💡Okay, so I have this problem where I'm given two logarithms: log base 2 of 9 is equal to a, and log base 3 of 5 is equal to b. I need to express lg 2, which I think is the same as log base 10 of 2, in terms of a and b. Hmm, let me see how I can approach this.First, I remember that logarithms can be converted using the change of base formula. The change of base formula says that log base m of n is equal to log base k of n divided by log base k of m, right? So maybe I can use that to rewrite these logarithms in terms of common logarithms, which are base 10.Given that log₂9 = a, I can rewrite this using the change of base formula as log₁₀9 divided by log₁₀2 equals a. Similarly, log₃5 = b can be rewritten as log₁₀5 divided by log₁₀3 equals b.Let me write that down:log₂9 = a ⇒ (log 9) / (log 2) = a log₃5 = b ⇒ (log 5) / (log 3) = bOkay, so now I have two equations:1. (log 9) / (log 2) = a 2. (log 5) / (log 3) = bI need to find log 2 in terms of a and b. Hmm, maybe I can express log 9 and log 5 in terms of other logarithms that I can relate to log 2 and log 3.I know that log 9 is the same as log 3 squared, so log 9 = 2 log 3. Similarly, log 5 is just log 5, which I don't know yet, but maybe I can relate it to log 3 through the second equation.Let me substitute log 9 in the first equation:2 log 3 / log 2 = a ⇒ (2 log 3) / (log 2) = aSo, from this, I can solve for log 3 in terms of log 2:log 3 = (a / 2) log 2Okay, so log 3 is (a/2) times log 2. That's useful.Now, looking at the second equation:(log 5) / (log 3) = b ⇒ log 5 = b log 3But I already have log 3 in terms of log 2, so I can substitute that in:log 5 = b * (a / 2) log 2 ⇒ log 5 = (ab / 2) log 2So, log 5 is (ab/2) times log 2.Now, I remember that the sum of log 2 and log 5 is log 10, because log 2 + log 5 = log (2*5) = log 10. And log 10 is 1 because it's base 10.So, log 2 + log 5 = 1But I have expressions for both log 2 and log 5 in terms of log 2. Let me substitute them:log 2 + (ab / 2) log 2 = 1Factor out log 2:log 2 (1 + ab / 2) = 1So, log 2 = 1 / (1 + ab / 2)Hmm, let me simplify the denominator:1 + ab / 2 = (2 + ab) / 2So, log 2 = 1 / ((2 + ab)/2) = 2 / (2 + ab)Therefore, log 2 is equal to 2 divided by (2 + ab).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Started with log₂9 = a ⇒ (log 9)/(log 2) = a ⇒ 2 log 3 / log 2 = a ⇒ log 3 = (a/2) log 2. That seems right.2. Then log₃5 = b ⇒ (log 5)/(log 3) = b ⇒ log 5 = b log 3. Substituted log 3 from above: log 5 = (ab/2) log 2. That also seems correct.3. Then, since log 2 + log 5 = log 10 = 1, substituted the expressions: log 2 + (ab/2) log 2 = 1 ⇒ log 2 (1 + ab/2) = 1 ⇒ log 2 = 1 / (1 + ab/2) = 2 / (2 + ab). Yep, that looks good.So, I think I did everything correctly. So, the answer should be 2 divided by (2 + ab), which is 2/(2 + ab).**Final Answer**The value of lg 2 expressed in terms of a and b is boxed{dfrac{2}{2 + ab}}.

