💡First, I need to determine the total number of baby wipes Randy has. He has 6 packs, and each pack contains 120 wipes.Next, I'll calculate the total number of wipes by multiplying the number of packs by the number of wipes per pack.After finding the total number of wipes, I'll divide this by the number of days (360 days) to find out how many wipes Randy uses each day.Since Randy uses one wipe per walk, the number of wipes used per day will directly indicate the number of times he walks his dog each day.

💡First, I need to identify the number(s) that satisfy the condition where the number squared equals the number itself. This can be represented by the equation ( x^2 = x ).To solve this equation, I'll start by moving all terms to one side to set the equation to zero:[x^2 - x = 0]Next, I'll factor out the common term ( x ):[x(x - 1) = 0]According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero. Therefore:[x = 0 quad text{or} quad x - 1 = 0]Solving these simple equations gives:[x = 0 quad text{or} quad x = 1]Thus, the numbers that satisfy the condition are 0 and 1.

💡Okay, so I have this problem where three points A(2,2), B(a,0), and C(0,b) are collinear, and I need to find the value of 1/a + 1/b. Alright, let me try to figure this out step by step.First, I remember that if three points are collinear, the slope between any two pairs of points should be the same. So, maybe I can use the slope formula here. The slope between points A and B should be equal to the slope between points A and C.Let me write down the coordinates again to make sure I have them right:- Point A is (2,2)- Point B is (a,0)- Point C is (0,b)So, the slope between A and B would be (0 - 2)/(a - 2) which simplifies to (-2)/(a - 2). Similarly, the slope between A and C would be (b - 2)/(0 - 2) which is (b - 2)/(-2) or (2 - b)/2.Since these slopes are equal, I can set them equal to each other:(-2)/(a - 2) = (2 - b)/2Hmm, let me solve this equation. First, cross-multiplying to get rid of the denominators:(-2) * 2 = (a - 2)(2 - b)-4 = (a - 2)(2 - b)Now, I need to expand the right side:(a - 2)(2 - b) = a*(2 - b) - 2*(2 - b) = 2a - ab - 4 + 2bSo, putting it back into the equation:-4 = 2a - ab - 4 + 2bHmm, let's see. I can add 4 to both sides to simplify:-4 + 4 = 2a - ab + 2b0 = 2a - ab + 2bLet me rearrange this equation:ab = 2a + 2bHmm, I need to find 1/a + 1/b. Maybe I can manipulate this equation to get that expression. Let me see.Starting from ab = 2a + 2b, I can divide both sides by ab to get:1 = (2a)/(ab) + (2b)/(ab)1 = 2/b + 2/aSo, 1 = 2/a + 2/bHmm, but I need 1/a + 1/b. Let me factor out the 2:1 = 2*(1/a + 1/b)So, if I divide both sides by 2:1/2 = 1/a + 1/bOh, so 1/a + 1/b equals 1/2. That seems straightforward.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Calculated the slopes between A and B, and A and C.2. Set the slopes equal because the points are collinear.3. Cross-multiplied to eliminate denominators.4. Expanded the equation and simplified.5. Rearranged to get ab = 2a + 2b.6. Divided both sides by ab to get 1 = 2/b + 2/a.7. Factored out 2 to get 1 = 2*(1/a + 1/b).8. Divided both sides by 2 to find 1/a + 1/b = 1/2.Everything seems to check out. I don't see any errors in my calculations. So, I think the value of 1/a + 1/b is indeed 1/2.**Final Answer**The value of frac{1}{a} + frac{1}{b} is equal to boxed{dfrac{1}{2}}.

