answer：Okay, so I have this function f(x) = log base 2 of (3 + x) minus log base 2 of (3 - x). I need to find the domain of this function and determine if it's odd or even. Then, I also have to find the value of alpha such that f(alpha) equals 1. Hmm, let me start with the first part.First, the domain. Since we're dealing with logarithms, the arguments inside the logs must be positive. So, for log base 2 of (3 + x) to be defined, 3 + x has to be greater than 0. That means x > -3. Similarly, for log base 2 of (3 - x) to be defined, 3 - x has to be greater than 0, which means x < 3. So, combining these two inequalities, x has to be greater than -3 and less than 3. Therefore, the domain of f(x) is all real numbers between -3 and 3, not including -3 and 3 themselves. So, in interval notation, that's (-3, 3).Now, determining if the function is odd or even. I remember that a function is even if f(-x) = f(x) for all x in the domain, and it's odd if f(-x) = -f(x) for all x in the domain. Let me compute f(-x) to see which one it is.So, f(-x) would be log base 2 of (3 + (-x)) minus log base 2 of (3 - (-x)). Simplifying that, it becomes log base 2 of (3 - x) minus log base 2 of (3 + x). Hmm, that looks familiar. If I factor out a negative sign from the second term, it becomes -[log base 2 of (3 + x) - log base 2 of (3 - x)], which is exactly -f(x). So, f(-x) = -f(x), which means the function is odd.Alright, that was part one. Now, moving on to part two: finding alpha such that f(alpha) = 1. Let me write that equation out:log base 2 of (3 + alpha) minus log base 2 of (3 - alpha) = 1.I can combine these two logarithms into a single logarithm using the logarithm subtraction rule, which states that log_b(A) - log_b(B) = log_b(A/B). So, applying that here, I get:log base 2 of [(3 + alpha)/(3 - alpha)] = 1.Now, to solve for alpha, I can rewrite this equation in exponential form. Remember that log_b(C) = D is equivalent to b^D = C. So, applying that here, we have:2^1 = (3 + alpha)/(3 - alpha).Simplifying that, 2 = (3 + alpha)/(3 - alpha). Now, I can solve this equation for alpha. Let me cross-multiply to get rid of the fraction:2*(3 - alpha) = 3 + alpha.Expanding the left side, I get:6 - 2*alpha = 3 + alpha.Now, let's get all the terms involving alpha on one side and the constants on the other. I'll add 2*alpha to both sides:6 = 3 + 3*alpha.Then, subtract 3 from both sides:3 = 3*alpha.Dividing both sides by 3:1 = alpha.So, alpha is 1. Let me check if this makes sense. Plugging alpha = 1 back into the original function:f(1) = log base 2 of (3 + 1) - log base 2 of (3 - 1) = log base 2 of 4 - log base 2 of 2.log base 2 of 4 is 2, and log base 2 of 2 is 1. So, 2 - 1 = 1, which matches the given condition f(alpha) = 1. Therefore, alpha = 1 is correct.Wait, just to make sure I didn't make any mistakes in my steps. Let me go through the algebra again:Starting from 2 = (3 + alpha)/(3 - alpha).Multiply both sides by (3 - alpha):2*(3 - alpha) = 3 + alpha.Which is 6 - 2*alpha = 3 + alpha.Adding 2*alpha to both sides:6 = 3 + 3*alpha.Subtracting 3:3 = 3*alpha.Divide by 3:1 = alpha.Yep, that seems consistent. So, I think I did that correctly.Just to recap, the steps were:1. Combine the logs into a single log expression.2. Convert the logarithmic equation to its exponential form.3. Solve the resulting linear equation for alpha.4. Verify the solution by plugging it back into the original function.Everything checks out, so I'm confident that alpha is indeed 1.**Final Answer**The value of α is boxed{1}.

answer：To find the value of ( n ) such that ( 10^n = 10^{-8} times sqrt{frac{10^{50}}{0.0001}} ), I start by expressing ( 0.0001 ) as a power of 10. Since ( 0.0001 = 10^{-4} ), the fraction inside the square root becomes ( frac{10^{50}}{10^{-4}} ).Next, I simplify this fraction by subtracting the exponents: ( 10^{50} div 10^{-4} = 10^{54} ).Taking the square root of ( 10^{54} ) involves halving the exponent, resulting in ( 10^{27} ).Now, the equation is ( 10^n = 10^{-8} times 10^{27} ). Combining the exponents on the right-hand side gives ( 10^{19} ).Since the bases are the same, the exponents must be equal, so ( n = 19 ).