💡Okay, so I'm looking at this problem here about a road sign that says the speed limit is 40 km/h. The question is asking me to figure out which inequality correctly represents the instruction given by the sign. The options are A: ( v < 40 ), B: ( v leq 40 ), C: ( v > 40 ), and D: ( v geq 40 ).Alright, let's break this down. First, I need to understand what the road sign is telling drivers. It says the speed limit is 40 km/h, which means drivers are not allowed to drive faster than that on that road section. So, the speed of their vehicle, which is denoted by ( v ), has to be at or below 40 km/h.Now, let's look at the inequalities provided. Option A is ( v < 40 ), which means the speed has to be strictly less than 40 km/h. But wait, the sign says "at or below," which includes 40 km/h itself. So, if I choose A, that would mean drivers can't even go exactly 40 km/h, which doesn't match the sign's instruction. So, A might not be the correct choice.Option B is ( v leq 40 ). This means the speed can be less than or equal to 40 km/h. That seems to fit perfectly because the sign allows drivers to go up to 40 km/h, including that exact speed. So, B looks like a good candidate for the right answer.Option C is ( v > 40 ), which means the speed has to be greater than 40 km/h. But the sign is about not exceeding 40 km/h, so this seems opposite of what the sign is saying. Drivers shouldn't be going faster than 40 km/h, so C doesn't make sense here.Option D is ( v geq 40 ), meaning the speed has to be greater than or equal to 40 km/h. Again, this is the opposite of the sign's instruction. The sign is telling drivers to slow down to 40 km/h or below, not to maintain or exceed that speed. So, D is also incorrect.To double-check, let's think about what each inequality represents. The sign says "at or below," which includes 40 km/h. So, the correct inequality should include 40 km/h as a possible value for ( v ). That's exactly what ( v leq 40 ) does. It allows ( v ) to be any number less than 40 or exactly 40.If I were to visualize this on a number line, 40 km/h would be a closed circle, indicating that it's included in the solution set. All numbers to the left of 40 (which are less than 40) would also be part of the solution. This matches perfectly with ( v leq 40 ).Another way to think about it is in terms of legal driving. If a sign says the speed limit is 40 km/h, you're allowed to drive at 40 km/h, but not faster. So, your speed can be 40 or anything below that. This again points to ( v leq 40 ) as the correct inequality.I also want to make sure I'm not confusing the inequality symbols. The less than symbol (<) doesn't include the number it's comparing to, while the less than or equal to symbol (≤) does include it. Since the sign includes 40 km/h as a permissible speed, I need an inequality that includes 40, which is why ( v leq 40 ) is the right choice.Just to cover all bases, let's consider what would happen if the sign had said something different, like "minimum speed" instead of "maximum speed." In that case, the inequality would have been different. For example, if it said "minimum speed 40 km/h," then the inequality would have been ( v geq 40 ), which is option D. But since it's a maximum speed limit, we're dealing with an upper bound, not a lower bound.Also, thinking about real-world scenarios, if a driver is going exactly 40 km/h, they're following the law. If they go any faster, they're breaking the law. So, the inequality needs to reflect that 40 is the highest allowable speed, which is exactly what ( v leq 40 ) does.In summary, after considering what each inequality means, visualizing it on a number line, and relating it to real-world driving rules, I'm confident that option B, ( v leq 40 ), is the correct answer.