💡Okay, so I have this problem here: Determine how many lattice points lie on the graph of the equation (x^2 - y^2 = 65) in the (x,y)-plane. Hmm, lattice points mean points where both (x) and (y) are integers, right? So, I need to find all integer solutions ((x, y)) that satisfy this equation.First, let me recall that (x^2 - y^2) can be factored as ((x + y)(x - y)). So, the equation becomes ((x + y)(x - y) = 65). That seems useful because 65 is a specific number, and maybe I can find pairs of integers that multiply to 65.So, I need to find all pairs of integers ((a, b)) such that (a times b = 65). Then, for each pair, I can set (a = x + y) and (b = x - y), and solve for (x) and (y). If both (x) and (y) turn out to be integers, then that pair ((x, y)) is a lattice point on the graph.Alright, let's list all the factor pairs of 65. Since 65 is positive, the pairs can be both positive and negative. The positive factor pairs are (1, 65) and (5, 13). The negative factor pairs would be (-1, -65) and (-5, -13). Also, I should consider the reverse of these pairs because multiplication is commutative, so (65, 1), (13, 5), (-65, -1), and (-13, -5) are also valid.So, in total, I have eight factor pairs:1. (1, 65)2. (65, 1)3. (5, 13)4. (13, 5)5. (-1, -65)6. (-65, -1)7. (-5, -13)8. (-13, -5)Now, for each of these pairs, I'll set (a = x + y) and (b = x - y), and solve for (x) and (y). Let's start with the first pair.**1. (a = 1), (b = 65)**So, we have:[x + y = 1][x - y = 65]To solve for (x) and (y), I can add these two equations together:[(x + y) + (x - y) = 1 + 65][2x = 66][x = 33]Now, substitute (x = 33) back into the first equation:[33 + y = 1][y = 1 - 33][y = -32]So, one solution is ((33, -32)). Both 33 and -32 are integers, so this is a valid lattice point.**2. (a = 65), (b = 1)**Now, let's take the reverse of the first pair:[x + y = 65][x - y = 1]Adding the equations:[2x = 66][x = 33]Substitute (x = 33) into the first equation:[33 + y = 65][y = 65 - 33][y = 32]So, another solution is ((33, 32)). Again, both are integers, so this is good.**3. (a = 5), (b = 13)**Next pair:[x + y = 5][x - y = 13]Adding the equations:[2x = 18][x = 9]Substitute (x = 9) into the first equation:[9 + y = 5][y = 5 - 9][y = -4]So, another solution is ((9, -4)). Both integers, so that's valid.**4. (a = 13), (b = 5)**Reverse of the previous pair:[x + y = 13][x - y = 5]Adding:[2x = 18][x = 9]Substitute (x = 9) into the first equation:[9 + y = 13][y = 13 - 9][y = 4]So, another solution is ((9, 4)). Good.**5. (a = -1), (b = -65)**Now, moving on to the negative pairs:[x + y = -1][x - y = -65]Adding:[2x = -66][x = -33]Substitute (x = -33) into the first equation:[-33 + y = -1][y = -1 + 33][y = 32]So, another solution is ((-33, 32)). Both integers, so that's valid.**6. (a = -65), (b = -1)**Reverse of the previous negative pair:[x + y = -65][x - y = -1]Adding:[2x = -66][x = -33]Substitute (x = -33) into the first equation:[-33 + y = -65][y = -65 + 33][y = -32]So, another solution is ((-33, -32)). Both integers, so that's good.**7. (a = -5), (b = -13)**Next negative pair:[x + y = -5][x - y = -13]Adding:[2x = -18][x = -9]Substitute (x = -9) into the first equation:[-9 + y = -5][y = -5 + 9][y = 4]So, another solution is ((-9, 4)). Both integers, so that's valid.**8. (a = -13), (b = -5)**Reverse of the previous negative pair:[x + y = -13][x - y = -5]Adding:[2x = -18][x = -9]Substitute (x = -9) into the first equation:[-9 + y = -13][y = -13 + 9][y = -4]So, another solution is ((-9, -4)). Both integers, so that's good.Let me recap all the solutions I found:1. ((33, -32))2. ((33, 32))3. ((9, -4))4. ((9, 4))5. ((-33, 32))6. ((-33, -32))7. ((-9, 4))8. ((-9, -4))So, that's eight solutions in total. Each of these corresponds to a lattice point on the graph of (x^2 - y^2 = 65). Wait, just to make sure I didn't miss any factor pairs, let me think again. 65 is 5 times 13, and both 5 and 13 are primes, so the only factors are 1, 5, 13, 65, and their negatives. So, the factor pairs I considered are indeed all possible integer pairs that multiply to 65. Also, when I solved each system, I made sure that both (x) and (y) came out as integers. If in any case, (a + b) or (a - b) were odd, then (x) or (y) would not be integers, but in all these cases, since 65 is odd, and all the factor pairs are either both odd or both even? Wait, actually, 65 is odd, so all its factor pairs must consist of two odd numbers because odd times odd is odd. So, in each case, (a) and (b) are both odd, so (a + b) and (a - b) are both even, meaning (x) and (y) will always be integers. So, that's why all these solutions are valid.Therefore, I can confidently say that there are 8 lattice points on the graph of (x^2 - y^2 = 65).**Final Answer**The number of lattice points on the graph is boxed{8}.

