💡Okay, so I have this logarithmic equation to solve: log base 8 of (x + 8) equals 3/2. Hmm, logarithms can be tricky, but I think I remember that they can be converted into exponential form. Let me try that.First, I know that log_b(a) = c means that b^c = a. So, applying that to my equation, log_8(x + 8) = 3/2 should mean that 8 raised to the power of 3/2 equals x + 8. So, I can write that as 8^(3/2) = x + 8.Now, I need to compute 8^(3/2). I think exponents with fractions can be broken down into roots and powers. The denominator of the exponent is the root, and the numerator is the power. So, 8^(3/2) is the same as the square root of 8 raised to the 3rd power, or (sqrt(8))^3.Wait, is that right? Let me think again. Actually, it's the same as 8^(1/2) raised to the 3rd power, which is (8^(1/2))^3. So, first, I find the square root of 8, and then cube that result.The square root of 8 is... well, 8 is 2^3, so sqrt(8) is sqrt(2^3). That can be written as 2^(3/2), which is 2^1 * 2^(1/2) = 2 * sqrt(2). So, sqrt(8) is 2*sqrt(2).Now, I need to cube that. So, (2*sqrt(2))^3. Let's compute that. First, cube the 2: 2^3 is 8. Then, cube the sqrt(2): (sqrt(2))^3 is (2^(1/2))^3 = 2^(3/2) = 2*sqrt(2). So, multiplying those together: 8 * 2*sqrt(2) = 16*sqrt(2).Wait, that seems a bit complicated. Maybe there's a simpler way. Let me try another approach. 8 is 2^3, so 8^(3/2) is (2^3)^(3/2). When you raise a power to a power, you multiply the exponents. So, 2^(3*(3/2)) = 2^(9/2). 9/2 is 4.5, which is 4 + 0.5. So, 2^4 is 16, and 2^0.5 is sqrt(2). So, 2^4.5 is 16*sqrt(2). Okay, that matches my earlier result.So, 8^(3/2) is 16*sqrt(2). Therefore, my equation now is 16*sqrt(2) = x + 8. To solve for x, I need to subtract 8 from both sides. So, x = 16*sqrt(2) - 8.Hmm, can I factor anything out to simplify this expression? Let's see, both terms have a factor of 8. So, factoring out 8, I get x = 8*(2*sqrt(2) - 1). That looks a bit cleaner.Let me double-check my steps to make sure I didn't make any mistakes. Starting with log_8(x + 8) = 3/2, I converted it to exponential form correctly: 8^(3/2) = x + 8. Then, I computed 8^(3/2) by expressing 8 as 2^3 and applying exponent rules, which gave me 16*sqrt(2). Subtracting 8 from both sides gave me x = 16*sqrt(2) - 8, and factoring out 8 resulted in x = 8*(2*sqrt(2) - 1). I think that's correct. To be thorough, maybe I should plug the value back into the original equation to verify. Let's compute log_8(x + 8) with x = 8*(2*sqrt(2) - 1). So, x + 8 would be 8*(2*sqrt(2) - 1) + 8 = 16*sqrt(2) - 8 + 8 = 16*sqrt(2). Then, log_8(16*sqrt(2)) should equal 3/2.Let me compute log_8(16*sqrt(2)). Since 16 is 2^4 and sqrt(2) is 2^(1/2), multiplying them gives 2^(4 + 1/2) = 2^(9/2). So, 16*sqrt(2) is 2^(9/2). Now, 8 is 2^3, so log_8(2^(9/2)) can be rewritten using the change of base formula or by expressing it in terms of base 2. Using the formula log_b(a) = log_c(a)/log_c(b), so log_8(2^(9/2)) = log_2(2^(9/2)) / log_2(8). log_2(2^(9/2)) is just 9/2, and log_2(8) is log_2(2^3) which is 3. So, the result is (9/2)/3 = (9/2)*(1/3) = 3/2. That's exactly what we started with, so it checks out.Therefore, my solution x = 8*(2*sqrt(2) - 1) is correct.

💡Okay, so I have this problem here about circles where each circle has a radius that's half the radius of the one before it. The first circle has a radius of 2 inches. I need to find the total area of all these circles. Hmm, let me think about how to approach this.First, I know that the area of a circle is calculated using the formula A = πr², where r is the radius. Since each subsequent circle has half the radius of the previous one, the radii form a geometric sequence. The first term of this sequence is 2 inches, and the common ratio is 1/2 because each radius is half of the previous one.So, the radii of the circles would be: 2, 1, 0.5, 0.25, and so on. Each time, the radius is getting smaller by half. Now, since I need the total area, I should calculate the area of each circle and then add them all up.Let me write down the areas:1. First circle: radius = 2 inches. Area = π*(2)² = 4π.2. Second circle: radius = 1 inch. Area = π*(1)² = π.3. Third circle: radius = 0.5 inches. Area = π*(0.5)² = π*(0.25) = 0.25π.4. Fourth circle: radius = 0.25 inches. Area = π*(0.25)² = π*(0.0625) = 0.0625π.5. And so on...So, the areas are: 4π, π, 0.25π, 0.0625π, etc. This looks like another geometric sequence, but now for the areas. Let me check:- The first term (a) is 4π.- The common ratio (r) is the factor by which we multiply each term to get the next one. So, π / 4π = 1/4. Similarly, 0.25π / π = 1/4. So, the common ratio is 1/4.Great, so the areas form an infinite geometric series with a = 4π and r = 1/4. I remember that the sum of an infinite geometric series is given by S = a / (1 - r), provided that |r| < 1. In this case, r = 1/4, which is less than 1, so the formula applies.Let me plug in the values:S = 4π / (1 - 1/4) = 4π / (3/4) = 4π * (4/3) = (16/3)π.So, the total area of all the circles is (16/3)π. Let me double-check my calculations to make sure I didn't make a mistake.Starting with the radii: 2, 1, 0.5, 0.25... Correct. Areas: 4π, π, 0.25π, 0.0625π... Yes, that's right. Common ratio for areas is 1/4, which is correct because each area is (1/2)^2 = 1/4 of the previous area. Sum formula: S = a / (1 - r) = 4π / (1 - 1/4) = 4π / (3/4) = 16π/3. Yep, that seems right.I think I've got it. The total area is 16π/3.

