💡Alright, so I've got this problem here: Let ( a ) be a constant and ( 0 < a < 1 ). If ( y = left( log_a frac{3}{4} right)^x ) is a strictly increasing function on ( mathbb{R} ), then I need to find the range of the real number ( a ). Hmm, okay, let's break this down step by step.First off, I know that for a function of the form ( y = b^x ) to be strictly increasing, the base ( b ) must be greater than 1. If ( b ) is between 0 and 1, the function is actually decreasing. So, in this case, the base of the exponent is ( log_a frac{3}{4} ). Therefore, for ( y ) to be strictly increasing, ( log_a frac{3}{4} ) must be greater than 1.Alright, so I can write that down as an inequality:[log_a frac{3}{4} > 1]Now, I need to solve this inequality for ( a ). Remembering the properties of logarithms, especially when dealing with bases between 0 and 1, which is the case here since ( 0 < a < 1 ).I recall that for logarithms with bases less than 1, the function is decreasing. That means if ( log_a frac{3}{4} > 1 ), then ( frac{3}{4} ) must be less than ( a^1 ), but since the logarithm is decreasing, the inequality flips when I exponentiate both sides. Wait, let me make sure I get this right.Actually, let's recall that ( log_a b = c ) is equivalent to ( a^c = b ). So, if ( log_a frac{3}{4} > 1 ), then ( a^1 < frac{3}{4} ) because the logarithm function with base ( a ) (which is less than 1) is decreasing. So, higher outputs correspond to lower inputs.Therefore, from ( log_a frac{3}{4} > 1 ), we get:[a < frac{3}{4}]But wait, hold on. The original condition is ( 0 < a < 1 ). So, combining this with ( a < frac{3}{4} ), we get ( 0 < a < frac{3}{4} ).But hold on a second, let me think again. If ( log_a frac{3}{4} > 1 ), and since ( a < 1 ), the logarithm is decreasing, so higher values of ( log_a frac{3}{4} ) correspond to lower values of ( a ). So, if ( log_a frac{3}{4} > 1 ), then ( a ) must be less than ( frac{3}{4} ). But does that make sense in terms of the function being increasing?Wait, let's consider the function ( y = left( log_a frac{3}{4} right)^x ). For this to be increasing, the base ( log_a frac{3}{4} ) must be greater than 1. So, ( log_a frac{3}{4} > 1 ). Since ( a < 1 ), the logarithm is decreasing, so ( frac{3}{4} < a ). Wait, now I'm confused because earlier I thought it was ( a < frac{3}{4} ), but now I'm getting ( a > frac{3}{4} ).Let me double-check. If ( log_a frac{3}{4} > 1 ), and ( a < 1 ), then because the logarithm is decreasing, ( frac{3}{4} < a^1 ), which is ( a > frac{3}{4} ). So, that means ( a ) must be greater than ( frac{3}{4} ) but still less than 1.Okay, that makes more sense. So, the range of ( a ) is ( frac{3}{4} < a < 1 ).Let me verify this with an example. Suppose ( a = frac{4}{5} ), which is greater than ( frac{3}{4} ). Then ( log_{frac{4}{5}} frac{3}{4} ) would be greater than 1 because ( frac{3}{4} ) is less than ( frac{4}{5} ), and since the logarithm is decreasing, the value is greater than 1. Therefore, ( y = left( log_{frac{4}{5}} frac{3}{4} right)^x ) would indeed be increasing.On the other hand, if ( a = frac{1}{2} ), which is less than ( frac{3}{4} ), then ( log_{frac{1}{2}} frac{3}{4} ) would be less than 1 because ( frac{3}{4} ) is greater than ( frac{1}{2} ), and since the logarithm is decreasing, the value is less than 1. Therefore, ( y = left( log_{frac{1}{2}} frac{3}{4} right)^x ) would be decreasing, which is not what we want.So, putting it all together, the range of ( a ) must be ( frac{3}{4} < a < 1 ) for the function ( y = left( log_a frac{3}{4} right)^x ) to be strictly increasing on ( mathbb{R} ).Just to make sure I haven't missed anything, let's consider the edge cases. If ( a = frac{3}{4} ), then ( log_{frac{3}{4}} frac{3}{4} = 1 ), so ( y = 1^x = 1 ), which is a constant function, not strictly increasing. Similarly, if ( a = 1 ), the logarithm is undefined because ( log_1 ) is not a valid function. So, the endpoints are excluded, and the range is indeed open interval ( left( frac{3}{4}, 1 right) ).Yeah, I think that's solid. I went through the properties, checked with examples, and considered the edge cases. So, I'm confident that the range of ( a ) is ( frac{3}{4} < a < 1 ).