💡Okay, so I need to prove that for any ( n geq 4 ), every cyclic quadrilateral can be decomposed into ( n ) cyclic quadrilaterals. Hmm, cyclic quadrilaterals are four-sided figures where all the vertices lie on a single circle. That means their opposite angles sum to 180 degrees. First, I should understand what it means to decompose a cyclic quadrilateral into smaller cyclic quadrilaterals. Decomposition here likely means dividing the original quadrilateral into smaller quadrilaterals, each of which is also cyclic. So, each smaller quadrilateral must also have all its vertices on some circle, possibly the same as the original or different ones.Let me start with the simplest case, ( n = 4 ). If I can figure out how to split a cyclic quadrilateral into four smaller cyclic quadrilaterals, maybe I can generalize the method for any ( n geq 4 ).Imagine a cyclic quadrilateral ( ABCD ) inscribed in a circle. The center of this circle is ( O ). If ( O ) is inside the quadrilateral, maybe I can draw lines from ( O ) to each side, creating four smaller quadrilaterals. But wait, if I draw perpendiculars from ( O ) to each side, those would be the shortest distances from the center to the sides. Would these perpendiculars create cyclic quadrilaterals?Let me visualize this. If I draw a perpendicular from ( O ) to side ( AB ), another to ( BC ), another to ( CD ), and another to ( DA ), these four perpendiculars would intersect the sides at midpoints if the quadrilateral is symmetric, but it might not be. However, each of these perpendiculars would create smaller quadrilaterals within ( ABCD ). But are these smaller quadrilaterals cyclic? For a quadrilateral to be cyclic, its opposite angles must sum to 180 degrees. If I have a smaller quadrilateral formed by two adjacent sides of ( ABCD ) and two perpendiculars from ( O ), would that satisfy the cyclic condition? I'm not sure. Maybe I need a different approach.Alternatively, what if I connect the center ( O ) to each vertex? That would divide the quadrilateral into four triangles, not quadrilaterals. So that doesn't help. Maybe I need to introduce more points.Another idea: if I choose points along the sides of ( ABCD ) such that the new points also lie on the circumcircle. But adding points on the sides might not necessarily create cyclic quadrilaterals unless the new points are chosen carefully.Wait, maybe I can use the fact that any triangle inscribed in a circle can be part of a cyclic quadrilateral. If I can create triangles within ( ABCD ) that can be extended to form cyclic quadrilaterals, that might work. But I'm not sure how to do that systematically.Let me think about the properties of cyclic quadrilaterals. One important property is that the product of the lengths of the diagonals can be related to the sum of the products of opposite sides. But I'm not sure if that helps here.Perhaps I should look for a way to inductively build up the decomposition. If I can show that if a cyclic quadrilateral can be decomposed into ( k ) cyclic quadrilaterals, then it can also be decomposed into ( k+1 ) cyclic quadrilaterals, then I can use induction to prove it for all ( n geq 4 ).So, starting with ( n = 4 ), I need a base case. Maybe I can split the original quadrilateral into four smaller ones by adding a point inside and connecting it to the vertices or midpoints. But again, ensuring that each smaller quadrilateral is cyclic is tricky.Wait, what if I use the fact that any two points on a circle can be connected to form a chord, and any chord can be part of a cyclic quadrilateral. If I add a point on the circumcircle and connect it appropriately, maybe I can create smaller cyclic quadrilaterals.Alternatively, maybe I can use the concept of similar triangles or some symmetry in the original quadrilateral to