2 = 25 ), so it's still less.Wait, but 24 is less than 25, so even for ( n = 5 ), the maximum is 24, which is less than ( n^2 ). So, the pattern holds.Therefore, I think I've got it. The maximum sum is ( n^2 ) when ( n ) is even, achieved by setting exactly half of the numbers to 0 and the other half to 2. For odd ( n ), the maximum is ( 4 times lfloor n/2 rfloor times lceil n/2 rceil ), which is less than ( n^2 ).So, to summarize, the inequality holds for any ( a_1, a_2, ldots, a_n in [0,2] ), and equality occurs if and only if ( n ) is even and exactly half of the numbers are 0 and the other half are 2.

💡First, I recognize that ( a ) is a solution to the quadratic equation ( x^2 - 2x - 1 = 0 ). This means that when ( x = a ), the equation holds true.I substitute ( x = a ) into the equation:[a^2 - 2a - 1 = 0]Next, I rearrange this equation to isolate the quadratic and linear terms:[a^2 - 2a = 1]Now, I need to evaluate the expression ( a^2 - 2a + 2022 ). Using the equation I derived, I substitute ( a^2 - 2a ) with 1:[a^2 - 2a + 2022 = 1 + 2022]Finally, I perform the addition:[1 + 2022 = 2023]Therefore, the value of the algebraic expression ( a^2 - 2a + 2022 ) is 2023.