💡Okay, so I have this geometry problem here, and I need to figure out which statement is incorrect. Let me read it again carefully.It says: Given two different lines a and b, and two different planes α and β, which of the following statements is incorrect?There are four options: A, B, C, and D. I need to analyze each one to see if they're correct or not.Starting with Option A: If a is perpendicular to α, and b is parallel to α, then a is perpendicular to b.Hmm, okay. So, if line a is perpendicular to plane α, that means a is orthogonal to every line in α. Now, if line b is parallel to α, then b lies within a plane that's parallel to α or is parallel to some line in α. But since a is perpendicular to α, it should also be perpendicular to any line parallel to α, right? Because if b is parallel to α, then it's essentially like a line in a parallel plane, and a is still orthogonal to it. So, I think A is correct.Moving on to Option B: If a is perpendicular to α, b is parallel to a, and b is contained in β, then α is perpendicular to β.Alright, so a is perpendicular to α. If b is parallel to a, then b is also perpendicular to α. Since b is in β, that means β contains a line (b) that's perpendicular to α. By the definition of perpendicular planes, if a plane contains a line perpendicular to another plane, then the two planes are perpendicular. So, yes, α and β should be perpendicular. So, B seems correct.Option C: If a is perpendicular to α, b is perpendicular to β, and α is parallel to β, then a is parallel to b.Okay, so a is perpendicular to α, and b is perpendicular to β. If α is parallel to β, then their normals (which are lines like a and b) should also be parallel. Because if two planes are parallel, their normals are parallel. So, if a and b are both normals to parallel planes, they must be parallel to each other. So, C is correct.Now, Option D: If a is parallel to α, and a is parallel to β, then α is parallel to β.Hmm, this one is tricky. So, if line a is parallel to both planes α and β, does that mean α and β are parallel? Let me think. If a line is parallel to two different planes, it doesn't necessarily mean the planes are parallel. They could intersect along some other line. For example, imagine two planes that intersect along a line c. If line a is parallel to both α and β, it just means a is parallel to the line of intersection c. So, the planes themselves aren't necessarily parallel; they can still intersect. Therefore, D might be incorrect.Wait, let me double-check. If a is parallel to α, that means a doesn't intersect α and lies in a direction parallel to α. Similarly, a is parallel to β. But two planes can still intersect even if they both have a line parallel to them. So, yes, α and β could intersect, meaning they aren't necessarily parallel. So, D is incorrect.So, after analyzing all options, D is the incorrect statement.

💡First, I need to determine how many shoes are in one pair. Typically, one pair of shoes consists of two individual shoes.Next, I'll calculate the total number of shoes Ezekiel bought by multiplying the number of pairs by the number of shoes per pair. So, 3 pairs multiplied by 2 shoes per pair equals 6 shoes.Therefore, Ezekiel has 6 new shoes in total.