create smaller cyclic quadrilaterals. But I'm not sure.Another approach: consider that any cyclic quadrilateral can be transformed into another cyclic quadrilateral by moving one vertex along the circumcircle. Maybe by doing this multiple times, I can create multiple smaller cyclic quadrilaterals within the original one.But I'm not sure how to formalize this idea. Maybe I need to look for specific constructions or known theorems about decomposing cyclic quadrilaterals.Wait, I recall that any convex quadrilateral can be divided into triangles, but here we need to divide into cyclic quadrilaterals. Maybe I can use a combination of triangles and other cyclic quadrilaterals, but I need to ensure that each piece is a cyclic quadrilateral.Perhaps I can start by dividing the original quadrilateral into two cyclic quadrilaterals and then further divide one of them into more cyclic quadrilaterals. But I need to figure out how to do that.Let me try to sketch this out. Suppose I have cyclic quadrilateral ( ABCD ). If I draw a diagonal ( AC ), it divides the quadrilateral into two triangles ( ABC ) and ( ACD ). But triangles aren't quadrilaterals, so that doesn't help. Maybe I need to draw a line that creates a smaller quadrilateral.What if I draw a line parallel to one of the sides? For example, draw a line parallel to ( AB ) inside the quadrilateral. This would create a smaller quadrilateral and a trapezoid. If the original quadrilateral is cyclic, would the smaller quadrilateral and the trapezoid also be cyclic?I think that if a trapezoid is cyclic, it must be isosceles. So, if I draw a line parallel to ( AB ) such that the resulting trapezoid is isosceles, then both the smaller quadrilateral and the trapezoid would be cyclic. That sounds promising.So, for ( n = 4 ), I can draw one such line, creating two cyclic quadrilaterals. Then, I can further decompose each of these into more cyclic quadrilaterals by drawing more parallel lines. But I need to ensure that each step maintains the cyclic property.Wait, but if I draw multiple parallel lines, each creating smaller trapezoids, and each trapezoid is isosceles, then each would be cyclic. So, starting from the original quadrilateral, I can draw ( n-1 ) parallel lines, creating ( n ) smaller cyclic quadrilaterals.But does this work for any ( n geq 4 )? Let me see. For ( n = 4 ), I draw three parallel lines, creating four smaller cyclic quadrilaterals. For ( n = 5 ), I draw four parallel lines, and so on. This seems scalable.However, I need to ensure that each of these smaller quadrilaterals is indeed cyclic. Since each trapezoid is isosceles and cyclic, and the smaller quadrilaterals formed by the parallel lines would also be cyclic because they inherit the cyclic property from the original.But wait, is it always possible to draw these parallel lines such that each resulting trapezoid is isosceles? I think so, because I can adjust the distance between the parallel lines to ensure that the non-parallel sides are equal in length, making the trapezoid isosceles.Therefore, by drawing ( n-1 ) parallel lines to one of the sides, I can decompose the original cyclic quadrilateral into ( n ) smaller cyclic quadrilaterals. This seems like a viable method.But I should verify this with an example. Let's take a simple cyclic quadrilateral, say a square, which is a special case of a cyclic quadrilateral. If I draw three parallel lines to one side, say the top side, equally spaced, they would divide the square into four smaller rectangles. Rectangles are cyclic quadrilaterals because all their angles are 90 degrees, and opposite angles sum to 180 degrees. So, this works for ( n = 4 ).Similarly, for ( n = 5 ), drawing four parallel lines would create five smaller rectangles, each of which is cyclic. This seems to hold.But