💡First, identify the probabilities of each event:- Probability of A passing the interview, P(A) = 3/4- Probability of B passing the interview, P(B) = 2/3Since the events are independent, calculate the probability of both A and B not passing:- Probability of A not passing, P(A') = 1 - 3/4 = 1/4- Probability of B not passing, P(B') = 1 - 2/3 = 1/3Multiply these probabilities to find the probability of both not passing:P(A' and B') = P(A') × P(B') = 1/4 × 1/3 = 1/12Finally, subtract this probability from 1 to find the probability of at least one passing:P(At least one passes) = 1 - P(A' and B') = 1 - 1/12 = 11/12

💡Okay, so I need to figure out how much Herman is going to spend on breakfast during this 16-week project. Let's see, he stops at the drive-through 5 days every week to buy breakfast combos for himself and 3 team members. Each meal costs 4.00. Hmm, okay, so first, I need to understand how many meals he's buying each week.He buys breakfast for himself and 3 team members, so that's 1 (himself) + 3 (team members) = 4 meals per day. And he does this 5 days a week. So, 4 meals/day * 5 days/week = 20 meals per week.Each meal costs 4.00, so the cost per week would be 20 meals * 4.00/meal = 80.00 per week.Now, the project lasts 16 weeks, so to find the total cost, I need to multiply the weekly cost by the number of weeks. That would be 80.00/week * 16 weeks = 1,280.00.Wait, let me double-check that. 4 meals a day for 5 days is indeed 20 meals. At 4 each, that's 80 a week. Over 16 weeks, that's 16 times 80, which is 1,280. Yeah, that seems right.I don't think I missed anything here. It's straightforward multiplication based on the given numbers. So, Herman will spend a total of 1,280.00 on breakfast during the 16-week project.

💡Okay, so I have this problem about calculating labor costs for a workplace. There are men, women, and boys working on a project, and I need to consider their efficiencies, overtime hours, and holiday pay rates. Let me try to break this down step by step.First, I need to understand the number of each group and their base daily wages. There are 5 men, 8 women, and 10 boys. The base daily wage for a man is 100, for a woman it's 80, and for a boy it's 50. So, without any overtime or holiday pay, the total daily wage would be:Men: 5 * 100 = 500Women: 8 * 80 = 640Boys: 10 * 50 = 500Adding these up: 500 + 640 + 500 = 1,640But wait, the problem mentions that the men are twice as efficient as the women and three times as efficient as the boys. Does this affect the wages? I think efficiency might relate to how much work they can do, but the wages are given as base daily wages regardless of efficiency. So maybe efficiency doesn't directly affect the wage calculation here. I'll keep that in mind.Next, there's overtime pay. Overtime is paid at time and a half for any hours worked over 8 hours in a day. On this particular day, 3 out of the 5 men worked 10 hours each. So, each of these 3 men worked 2 hours of overtime. Calculating overtime pay for these men:Overtime pay rate is 1.5 times the base rate. So, for a man, it's 1.5 * 100 = 150 per hour.Each of these 3 men worked 2 hours overtime, so per man: 2 * 150 = 300Total overtime pay for 3 men: 3 * 300 = 900Now, the team worked on a holiday, and holiday pay is double the base rate. So, for everyone, their base pay is doubled.Let's calculate the holiday pay for each group:Men: 5 * 100 * 2 = 1,000Women: 8 * 80 * 2 = 1,280Boys: 10 * 50 * 2 = 1,000Adding these up: 1,000 + 1,280 + 1,000 = 3,280But wait, we also have the overtime pay for the 3 men. So, we need to add that 900 to the total.Total labor costs: 3,280 + 900 = 4,180Hmm, that seems straightforward, but let me double-check if I missed anything.- Base wages: 1,640- Holiday pay doubles everything: 1,640 * 2 = 3,280- Overtime for 3 men: 3 * 2 hours * 1.5 * 100 = 900- Total: 3,280 + 900 = 4,180Yes, that matches what I calculated earlier. So, the total labor cost for the day is 4,180.

💡First, I need to determine the total amount Agatha has already spent and the amount she plans to spend on the listed components. She has spent 85 on the frame and intends to purchase a front wheel for 35, a rear wheel for 40, a seat for 25, handlebar tape for 15, a water bottle cage for 10, and a bike lock for 20. Additionally, she wants to keep 10 aside for future expenses.Next, I'll calculate the sum of all these expenses:- Frame: 85- Front wheel: 35- Rear wheel: 40- Seat: 25- Handlebar tape: 15- Water bottle cage: 10- Bike lock: 20- Future expenses: 10Adding these amounts together:85 + 35 + 40 + 25 + 15 + 10 + 20 + 10 = 240Agatha's total budget is 250. To find out how much money she has left for additional accessories, I'll subtract the total expenses from her budget:250 - 240 = 10Therefore, Agatha has 10 left for additional accessories.