💡First, I need to understand the original ratio of flour to water to sugar, which is 8:4:3. This means for every 8 parts of flour, there are 4 parts of water and 3 parts of sugar.Next, the new recipe calls for doubling the ratio of flour to water. The original ratio of flour to water is 8:4, which simplifies to 2:1. Doubling this ratio gives us 4:1. Since the new recipe requires 2 cups of water, the amount of flour needed would be 4 times the amount of water, resulting in 8 cups of flour.The new recipe also requires 6 cups of sugar. To find the new ratio of flour to sugar, I divide the amount of flour by the amount of sugar: 8 cups of flour divided by 6 cups of sugar equals 4/3 or approximately 1.33.The original ratio of flour to sugar was 8:3, which simplifies to 8/3 or approximately 2.67. To determine the change in the ratio of flour to sugar, I subtract the new ratio from the original ratio: 2.67 minus 1.33 equals approximately 1.34.Therefore, the change in the ratio of flour to sugar from the original recipe is approximately 1.34.

💡Okay, so I have this inequality to prove: for any positive real numbers (a_i > 0) and any positive integers (M, P > 0), the following holds:[M cdotleft(sum_{i=1}^{M} a_{i}^{P}right) leqslantleft(sum_{i=1}^{M} a_{i}^{P+1}right)left(sum_{i=1}^{M} a_{i}^{-1}right)]Hmm, let me try to understand what this inequality is saying. It involves sums of powers of (a_i), specifically the sum of (a_i^P), the sum of (a_i^{P+1}), and the sum of (a_i^{-1}). The left side is (M) times the sum of (a_i^P), and the right side is the product of the sum of (a_i^{P+1}) and the sum of (a_i^{-1}). I need to show that this inequality is always true for positive (a_i), (M), and (P). Maybe I can start by looking at some specific cases to get an intuition.Let's take (M = 1). Then the inequality becomes:[1 cdot a_1^P leqslant a_1^{P+1} cdot a_1^{-1}]Simplifying the right side:[a_1^{P+1} cdot a_1^{-1} = a_1^{P}]So, the inequality becomes:[a_1^P leqslant a_1^P]Which is just an equality. So, for (M = 1), the inequality holds as an equality.What about (M = 2) and (P = 1)? Let's see:Left side:[2 cdot (a_1 + a_2)]Right side:[(a_1^2 + a_2^2) cdot left(frac{1}{a_1} + frac{1}{a_2}right)]Let me compute the right side:[(a_1^2 + a_2^2) cdot left(frac{1}{a_1} + frac{1}{a_2}right) = (a_1^2 + a_2^2) cdot left(frac{a_1 + a_2}{a_1 a_2}right) = frac{(a_1^2 + a_2^2)(a_1 + a_2)}{a_1 a_2}]So, the inequality becomes:[2(a_1 + a_2) leqslant frac{(a_1^2 + a_2^2)(a_1 + a_2)}{a_1 a_2}]Assuming (a_1 + a_2 neq 0), we can divide both sides by (a_1 + a_2):[2 leqslant frac{a_1^2 + a_2^2}{a_1 a_2}]Which simplifies to:[2 a_1 a_2 leqslant a_1^2 + a_2^2]This is a known inequality, the arithmetic mean-geometric mean inequality (AM-GM). Specifically, for two positive numbers, the arithmetic mean is greater than or equal to the geometric mean:[frac{a_1^2 + a_2^2}{2} geqslant a_1 a_2]Multiplying both sides by 2 gives:[a_1^2 + a_2^2 geqslant 2 a_1 a_2]Which is exactly what we have. So, for (M = 2) and (P = 1), the inequality holds.Maybe I can try another case, say (M = 3) and (P = 2). Let's see:Left side:[3 cdot (a_1^2 + a_2^2 + a_3^2)]Right side:[(a_1^3 + a_2^3 + a_3^3) cdot left(frac{1}{a_1} + frac{1}{a_2} + frac{1}{a_3}right)]Expanding the right side:[(a_1^3 + a_2^3 + a_3^3) cdot left(frac{1}{a_1} + frac{1}{a_2} + frac{1}{a_3}right) = a_1^3 cdot frac{1}{a_1} + a_1^3 cdot frac{1}{a_2} + a_1^3 cdot frac{1}{a_3} + a_2^3 cdot frac{1}{a_1} + a_2^3 cdot frac{1}{a_2} + a_2^3 cdot frac{1}{a_3} + a_3^3 cdot frac{1}{a_1} + a_3^3 cdot frac{1}{a_2} + a_3^3 cdot frac{1}{a_3}]Simplifying each term:[a_1^2 + frac{a_1^3}{a_2} + frac{a_1^3}{a_3} + frac{a_2^3}{a_1} + a_2^2 + frac{a_2^3}{a_3} + frac{a_3^3}{a_1} + frac{a_3^3}{a_2} + a_3^2]So, the right side becomes:[a_1^2 + a_2^2 + a_3^2 + frac{a_1^3}{a_2} + frac{a_1^3}{a_3} + frac{a_2^3}{a_1} + frac{a_2^3}{a_3} + frac{a_3^3}{a_1} + frac{a_3^3}{a_2}]Therefore, the inequality is:[3(a_1^2 + a_2^2 + a_3^2) leqslant a_1^2 + a_2^2 + a_3^2 + frac{a_1^3}{a_2} + frac{a_1^3}{a_3} + frac{a_2^3}{a_1} + frac{a_2^3}{a_3} + frac{a_3^3}{a_1} + frac{a_3^3}{a_2}]Subtracting (a_1^2 + a_2^2 + a_3^2) from both sides:[2(a_1^2 + a_2^2 + a_3^2) leqslant frac{a_1^3}{a_2} + frac{a_1^3}{a_3} + frac{a_2^3}{a_1} + frac{a_2^3}{a_3} + frac{a_3^3}{a_1} + frac{a_3^3}{a_2}]Hmm, this seems more complicated. Maybe I can use the AM-GM inequality again. For each pair of