what if the original cyclic quadrilateral isn't a square or a rectangle? For example, consider a kite-shaped cyclic quadrilateral, which is actually a rhombus if it's cyclic. Drawing parallel lines to one of the sides would still create smaller cyclic quadrilaterals, as the properties would be preserved.Wait, but not all cyclic quadrilaterals are symmetric like squares or rhombuses. What about a general cyclic quadrilateral where the sides are of different lengths and the angles are not necessarily equal?In that case, drawing parallel lines might not result in isosceles trapezoids unless I carefully adjust the spacing. But I think it's still possible because I can choose the points where the parallel lines intersect the sides such that the resulting trapezoids are isosceles.Alternatively, maybe I can use a different approach, such as adding points on the circumcircle and connecting them appropriately to form smaller cyclic quadrilaterals. But I need to ensure that each new quadrilateral shares the same circumcircle or has its own.Wait, if I add points on the circumcircle, the new quadrilaterals would automatically be cyclic because all their vertices lie on the same circle. So, maybe that's a better approach.For example, starting with cyclic quadrilateral ( ABCD ), I can add a point ( E ) on the circumcircle between ( A ) and ( B ). Then, connecting ( E ) to ( C ) and ( D ) might create smaller cyclic quadrilaterals ( AEDC ) and ( EBCD ). But I need to check if these are indeed cyclic.Since ( E ) lies on the circumcircle of ( ABCD ), any quadrilateral formed by connecting ( E ) to other vertices will also be cyclic. So, ( AEDC ) and ( EBCD ) are both cyclic. Thus, by adding one point, I've decomposed the original quadrilateral into two smaller cyclic quadrilaterals.To get more than two, I can add more points. For ( n = 4 ), I need three additional points, but that might complicate things. Alternatively, I can use a combination of adding points and drawing lines.Wait, maybe a better way is to use the fact that any cyclic quadrilateral can be divided into triangles, and then those triangles can be combined with other points to form cyclic quadrilaterals. But I need to ensure that each step maintains the cyclic property.Another idea: use the concept of similar cyclic quadrilaterals. If I can create smaller cyclic quadrilaterals similar to the original, then I can fit multiple of them within the original one. But similarity might not be necessary here.Alternatively, maybe I can use the fact that the intersection of two chords in a circle creates cyclic quadrilaterals. If I draw multiple chords intersecting inside the circle, each intersection can form a cyclic quadrilateral. But I need to ensure that the decomposition covers the entire original quadrilateral without overlapping.This seems complicated, but perhaps manageable. For example, starting with ( ABCD ), draw a chord ( AC ). Then, draw another chord ( BD ). These intersect at point ( E ), creating four smaller quadrilaterals: ( AEB ), ( BEC ), ( CED ), and ( DEA ). But these are triangles, not quadrilaterals. So, that doesn't help.Wait, but if I draw more chords, maybe I can create quadrilaterals. For instance, draw chords ( AC ) and ( BD ), and then draw another chord ( EG ) intersecting ( AC ) and ( BD ) at some points. This might create smaller quadrilaterals, but I need to ensure they are cyclic.This approach might not be straightforward. Maybe I should stick with the earlier idea of drawing parallel lines to one side, creating isosceles trapezoids, which are cyclic.So, to summarize my thoughts so far: I can decompose a cyclic quadrilateral into ( n ) cyclic quadrilaterals by drawing ( n-1 ) parallel lines to one of its sides, creating ( n ) smaller isosceles trapezoids, each of which