💡To solve the first part, I need to determine the value of (log_{3}{9}). Since (9) is (3) squared, (log_{3}{9}) equals (2).For the second part, I am given that (a = log_{4}{3}). I need to find (2^{a}). Using the definition of logarithms, (4^{a} = 3). Since (4) is (2) squared, I can rewrite this as ((2^{2})^{a} = 3), which simplifies to (2^{2a} = 3). Taking the square root of both sides gives (2^{a} = sqrt{3}).Thus, the solutions are (log_{3}{9} = 2) and (2^{a} = sqrt{3}).

💡Alright, so I've got this problem here about an odd function ( y = f(x) ) that's monotonically increasing on the interval ( (0, +infty) ). They also tell me that ( f(-3) = 0 ), and I need to find the solution set for the inequality ( x f(x) > 0 ). The options are given as A, B, C, D, each describing different intervals for ( x ).First, I need to recall what it means for a function to be odd. An odd function satisfies the condition ( f(-x) = -f(x) ) for all ( x ) in its domain. This means that the graph of the function is symmetric about the origin. So, if I know the behavior of the function for positive ( x ), I can determine it for negative ( x ) as well.Next, the function is monotonically increasing on ( (0, +infty) ). Monotonically increasing means that as ( x ) increases, ( f(x) ) also increases. So, for ( x_1 < x_2 ) in ( (0, +infty) ), we have ( f(x_1) leq f(x_2) ). Since it's strictly increasing, the inequality should be strict: ( f(x_1) < f(x_2) ).Given that ( f(-3) = 0 ), and since ( f ) is odd, I can find ( f(3) ). Using the property of odd functions:[f(-3) = -f(3) implies 0 = -f(3) implies f(3) = 0]So, ( f(3) = 0 ) as well.Now, I need to analyze the inequality ( x f(x) > 0 ). This inequality will hold true when both ( x ) and ( f(x) ) are positive or when both are negative. So, I need to find the intervals where ( x ) and ( f(x) ) have the same sign.Let's break this down into cases based on the value of ( x ):**Case 1: ( x > 0 )**In this case, ( x ) is positive. Since ( f ) is monotonically increasing on ( (0, +infty) ) and ( f(3) = 0 ), the behavior of ( f(x) ) around ( x = 3 ) is crucial.- For ( 0 < x < 3 ): Since ( f ) is increasing and ( f(3) = 0 ), ( f(x) ) must be less than 0 in this interval. Therefore, ( f(x) < 0 ) when ( 0 < x < 3 ). - For ( x > 3 ): Similarly, since ( f ) is increasing, ( f(x) ) must be greater than 0 when ( x > 3 ).So, in the interval ( (0, 3) ), ( x ) is positive, but ( f(x) ) is negative, making ( x f(x) < 0 ). In the interval ( (3, +infty) ), both ( x ) and ( f(x) ) are positive, so ( x f(x) > 0 ).**Case 2: ( x < 0 )**Here, ( x ) is negative. Since ( f ) is an odd function, ( f(x) = -f(-x) ). Let's analyze the behavior of ( f(x) ) for negative ( x ):- For ( x < -3 ): Let's consider ( x = -a ) where ( a > 3 ). Then, ( f(x) = f(-a) = -f(a) ). Since ( a > 3 ), and ( f ) is increasing on ( (0, +infty) ), ( f(a) > f(3) = 0 ). Therefore, ( f(x) = -f(a) < 0 ). So, ( f(x) < 0 ) when ( x < -3 ).- For ( -3 < x < 0 ): Let ( x = -a ) where ( 0 < a < 3 ). Then, ( f(x) = f(-a) = -f(a) ). Since ( 0 < a < 3 ), and ( f ) is increasing on ( (0, +infty) ), ( f(a) < f(3) = 0 ). Therefore, ( f(x) = -f(a) > 0 ). So, ( f(x) > 0 ) when ( -3 < x < 0 ).Now, let's see the product ( x f(x) ) in these intervals:- For ( x < -3 ): ( x ) is negative, and ( f(x) ) is negative. The product of two negatives is positive, so ( x f(x) > 0 ).