terms like (frac{a_i^3}{a_j}) and (frac{a_j^3}{a_i}), perhaps?Let's consider two terms: (frac{a_i^3}{a_j}) and (frac{a_j^3}{a_i}). Applying AM-GM to these two terms:[frac{frac{a_i^3}{a_j} + frac{a_j^3}{a_i}}{2} geqslant sqrt{frac{a_i^3}{a_j} cdot frac{a_j^3}{a_i}} = sqrt{frac{a_i^3 a_j^3}{a_j a_i}} = sqrt{a_i^2 a_j^2} = a_i a_j]Multiplying both sides by 2:[frac{a_i^3}{a_j} + frac{a_j^3}{a_i} geqslant 2 a_i a_j]So, for each pair ((i, j)), we have (frac{a_i^3}{a_j} + frac{a_j^3}{a_i} geqslant 2 a_i a_j). In our case, we have three such pairs: ((1,2)), ((1,3)), and ((2,3)). So, summing over all pairs:[left(frac{a_1^3}{a_2} + frac{a_2^3}{a_1}right) + left(frac{a_1^3}{a_3} + frac{a_3^3}{a_1}right) + left(frac{a_2^3}{a_3} + frac{a_3^3}{a_2}right) geqslant 2(a_1 a_2 + a_1 a_3 + a_2 a_3)]So, the right side of our inequality is at least (2(a_1 a_2 + a_1 a_3 + a_2 a_3)). Therefore, we have:[2(a_1^2 + a_2^2 + a_3^2) leqslant 2(a_1 a_2 + a_1 a_3 + a_2 a_3)]Dividing both sides by 2:[a_1^2 + a_2^2 + a_3^2 leqslant a_1 a_2 + a_1 a_3 + a_2 a_3]Wait, but this is not true in general. For example, take (a_1 = a_2 = a_3 = 1). Then both sides are equal to 3. But if (a_1 = 2), (a_2 = 1), (a_3 = 1), then the left side is (4 + 1 + 1 = 6), and the right side is (2*1 + 2*1 + 1*1 = 2 + 2 + 1 = 5). So, (6 leqslant 5) is false. Hmm, that suggests that my approach might be flawed. Maybe I need a different strategy. Let me think again. The original inequality is:[M cdotleft(sum_{i=1}^{M} a_{i}^{P}right) leqslantleft(sum_{i=1}^{M} a_{i}^{P+1}right)left(sum_{i=1}^{M} a_{i}^{-1}right)]Perhaps I can use the Cauchy-Schwarz inequality or Holder's inequality. Holder's inequality might be more appropriate here because it deals with products of sums of different powers.Holder's inequality states that for positive real numbers (a_i), (b_i), and exponents (p) and (q) such that (frac{1}{p} + frac{1}{q} = 1), we have:[sum_{i=1}^{M} a_i b_i leqslant left(sum_{i=1}^{M} a_i^pright)^{1/p} left(sum_{i=1}^{M} b_i^qright)^{1/q}]But in our case, the right side of the inequality is a product of two sums, not a sum of products. Maybe I need to manipulate the inequality to fit into Holder's framework.Alternatively, maybe I can use the AM-GM inequality in a different way. Let's consider the terms in the sums. Let me try to write the right side as:[left(sum_{i=1}^{M} a_i^{P+1}right) left(sum_{j=1}^{M} a_j^{-1}right) = sum_{i=1}^{M} sum_{j=1}^{M} a_i^{P+1} a_j^{-1}]Which is:[sum_{i=1}^{M} sum_{j=1}^{M} frac{a_i^{P+1}}{a_j}]So, the right side is a double sum. The left side is (M) times the sum of (a_i^P). Maybe I can compare each term in the double sum to something related to (a_i^P).Let me fix (i) and (j). For each pair ((i, j)), we have the term (frac{a_i^{P+1}}{a_j}). If I can relate this to (a_i^P) and (a_j^{-1}), perhaps I can find a way to bound the double sum.Wait, maybe I can use the AM-GM inequality on each term (frac{a_i^{P+1}}{a_j}). Let's see:Consider the term (frac{a_i^{P+1}}{a_j}). Let me write this as (a_i^{P} cdot frac{a_i}{a_j}). Hmm, not sure if that helps.Alternatively, think of (frac{a_i^{P+1}}{a_j}) as (a_i^{P} cdot a_i cdot frac{1}{a_j}). Maybe I can apply AM-GM to these factors.But I have three factors here: (a_i^P), (a_i), and (frac{1}{a_j}). Maybe not straightforward.Alternatively, perhaps I can use the Cauchy-Schwarz inequality on the sums. Let me recall that:[left(sum_{i=1}^{M} x_i y_iright)^2 leqslant left(sum_{i=1}^{M} x_i^2right) left(sum_{i=1}^{M} y_i^2right)]But again, not sure how to apply this directly.Wait, maybe I can think of the right side as the product of two sums, which is similar to the Cauchy-Schwarz setup. Let me try to set (x_i = a_i^{(P+1)/2}) and (y_i = a_i^{-1/2}). Then:[sum_{i=1}^{M} x_i y_i = sum_{i=1}^{M} a_i^{(P+1)/2} cdot a_i^{-1/2} = sum_{i=1}^{M} a_i^{P/2}]And by Cauchy-Schwarz:[left(sum_{i=1}^{M} a_i^{P/2}right)^2 leqslant left(sum_{i=1}^{M} a_i^{P+1}right) left(sum_{i=1}^{M} a_i^{-1}right)]So, we have:[left(sum_{i=1}^{M} a_i^{P/2}right)^2 leqslant left(sum_{i=1}^{M} a_i^{P+1}right) left(sum_{i=1}^{M} a_i^{-1}right)]But the left