is cyclic. This method seems scalable for any ( n geq 4 ).But I need to make sure that this works for any cyclic quadrilateral, not just symmetric ones. Let me think about a general cyclic quadrilateral ( ABCD ) with sides ( AB ), ( BC ), ( CD ), and ( DA ). If I draw a line parallel to ( AB ) inside the quadrilateral, it will intersect ( BC ) and ( DA ) at some points ( E ) and ( F ), respectively. The resulting figure ( ABEF ) would be a trapezoid. For it to be cyclic, it must be isosceles, meaning ( BE = CF ).To ensure that ( BE = CF ), I need to choose the position of the parallel line such that the distances from ( E ) and ( F ) to the respective sides are equal. This might require some calculation, but it's possible.Once I have one isosceles trapezoid ( ABEF ), the remaining part of the original quadrilateral is another quadrilateral ( EFCD ). I need to check if ( EFCD ) is also cyclic. Since ( EF ) is parallel to ( AB ), and ( ABCD ) is cyclic, ( EFCD ) should also be cyclic because it shares the same circumcircle.Wait, is that necessarily true? If ( EF ) is parallel to ( AB ), and ( ABCD ) is cyclic, does that guarantee that ( EFCD ) is cyclic? I'm not entirely sure, but I think it does because the angles subtended by the same chord should still hold.Let me verify. In cyclic quadrilateral ( ABCD ), angles ( A ) and ( C ) are supplementary. If I draw ( EF ) parallel to ( AB ), then angle ( E ) in ( EFCD ) should be equal to angle ( A ), and angle ( F ) should be equal to angle ( B ). Since ( A ) and ( C ) are supplementary, and ( E ) is equal to ( A ), then ( E ) and ( C ) should also be supplementary, making ( EFCD ) cyclic.Yes, that makes sense. So, by drawing a single parallel line, I can decompose ( ABCD ) into two cyclic quadrilaterals ( ABEF ) and ( EFCD ). Repeating this process by drawing more parallel lines would allow me to decompose the original quadrilateral into more cyclic quadrilaterals.For example, to get ( n = 4 ), I can draw three parallel lines, creating four smaller cyclic quadrilaterals. Each time I draw a parallel line, I ensure that the new trapezoid is isosceles by adjusting the position of the line accordingly. This way, each new quadrilateral is cyclic, and the entire original quadrilateral is decomposed into ( n ) cyclic quadrilaterals.This seems like a solid approach. I just need to formalize it and ensure that it works for any ( n geq 4 ). Maybe I can use induction: assume that a cyclic quadrilateral can be decomposed into ( k ) cyclic quadrilaterals, then show that it can also be decomposed into ( k+1 ) cyclic quadrilaterals by adding another parallel line.Yes, that would work. The base case ( n = 4 ) is established by drawing three parallel lines, and the inductive step involves adding one more parallel line to increase the number of cyclic quadrilaterals by one. Therefore, by induction, the decomposition is possible for all ( n geq 4 ).I think I've got a good handle on this now. The key idea is to use parallel lines to create isosceles trapezoids, which are cyclic, and by adding more such lines, we can achieve the desired number of cyclic quadrilaterals.

💡Okay, so I have this integral here that I need to minimize with respect to ( a ), where ( a > 1 ). The integral is from ( a ) to ( a^2 ) of ( frac{1}{x} log frac{x-1}{32} , dx ). Hmm, that looks a bit complicated, but let me try to break it down.First, I remember that to find the minimum of a function, I need to take its derivative and set it equal to zero. Since this is an integral with variable limits, I think I can use the Fundamental Theorem of Calculus and the Leibniz rule for differentiation under the integral sign. Let me recall how that works.So, if I have a function defined as ( f(a) = int_{u(a)}^{v(a)} g(x) , dx ), then the derivative ( f'(a) ) is ( g(v(a)) cdot v'(a) - g(u(a)) cdot u'(a) ). Yeah, that sounds right. So in this case, ( u(a) = a ) and ( v(a) = a^2 ), and ( g(x) = frac{1}{x} log frac{x-1}{32} ).Let me write that out:[ f'(a) = g(a^2) cdot frac{d}{da}(a^2) - g(a) cdot frac{d}{da}(a) ]Calculating each part:First, ( g(a^2) = frac{1}{a^2} log frac{a^2 - 1}{32} ), and ( frac{d}{da}(a^2) = 2a ).Then, ( g(a) = frac{1}{a} log frac{a - 1}{32} ), and ( frac{d}{da}(a) = 1 ).Putting it all together:[ f'(a) = left( frac{1}{a^2} log frac{a^2 - 1}{32} right) cdot 2a - left( frac{1}{a} log frac{a - 1}{32} right) cdot 1 ]Simplify this expression:[ f'(a) = frac{2}{a} log frac{a^2 - 1}{32} - frac{1}{a} log frac{a - 1}{32} ]Okay, so I need to set this derivative equal to zero and solve for ( a ):[ frac{2}{a} log frac{a^2 - 1}{32} - frac{1}{a} log frac{a - 1}{32} = 0 ]Let me factor out ( frac{1}{a} ):[ frac{1}{a} left( 2 log frac{a^2 - 1}{32} - log frac{a - 1}{32} right) = 0 ]Since ( a > 1 ), ( frac{1}{a} ) is never zero, so I can focus on the expression inside the parentheses:[ 2 log frac{a^2 - 1}{32} - log frac{a - 1}{32} = 0 ]Hmm, let's use logarithm properties to simplify this. Remember that ( 2 log x = log x^2 ) and ( log x - log y = log frac{x}{y} ). So applying that:[ log left( left( frac{a^2 - 1}{32} right)^2 right) - log left( frac{a - 1}{32} right) = 0 ]Which can be written as:[ log left( frac{left( frac{a^2 - 1}{32} right)^2}{frac{a - 1}{32}} right) = 0 ]Simplify the fraction inside the log:First, ( left( frac{a^2 - 1}{32} right)^2 = frac{(a^2 - 1)^2}{32^2} ), and ( frac{a - 1}{32} ) is as is. So dividing these:[ frac{(a^2 - 1)^2}{32^2} div frac{a - 1}{32} = frac{(a^2 - 1)^2}{32^2} times frac{32}{a - 1} = frac{(a^2 - 1)^2}{32(a - 1)} ]So now the equation is:[ log left( frac{(a^2 - 1)^2}{32(a - 1)} right) = 0 ]Since ( log 1 = 0 ), this implies:[ frac{(a^2 - 1)^2}{32(a - 1)} = 1 ]Multiply both sides by ( 32(a - 1) ):[ (a^2 - 1)^2 = 32(a - 1) ]Hmm, let's expand ( (a^2 - 1)^2 ). Remember that ( a^2 - 1 = (a - 1)(a + 1) ), so:[ (a^2 - 1)^2 = (a - 1)^2(a + 1)^2 ]So plugging that back in:[ (a - 1)^2(a + 1)^2 = 32(a - 1) ]We can factor out ( (a - 1) ) from both sides, but we have to be careful because ( a > 1 ), so ( a - 1 neq 0 ). So divide both sides by ( (a - 1) ):[ (a - 1)(a + 1)^2 = 32 ]So now we have:[ (a - 1)(a + 1)^2 = 32 ]This looks like a cubic equation. Let me expand it:First, expand ( (a + 1)^2 = a^2 + 2a + 1 ). Then multiply by ( (a - 1) ):[ (a - 1)(a^2 + 2a + 1) = a(a^2 + 2a + 1) - 1(a^2 + 2a + 1) ][ = a^3 + 2a^2 + a - a^2 - 2a - 1 ][ = a^3 + (2a^2 - a^2) + (a - 2a) - 1 ][ = a^3 + a^2 - a - 1 ]So the equation becomes:[ a^3 + a^2 - a - 1 = 32 ][ a^3 + a^2 - a - 33 = 0 ]Now, I need to solve this cubic equation: ( a^3 + a^2 - a - 33 = 0 ). Let me try to find rational roots using the Rational Root Theorem. The possible rational roots are factors of 33 over factors of 1, so possible roots are ( pm1, pm3, pm11, pm33 ).Let me test ( a = 3 ):[ 3^3 + 3^2 - 3 - 33 = 27 + 9 - 3 - 33 = 36 - 36 = 0 ]Oh, that works! So ( a = 3 ) is a root. Therefore, we can factor the cubic as:[ (a - 3)(a^2 + 4a + 11) = 0 ]Now, set each factor equal to zero:1. ( a - 3 = 0 ) gives ( a = 3 ).2. ( a^2 + 4a + 11 = 0 ). Let's compute the discriminant: ( 16 - 44 = -28 ). Since the discriminant is negative, there are no real roots from this quadratic.Therefore, the only real solution is ( a = 3 ).But wait, I should verify if this is indeed a minimum. Maybe I should check the second derivative or analyze the behavior of the function around ( a = 3 ). Alternatively, I can consider the values of the integral at ( a = 3 ) and nearby points to see if it's a minimum.Alternatively, since the integral is from ( a ) to ( a^2 ), and ( a > 1 ), as ( a ) increases beyond 3, the upper limit ( a^2 ) grows much faster