- For ( -3 < x < 0 ): ( x ) is negative, and ( f(x) ) is positive. The product of a negative and a positive is negative, so ( x f(x) < 0 ).**Case 3: ( x = 0 )**At ( x = 0 ), ( f(0) ) must be 0 because ( f ) is odd: ( f(-0) = -f(0) implies f(0) = -f(0) implies f(0) = 0 ). So, ( x f(x) = 0 ), which doesn't satisfy the inequality ( x f(x) > 0 ).**Case 4: ( x = 3 ) and ( x = -3 )**At ( x = 3 ) and ( x = -3 ), ( f(x) = 0 ). Therefore, ( x f(x) = 0 ) at these points, which again doesn't satisfy the inequality.Putting it all together:- ( x f(x) > 0 ) when ( x > 3 ) (both positive) and when ( x < -3 ) (both negative).So, the solution set is ( x < -3 ) or ( x > 3 ).Looking back at the options:A: ( {x | x < -3 text{ or } 0 < x < 3} ) B: ( {x | -3 < x < 0 text{ or } x > 3} ) C: ( {x | x < -3 text{ or } x > 3} ) D: ( {x | -3 < x < 0 text{ or } 0 < x < 3} )Option C matches our solution set.**Final Answer**boxed{C}

💡First, I need to determine how many days Raja takes to complete the work alone. Let's denote the number of days Raja takes as ( x ).Raja's work rate is ( frac{1}{x} ) of the work per day, and Ram's work rate is ( frac{1}{6} ) of the work per day since Ram can complete the work in 6 days.When working together, their combined work rate is the sum of their individual work rates:[frac{1}{x} + frac{1}{6}]They can complete the work together in 4 days, so their combined work rate is also:[frac{1}{4}]Setting the combined work rate equal to the sum of their individual work rates gives:[frac{1}{x} + frac{1}{6} = frac{1}{4}]To solve for ( x ), I'll find a common denominator and combine the fractions:[frac{6}{6x} + frac{x}{6x} = frac{3}{6x}][frac{6 + x}{6x} = frac{3}{6x}]Now, I'll cross-multiply to solve for ( x ):[6 + x = 3x][6 = 2x][x = frac{6}{2} = 3]Therefore, Raja can complete the work alone in 3 days.

💡Okay, so I have this problem where I need to find the value of α³ + 14β + 5, given that α and β are the real roots of the quadratic equation x² + 4x + 2 = 0. Hmm, let me think about how to approach this step by step.First, I remember that for any quadratic equation of the form x² + bx + c = 0, the sum of the roots is -b, and the product of the roots is c. So in this case, the equation is x² + 4x + 2 = 0, which means the sum of the roots α + β should be -4, and the product αβ should be 2. Let me write that down:α + β = -4 αβ = 2Okay, that's straightforward. Now, the expression I need to evaluate is α³ + 14β + 5. Hmm, α³ is a bit tricky because it's a cubic term. I don't have a direct formula for that, but maybe I can express α³ in terms of α and constants using the original equation.Since α is a root of the equation x² + 4x + 2 = 0, it must satisfy the equation. That means:α² + 4α + 2 = 0 => α² = -4α - 2Alright, so I can use this to express higher powers of α in terms of lower powers. Let's try to find α³. To get α³, I can multiply both sides of the equation α² = -4α - 2 by α:α³ = α * α² = α * (-4α - 2) = -4α² - 2αBut wait, I already have an expression for α², which is -4α - 2. Let me substitute that into the equation:α³ = -4*(-4α - 2) - 2α = 16α + 8 - 2α = (16α - 2α) + 8 = 14α + 8Okay, so α³ simplifies to 14α + 8. That's helpful. Now, let's substitute this back into the original expression we need to evaluate:α³ + 14β + 5 = (14α + 8) + 14β + 5 = 14α + 14β + 8 + 5 = 14(α + β) + 13Wait a minute, I remember from earlier that α + β = -4. So, substituting that in:= 14*(-4) + 13 = -56 + 13 = -43So, the value of α³ + 14β + 5 is -43.But let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning:1. Identified that α and β are roots, so sum is -4 and product is 2. 