side of our original inequality is (M cdot sum a_i^P). How does this relate to (left(sum a_i^{P/2}right)^2)?Well, by the Cauchy-Schwarz inequality, we also have:[left(sum_{i=1}^{M} a_i^{P/2}right)^2 geqslant M cdot sum_{i=1}^{M} a_i^P]Wait, is that true? Let me recall that for any positive numbers, the square of the sum is greater than or equal to (M) times the sum of squares. Specifically:[left(sum_{i=1}^{M} x_iright)^2 geqslant M cdot sum_{i=1}^{M} x_i^2]This is known as the Cauchy-Schwarz inequality in the form of the Titu's lemma or the Engel form. So, if I set (x_i = a_i^{P/2}), then:[left(sum_{i=1}^{M} a_i^{P/2}right)^2 geqslant M cdot sum_{i=1}^{M} a_i^P]Therefore, combining this with the earlier inequality from Cauchy-Schwarz:[M cdot sum_{i=1}^{M} a_i^P leqslant left(sum_{i=1}^{M} a_i^{P/2}right)^2 leqslant left(sum_{i=1}^{M} a_i^{P+1}right) left(sum_{i=1}^{M} a_i^{-1}right)]Putting it all together:[M cdot sum_{i=1}^{M} a_i^P leqslant left(sum_{i=1}^{M} a_i^{P+1}right) left(sum_{i=1}^{M} a_i^{-1}right)]Which is exactly the inequality we wanted to prove. So, by applying the Cauchy-Schwarz inequality twice, once to relate the sum of (a_i^{P/2}) to the sum of (a_i^P), and once more to relate the sum of (a_i^{P/2}) to the product of the sums of (a_i^{P+1}) and (a_i^{-1}), we have established the desired inequality.I think this approach works because it uses fundamental inequalities (Cauchy-Schwarz) to bridge the gap between the left and right sides of the original inequality. It's interesting how breaking down the problem into smaller parts and applying known inequalities step by step can lead to the solution.Just to recap:1. Recognize that the right side of the inequality is a product of two sums, which suggests using Cauchy-Schwarz or Holder's inequality.2. Apply Cauchy-Schwarz to the product of sums by choosing appropriate terms.3. Use another application of Cauchy-Schwarz (or Titu's lemma) to relate the resulting expression to the left side of the original inequality.4. Combine the results to show the desired inequality holds.This method seems solid and generalizes well for any positive (a_i), (M), and (P). I don't see any flaws in this reasoning, and the specific cases I tested earlier support this conclusion.

💡To find ( f^{-1}(10) ) for the function ( f(x) = 1 + 3^{-x} ), I start by setting ( f(x) = 10 ). This gives the equation:[1 + 3^{-x} = 10]Subtracting 1 from both sides:[3^{-x} = 9]Recognizing that 9 is ( 3^2 ), I rewrite the equation as:[3^{-x} = 3^2]Since the bases are the same, the exponents must be equal:[-x = 2]Solving for ( x ):[x = -2]Therefore, ( f^{-1}(10) = -2 ), which corresponds to option B.

💡Okay, so I have this arithmetic sequence problem here. It says that in an arithmetic sequence {a_n}, a_{2015} equals a_{2013} plus 6. I need to find the common difference d. Hmm, arithmetic sequences, right? They have a common difference between consecutive terms. So, each term is the previous term plus d.Let me recall the formula for the nth term of an arithmetic sequence. It's a_n = a_1 + (n - 1)d, where a_1 is the first term and d is the common difference. So, if I want to find a_{2015}, it would be a_1 + (2015 - 1)d, which simplifies to a_1 + 2014d. Similarly, a_{2013} would be a_1 + (2013 - 1)d, which is a_1 + 2012d.Now, according to the problem, a_{2015} = a_{2013} + 6. So, substituting the expressions I just wrote, that would be:a_1 + 2014d = (a_1 + 2012d) + 6.Hmm, okay. Let me write that out:a_1 + 2014d = a_1 + 2012d + 6.Now, I can subtract a_1 from both sides to simplify:2014d = 2012d + 6.Subtracting 2012d from both sides gives:2d = 6.Oh, so then d must be 6 divided by 2, which is 3. So, the common difference d is 3.Wait, let me double-check. If d is 3, then each term increases by 3. So, from a_{2013} to a_{2014}, it's +3, and from a_{2014} to a_{2015}, it's another +3. So, from a_{2013} to a_{2015}, it's two steps, which would be 3 + 3 = 6. That matches the given condition that a_{2015} is 6 more than a_{2013}. So, yeah, that makes sense.I think that's correct. So, the common difference d is 3.