than the lower limit ( a ). So maybe the integral tends to increase as ( a ) moves away from 3 in either direction, making ( a = 3 ) the point of minimum.Also, considering the behavior as ( a ) approaches 1 from the right, the integral becomes:[ int_{1}^{1} frac{1}{x} log frac{x-1}{32} , dx = 0 ]But wait, actually, as ( a ) approaches 1, the lower limit approaches 1, and the upper limit approaches 1 as well, so the integral approaches zero. However, the integrand near ( x = 1 ) is ( frac{1}{x} log frac{x-1}{32} ), which tends to ( -infty ) because ( log frac{x-1}{32} ) approaches ( log 0 ), which is ( -infty ). But since the interval is shrinking to zero, it's not clear if the integral approaches zero or some finite value. Maybe I should compute the limit as ( a ) approaches 1.Alternatively, let's test ( a = 2 ) and ( a = 4 ) to see how the integral behaves.For ( a = 2 ):The integral is from 2 to 4 of ( frac{1}{x} log frac{x - 1}{32} , dx ).Compute this numerically:Let me approximate it. Let's take some sample points.At ( x = 2 ): ( frac{1}{2} log frac{1}{32} = frac{1}{2} log frac{1}{32} = frac{1}{2} (-log 32) approx frac{1}{2} (-3.4657) approx -1.7328 )At ( x = 3 ): ( frac{1}{3} log frac{2}{32} = frac{1}{3} log frac{1}{16} = frac{1}{3} (-2.7726) approx -0.9242 )At ( x = 4 ): ( frac{1}{4} log frac{3}{32} approx frac{1}{4} (-3.0445) approx -0.7611 )So the integrand is negative throughout the interval, and the area under the curve is negative. But how does it compare to when ( a = 3 )?For ( a = 3 ):Integral from 3 to 9 of ( frac{1}{x} log frac{x - 1}{32} , dx ).Again, approximate:At ( x = 3 ): ( frac{1}{3} log frac{2}{32} = frac{1}{3} log frac{1}{16} approx -0.9242 )At ( x = 5 ): ( frac{1}{5} log frac{4}{32} = frac{1}{5} log frac{1}{8} approx frac{1}{5} (-2.0794) approx -0.4159 )At ( x = 9 ): ( frac{1}{9} log frac{8}{32} = frac{1}{9} log frac{1}{4} approx frac{1}{9} (-1.3863) approx -0.1540 )So the integrand is still negative, but the area might be less negative than when ( a = 2 ), or more? Hmm, it's hard to tell without exact computation.Alternatively, maybe I can compute the integral at ( a = 3 ) and see if it's indeed the minimum.But perhaps a better approach is to consider the second derivative test. If ( f'(a) = 0 ) at ( a = 3 ), and if ( f''(a) > 0 ) there, then it's a local minimum.But computing the second derivative might be complicated. Alternatively, since we found that ( a = 3 ) is the only critical point, and as ( a ) approaches 1, the integral approaches zero (but from negative values, since the integrand is negative), and as ( a ) increases beyond 3, the integral might start increasing again because the upper limit grows faster, but the integrand's behavior is not straightforward.Wait, actually, as ( a ) increases, the interval ( [a, a^2] ) becomes larger, but the integrand ( frac{1}{x} log frac{x - 1}{32} ) becomes less negative because ( frac{x - 1}{32} ) increases, so ( log frac{x - 1}{32} ) increases (becomes less negative). So the integrand becomes less negative as ( x ) increases, meaning the integral might start increasing (becoming less negative) as ( a ) increases beyond 3.Similarly, as ( a ) approaches 1, the integral becomes more negative because the integrand is more negative near ( x = 1 ), but the interval shrinks. It's a bit unclear without precise computation.But given that ( a = 3 ) is the only critical point, and since the integral is negative there, and the integral tends to zero as ( a ) approaches 1, but is negative, and tends to some positive value as ( a ) approaches infinity (since the integrand becomes positive when ( x > 32 )), it's likely that ( a = 3 ) is indeed the point where the integral attains its minimum value.Therefore, after all this, I think the value of ( a ) that minimizes the integral is 3.