2. Expressed α² in terms of α: α² = -4α - 2. 3. Calculated α³ by multiplying α² by α: α³ = -4α² - 2α. 4. Substituted α² again: α³ = -4*(-4α - 2) - 2α = 16α + 8 - 2α = 14α + 8. 5. Plugged α³ back into the expression: 14α + 8 + 14β + 5 = 14(α + β) + 13. 6. Substituted α + β = -4: 14*(-4) + 13 = -56 + 13 = -43.Everything seems to check out. I don't see any errors in the calculations. So, I'm confident that the answer is -43.**Final Answer**The value of α³ + 14β + 5 is boxed{-43}.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. Part (Ⅰ) says that {a_n} is a geometric sequence, and the sum of the first n terms is S_n, which is given by S_n = 2^n + a, where n is a positive integer. I need to find the value of a and the general formula for the sequence {a_n}.Alright, let's start by recalling what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as q. So, if the first term is a_1, then the sequence goes a_1, a_1*q, a_1*q^2, and so on.The sum of the first n terms of a geometric sequence can be calculated using the formula S_n = a_1*(1 - q^n)/(1 - q) if q ≠ 1. But in this problem, the sum is given as S_n = 2^n + a. So, I need to reconcile these two expressions.Let me write down what I know:1. S_1 = a_1 = 2^1 + a = 2 + a2. S_2 = a_1 + a_2 = 2^2 + a = 4 + a3. S_3 = a_1 + a_2 + a_3 = 2^3 + a = 8 + aSince {a_n} is a geometric sequence, the ratio between consecutive terms should be constant. So, a_2 / a_1 = a_3 / a_2 = q.Let me compute a_2 and a_3 using the sums:a_2 = S_2 - S_1 = (4 + a) - (2 + a) = 2a_3 = S_3 - S_2 = (8 + a) - (4 + a) = 4So, a_2 = 2 and a_3 = 4. Therefore, the common ratio q = a_3 / a_2 = 4 / 2 = 2.Now, since it's a geometric sequence with common ratio 2, the general term should be a_n = a_1 * 2^{n-1}.But I also know that a_1 = 2 + a. So, let's use the fact that a_2 = 2 to find a_1.From the geometric sequence, a_2 = a_1 * q = (2 + a) * 2 = 2*(2 + a). But we already found a_2 = 2, so:2*(2 + a) = 2Divide both sides by 2:2 + a = 1Subtract 2:a = -1So, a = -1. Therefore, a_1 = 2 + (-1) = 1.Thus, the general term is a_n = 1 * 2^{n-1} = 2^{n-1}.Let me double-check this. If a_n = 2^{n-1}, then:a_1 = 1, a_2 = 2, a_3 = 4, which matches what we found earlier.And the sum S_n should be 2^n + a. Let's verify for n=1,2,3.For n=1: S_1 = 1 = 2^1 + (-1) = 2 - 1 = 1. Correct.For n=2: S_2 = 1 + 2 = 3. 2^2 + (-1) = 4 - 1 = 3. Correct.For n=3: S_3 = 1 + 2 + 4 = 7. 2^3 + (-1) = 8 - 1 = 7. Correct.Good, so part (Ⅰ) seems solved: a = -1 and a_n = 2^{n-1}.Moving on to part (Ⅱ): If b_n = (2n - 1)a_n, find the sum of the first n terms of the sequence {b_n}, denoted as T_n.So, b_n is given by (2n - 1) * a_n. Since we already found a_n = 2^{n-1}, then b_n = (2n - 1)*2^{n-1}.We need to find T_n = sum_{k=1}^n b_k = sum_{k=1}^n (2k - 1)*2^{k-1}.Hmm, this looks like a sum that involves both a linear term (2k - 1) and an exponential term (2^{k-1}). I remember that when dealing with sums of the form sum_{k=1}^n k*r^{k}, we can use differentiation of generating functions or some standard summation techniques.Let me write out the sum:T_n = sum_{k=1}^n (2k - 