💡Okay, so I have this problem where I need to find the value of 'a' for a cubic equation. The equation is given as 8x³ + 3a x² + 6b x + a = 0. It says that this equation has three distinct positive roots. Also, the sum of the base-3 logarithms of the roots is 3. Hmm, okay, let me try to unpack this step by step.First, I remember that for a cubic equation of the form ax³ + bx² + cx + d = 0, the sum of the roots, the sum of the product of roots two at a time, and the product of all three roots can be found using Vieta's formulas. So, maybe I can apply that here.Let me denote the roots as r, s, and t. Since all roots are positive and distinct, that might help later on, but for now, I'll just note that they are positive.The equation given is 8x³ + 3a x² + 6b x + a = 0. Comparing this to the standard cubic equation, I can see that the coefficients are:- Leading coefficient (for x³): 8- Coefficient for x²: 3a- Coefficient for x: 6b- Constant term: aUsing Vieta's formulas, for a cubic equation ax³ + bx² + cx + d = 0, the sum of the roots is -b/a, the sum of the product of roots two at a time is c/a, and the product of the roots is -d/a.So, applying that here:1. Sum of the roots: r + s + t = - (3a)/82. Sum of the product of roots two at a time: rs + rt + st = (6b)/8 = (3b)/43. Product of the roots: rst = -a/8Okay, so that's the Vieta's part.Now, the problem also mentions that the sum of the base-3 logarithms of the roots is 3. So, mathematically, that would be:log₃(r) + log₃(s) + log₃(t) = 3I remember that one of the logarithm properties is that the sum of logs is the log of the product. So, log₃(r) + log₃(s) + log₃(t) = log₃(rst) = 3.So, that means:log₃(rst) = 3Which implies that:rst = 3³ = 27So, the product of the roots rst is 27.But from Vieta's formula, we already have that rst = -a/8. So, setting that equal to 27:-a/8 = 27Now, solving for 'a':Multiply both sides by 8:-a = 27 * 827 * 8 is 216, so:-a = 216Multiply both sides by -1:a = -216Wait, so is that it? Did I just find 'a'?Let me double-check my steps to make sure I didn't make a mistake.1. I used Vieta's formula correctly for the product of the roots. The product is indeed -a/8 because the constant term is 'a' and the leading coefficient is 8.2. The logarithm property: sum of logs is log of the product. That seems correct.3. Calculated 3³ as 27. That's right.4. Set -a/8 equal to 27, solved for 'a' and got -216. That seems correct.Is there anything else I need to consider? The problem mentions that the equation has three distinct positive roots. So, I should make sure that with a = -216, the equation indeed has three distinct positive roots.Let me substitute a = -216 into the equation:8x³ + 3*(-216)x² + 6b x + (-216) = 0Simplify:8x³ - 648x² + 6b x - 216 = 0Hmm, but I don't know the value of 'b', so maybe I can't directly check the roots. However, since the problem states that there are three distinct positive roots, and we've found 'a' based on the product of the roots, which is a necessary condition, I think that's sufficient for the answer.Alternatively, maybe I can check if the cubic equation with a = -216 has three positive roots. But without knowing 'b', it's tricky. However, since the problem only asks for 'a', and we've used the given condition about the logarithms, which led us directly to 'a', I think that's the answer they're looking for.So, I think I'm confident that a = -216 is the correct answer.**Final Answer**The value of ( a ) is boxed{-216}.