1)*2^{k-1}I can split this into two separate sums:T_n = 2*sum_{k=1}^n k*2^{k-1} - sum_{k=1}^n 2^{k-1}Let me compute each sum separately.First, compute sum_{k=1}^n 2^{k-1}. That's a geometric series with first term 1 and ratio 2, so the sum is (2^n - 1)/(2 - 1) = 2^n - 1.Second, compute sum_{k=1}^n k*2^{k-1}. Let's denote this sum as S.I recall that sum_{k=1}^n k*r^{k} can be found using the formula:sum_{k=1}^n k*r^{k} = r*(1 - (n+1)*r^n + n*r^{n+1}) / (1 - r)^2But in our case, the exponent is k-1, so let me adjust accordingly.Let me write S = sum_{k=1}^n k*2^{k-1}.Let me make a substitution: let m = k - 1, then when k=1, m=0, and when k=n, m = n-1.So, S = sum_{m=0}^{n-1} (m + 1)*2^{m} = sum_{m=0}^{n-1} m*2^{m} + sum_{m=0}^{n-1} 2^{m}We know that sum_{m=0}^{n-1} 2^{m} = 2^{n} - 1.Now, for sum_{m=0}^{n-1} m*2^{m}, I can use the formula for sum_{m=0}^{n-1} m*r^{m}.The formula is:sum_{m=0}^{n-1} m*r^{m} = r*(1 - (n)*r^{n-1} + (n - 1)*r^{n}) / (1 - r)^2But let me verify this.Alternatively, I can derive it.Let me denote S = sum_{m=0}^{n-1} m*2^{m}Multiply both sides by 2:2S = sum_{m=0}^{n-1} m*2^{m+1} = sum_{m=1}^{n} (m - 1)*2^{m}Subtract the original S:2S - S = S = sum_{m=1}^{n} (m - 1)*2^{m} - sum_{m=0}^{n-1} m*2^{m}Let's write out the terms:sum_{m=1}^{n} (m - 1)*2^{m} = sum_{m=1}^{n} m*2^{m} - sum_{m=1}^{n} 2^{m}Similarly, sum_{m=0}^{n-1} m*2^{m} = sum_{m=1}^{n-1} m*2^{m}So, subtracting:S = [sum_{m=1}^{n} m*2^{m} - sum_{m=1}^{n} 2^{m}] - [sum_{m=1}^{n-1} m*2^{m}]Simplify:S = [sum_{m=1}^{n} m*2^{m} - sum_{m=1}^{n} 2^{m} - sum_{m=1}^{n-1} m*2^{m}]Notice that sum_{m=1}^{n} m*2^{m} - sum_{m=1}^{n-1} m*2^{m} = n*2^{n}And sum_{m=1}^{n} 2^{m} = 2^{n+1} - 2Therefore:S = n*2^{n} - (2^{n+1} - 2)Simplify:S = n*2^{n} - 2^{n+1} + 2 = n*2^{n} - 2*2^{n} + 2 = (n - 2)*2^{n} + 2So, sum_{m=0}^{n-1} m*2^{m} = (n - 2)*2^{n} + 2Therefore, going back to S:S = sum_{m=0}^{n-1} m*2^{m} + sum_{m=0}^{n-1} 2^{m} = [(n - 2)*2^{n} + 2] + [2^{n} - 1] = (n - 2)*2^{n} + 2 + 2^{n} - 1 = (n - 2 + 1)*2^{n} + (2 - 1) = (n - 1)*2^{n} + 1Wait, let me check the calculation step by step.We had:sum_{m=0}^{n-1} m*2^{m} = (n - 2)*2^{n} + 2sum_{m=0}^{n-1} 2^{m} = 2^{n} - 1So, S = (n - 2)*2^{n} + 2 + 2^{n} - 1 = (n - 2 + 1)*2^{n} + (2 - 1) = (n - 1)*2^{n} + 1Yes, that's correct.So, sum_{k=1}^n k*2^{k-1} = S = (n - 1)*2^{n} + 1Therefore, going back to T_n:T_n = 2*S - sum_{k=1}^n 2^{k-1} = 2*[(n - 1)*2^{n} + 1] - (2^{n} - 1)Compute this:First, expand 2*S:2*(n - 1)*2^{n} + 2*1 = (n - 1)*2^{n+1} + 2Then subtract (2^{n} - 1):T_n = (n - 1)*2^{n+1} + 2 - 2^{n} + 1 = (n - 1)*2^{n+1} - 2^{n} + 3Factor out 2^{n}:= 2^{n}*(2*(n - 1) - 1) + 3 = 2^{n}*(2n - 2 - 1) + 3 = 2^{n}*(2n - 3) + 3So, T_n = (2n - 3)*2^{n} + 3Let me verify this formula with small n.For n=1:b_1 = (2*1 - 1)*2^{0} = 1*1 = 1T_1 = 1Using the formula: (2*1 - 3)*2^{1} + 3 = (-1)*2 + 3 = -2 + 3 = 1. Correct.For n=2:b_1 = 1, b_2 = (4 - 1)*2^{1} = 3*2 = 6T_2 = 1 + 6 = 7Using the formula: (4 - 3)*4 + 3 = 1*4 + 3 = 7. Correct.For n=3:b_3 = (6 - 1)*4 = 5*4 = 20T_3 = 1 + 6 + 20 = 27Using the formula: (6 - 3)*8 + 3 = 3*8 + 3 = 24 + 3 = 27. Correct.Good, seems to work.So, the sum T_n is (2n - 3)*2^{n} + 3.**Final Answer**Ⅰ. The value of ( a ) is (boxed{-1}) and the general term is (boxed{a_n = 2^{n-1}}).Ⅱ. The sum of the first ( n ) terms of the sequence ({b_n}) is (boxed{T_n = (2n - 3) cdot 2^n + 3}).

💡Okay, so I have this problem about a convex quadrilateral ABCD. Each side is extended past a vertex to a new point, creating another quadrilateral A'B'C'D'. The given information is that each extension doubles the length of the original side: AB = BB' = 8, BC = CC' = 9, CD = DD' = 10, and DA = AA' = 11. Also, the area of ABCD is 20. I need to find the area of A'B'C'D'.Hmm, let me visualize this. ABCD is a convex quadrilateral, so all its interior angles are less than 180 degrees, and the extensions of its sides will form another quadrilateral outside of ABCD. Each extension is equal in length to the original side, so each new point (B', C', D', A') is such that the segment from the original vertex to the new point is equal to the original side.I remember that when you extend sides of a quadrilateral and form triangles outside, the areas of those triangles can be related to the area of the original quadrilateral. Maybe I can use some properties of similar triangles or area ratios here.Let me think about each triangle formed by the extensions. For example, triangle A'AB is formed by extending DA past A to A', such that DA = AA' = 11. Similarly, triangle BB'C is formed by extending AB past B to B', with AB = BB' = 8. The same goes for the other triangles.Since each extension is equal to the original side, each of these triangles is similar to the triangles formed within the original quadrilateral. For instance, triangle A'AB is similar to triangle ABD because they share angle at A, and the sides are proportional. Wait, is that correct?Actually, maybe it's not exactly similar because the angles might not be the same. But perhaps the areas can be related by the ratio of the sides. Since each extension is equal to the original side, the ratio of the areas might be 2:1 or something like that.Let me consider triangle A'AB. Since AA' = DA = 11, and DA is a side of the quadrilateral, maybe triangle A'AB has an area twice that of triangle ABD. Similarly, triangle BB'C would have an area twice that of triangle ABC, and so on for the other triangles.If that's the case, then each of these four triangles (A'AB, BB'C, CC'D, DD'A) would each have an area twice the area of their corresponding smaller triangles within ABCD. Since the total area of ABCD is 20, which is the sum of the areas of triangles ABD and BCD, or alternatively, ABC and ADC, depending on how you split the quadrilateral.Wait, actually, the area of ABCD can be split into two triangles in two different ways: either into triangles ABC and ADC, or into triangles ABD and BCD. Either way, each of these triangles would have areas that add up to 20.If each of the four extended triangles has an area twice that of the corresponding smaller triangles, then the total area of the four extended triangles would be 2 times the sum of the areas of the smaller triangles. But the sum of the areas of the smaller triangles is 20, so 2 times 20 is 40.Therefore, the area of the larger quadrilateral A'B'C'D' would be the area of ABCD plus the areas of these four extended triangles. That would be 20 + 40 = 60.Wait, does that make sense? Let me double-check. If each extension creates a triangle with double the area of the corresponding smaller triangle, then four such triangles would add 40 to the original 20, giving a total of 60. That seems plausible.Alternatively, maybe I can think about the homothety transformation. Extending each side by its own length is equivalent to a homothety with a factor of 2 centered at each vertex. But since each extension is done from a different vertex, it's not a uniform scaling of the entire quadrilateral, so the area might not just be scaled by a factor of 4.But in this case, since each triangle added is twice the area of the corresponding smaller triangle, and there are four such triangles, adding up to 40, and the original area is 20, the total area becomes 60.I think that's the correct approach. So, the area of A'B'C'D' is 60.