💡First, substitute the given values of ( x = 5 ) and ( y = 8 ) into the expression:[frac{frac{1}{5} - frac{1}{8}}{frac{1}{5}}]Next, find a common denominator for the fractions in the numerator. The least common denominator of 5 and 8 is 40:[frac{8}{40} - frac{5}{40} = frac{3}{40}]Now, divide this result by (frac{1}{5}):[frac{frac{3}{40}}{frac{1}{5}} = frac{3}{40} times frac{5}{1} = frac{15}{40}]Finally, simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 5:[frac{15}{40} = frac{3}{8}]

💡Okay, so I have this problem where I need to prove that at least one of the numbers ( x, y, z ) is equal to ( a ). The system of equations given is:[left{begin{array}{l}x + y + z = a frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{1}{a}end{array}right.]Hmm, let me think about how to approach this. I remember that when dealing with systems of equations involving sums and reciprocals, sometimes it's helpful to consider symmetric polynomials or maybe even think about the roots of a polynomial. Let me try to recall Vieta's formulas. For a cubic equation ( t^3 - pt^2 + qt - r = 0 ), the sum of the roots is ( p ), the sum of the products of the roots two at a time is ( q ), and the product of the roots is ( r ). Maybe I can use this idea here.So, if I consider ( x, y, z ) as roots of some cubic polynomial, then the polynomial would be:[t^3 - (x + y + z)t^2 + (xy + yz + zx)t - xyz = 0]From the first equation, I know that ( x + y + z = a ). So, substituting that in, the polynomial becomes:[t^3 - a t^2 + (xy + yz + zx)t - xyz = 0]Now, let's look at the second equation: ( frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{1}{a} ). I can rewrite this as:[frac{xy + yz + zx}{xyz} = frac{1}{a}]Let me denote ( S = x + y + z = a ), ( P = xy + yz + zx ), and ( Q = xyz ). Then, the second equation becomes:[frac{P}{Q} = frac{1}{a}]Which implies that:[P = frac{Q}{a}]So, substituting ( P = frac{Q}{a} ) back into the polynomial, we get:[t^3 - a t^2 + frac{Q}{a} t - Q = 0]Hmm, now I have this polynomial in terms of ( Q ). I wonder if I can factor this polynomial or find a root that is equal to ( a ). Let me try plugging ( t = a ) into the polynomial:[a^3 - a cdot a^2 + frac{Q}{a} cdot a - Q = 0]Simplifying each term:- ( a^3 ) stays as it is.- ( -a cdot a^2 = -a^3 )- ( frac{Q}{a} cdot a = Q )- ( -Q ) stays as it is.So, putting it all together:[a^3 - a^3 + Q - Q = 0]Which simplifies to:[0 = 0]Oh! So, ( t = a ) is indeed a root of the polynomial. That means that ( a ) is one of the roots, which are ( x, y, z ). Therefore, at least one of ( x, y, z ) must be equal to ( a ).Wait, let me double-check that. If ( t = a ) is a root, then the polynomial can be factored as ( (t - a)(t^2 + bt + c) ), right? Let me perform the division to see if that's the case.Dividing ( t^3 - a t^2 + frac{Q}{a} t - Q ) by ( t - a ):Using polynomial long division or synthetic division. Let me try synthetic division with root ( a ):Coefficients: 1 (for ( t^3 )), -a (for ( t^2 )), ( frac{Q}{a} ) (for ( t )), -Q (constant term).Bring down the 1.Multiply 1 by ( a ): ( a ). Add to the next coefficient: ( -a + a = 0 ).Multiply 0 by ( a ): 0. Add to the next coefficient: ( frac{Q}{a} + 0 = frac{Q}{a} ).Multiply ( frac{Q}{a} ) by ( a ): ( Q ). Add to the last coefficient: ( -Q + Q = 0 ).So, the division yields ( t^2 + 0t + frac{Q}{a} ), meaning:[t^3 - a t^2 + frac{Q}{a} t - Q = (t - a)(t^2 + frac{Q}{a})]So, the quadratic factor is ( t^2 + frac{Q}{a} ). For real numbers, the roots of the quadratic would be ( t = pm sqrt{-frac{Q}{a}} ). But since we're dealing with real numbers, ( -frac{Q}{a} ) must be non-negative, so ( Q ) must be non-positive if ( a ) is positive, or non-negative if ( a ) is negative. But regardless, the key point is that ( t = a ) is a root, so one of ( x, y, z ) must be ( a ). Therefore, the proof is complete.I think that makes sense. I started by considering the polynomial whose roots are ( x, y, z ), used Vieta's formulas to relate the given equations to the coefficients of the polynomial, substituted the known values, and then tested ( t = a ) as a root, which worked out. So, yeah, at least one of the numbers has to be ( a ).**Final Answer**boxed{a}