answer：Okay, so I have this problem about a geometric sequence. Let me try to figure it out step by step. First, the problem says that the geometric sequence {a_n} satisfies a_1 * a_6 = 32 * a_2 * a_{10}, and the sum of the first 3 terms, S_3, is 21/4. I need to find the general formula for the sequence and then find the sum of the first n terms of another sequence {b_n}, where b_n is defined as log base 2 of (a_n / 3).Alright, starting with part (1). So, it's a geometric sequence, which means each term is the previous term multiplied by a common ratio, q. So, the nth term is a_n = a_1 * q^{n-1}.Given that a_1 * a_6 = 32 * a_2 * a_{10}. Let me write down what each of these terms is in terms of a_1 and q.a_1 is just a_1.a_6 is a_1 * q^{5}.Similarly, a_2 is a_1 * q^{1}, and a_{10} is a_1 * q^{9}.So, substituting these into the equation:a_1 * (a_1 * q^5) = 32 * (a_1 * q) * (a_1 * q^9)Let me compute both sides.Left side: a_1 * a_1 * q^5 = a_1^2 * q^5Right side: 32 * (a_1 * q) * (a_1 * q^9) = 32 * a_1^2 * q^{10}So, setting left side equal to right side:a_1^2 * q^5 = 32 * a_1^2 * q^{10}Hmm, I can divide both sides by a_1^2, assuming a_1 is not zero, which it can't be in a geometric sequence because all terms would be zero otherwise.So, dividing both sides by a_1^2:q^5 = 32 * q^{10}Now, I can write this as:q^5 = 32 * q^{10}Let me divide both sides by q^5 (assuming q ≠ 0, which it can't be in a geometric sequence because otherwise all terms after the first would be zero):1 = 32 * q^{5}So, 32 * q^5 = 1Therefore, q^5 = 1/32Hmm, 32 is 2^5, so 1/32 is 2^{-5}Therefore, q^5 = 2^{-5}So, taking the fifth root of both sides, q = 2^{-1} = 1/2So, the common ratio q is 1/2.Alright, so now we know q = 1/2.Now, the sum of the first 3 terms, S_3, is given as 21/4.In a geometric sequence, the sum of the first n terms is S_n = a_1 * (1 - q^n) / (1 - q), when q ≠ 1.So, S_3 = a_1 * (1 - q^3) / (1 - q) = 21/4We already know q = 1/2, so let's plug that in.First, compute 1 - q^3:1 - (1/2)^3 = 1 - 1/8 = 7/8Then, 1 - q = 1 - 1/2 = 1/2So, S_3 = a_1 * (7/8) / (1/2) = a_1 * (7/8) * (2/1) = a_1 * (7/4)So, S_3 = (7/4) * a_1 = 21/4Therefore, (7/4) * a_1 = 21/4Multiplying both sides by 4:7 * a_1 = 21Divide both sides by 7:a_1 = 3So, the first term is 3, and the common ratio is 1/2.Therefore, the general formula for the sequence is:a_n = a_1 * q^{n-1} = 3 * (1/2)^{n-1}So, that's part (1) done.Now, moving on to part (2). We have another sequence {b_n} defined by b_n = log_2 (a_n / 3). We need to find the sum of the first n terms, T_n.First, let's write out b_n in terms of a_n.Given that a_n = 3 * (1/2)^{n-1}, so a_n / 3 = (1/2)^{n-1}Therefore, b_n = log_2 ( (1/2)^{n-1} )We know that log_b (x^k) = k * log_b x, so:b_n = (n - 1) * log_2 (1/2)But log_2 (1/2) is log_2 (2^{-1}) = -1So, b_n = (n - 1) * (-1) = - (n - 1) = 1 - nSo, b_n = 1 - nWait, let me check that again.Wait, a_n / 3 is (1/2)^{n - 1}, so log base 2 of that is log_2 ( (1/2)^{n - 1} ) = (n - 1) * log_2 (1/2) = (n - 1) * (-1) = - (n - 1) = 1 - nYes, that's correct.So, b_n = 1 - nSo, the sequence {b_n} is 1 - n for each term.So, let's write out the first few terms:b_1 = 1 - 1 = 0b_2 = 1 - 2 = -1b_3 = 1 - 3 = -2b_4 = 1 - 4 = -3And so on.So, this is an arithmetic sequence where each term decreases by 1.In an arithmetic sequence, the difference between consecutive terms is constant. Let's check:b_2 - b_1 = -1 - 0 = -1b_3 - b_2 = -2 - (-1) = -1b_4 - b_3 = -3 - (-2) = -1Yes, so the common difference is -1.So, the sequence {b_n} is an arithmetic sequence with first term b_1 = 0 and common difference d = -1.Now, we need to find the sum of the first n terms, T_n.The formula for the sum of the first n terms of an arithmetic sequence is:T_n = n/2 * [2 * b_1 + (n - 1) * d]Alternatively, it's also written as T_n = n/2 * (b_1 + b_n)Since we know b_n = 1 - n, and b_1 = 0, let's use the second formula.So, T_n = n/2 * (b_1 + b_n) = n/2 * (0 + (1 - n)) = n/2 * (1 - n) = (n * (1 - n)) / 2 = (n - n^2)/2 = (-n^2 + n)/2Alternatively, we can write it as (-n^2 + n)/2 or factor out a negative sign: - (n^2 - n)/2, but both are correct.So, T_n = (-n^2 + n)/2Let me check this formula with the first few terms.For n = 1: T_1 = (-1 + 1)/2 = 0/2 = 0, which is correct because b_1 = 0.For n = 2: T_2 = (-4 + 2)/2 = (-2)/2 = -1, which is b_1 + b_2 = 0 + (-1) = -1, correct.For n = 3: T_3 = (-9 + 3)/2 = (-6)/2 = -3, which is 0 + (-1) + (-2) = -3, correct.For n = 4: T_4 = (-16 + 4)/2 = (-12)/2 = -6, which is 0 + (-1) + (-2) + (-3) = -6, correct.So, the formula seems to be working.Therefore, the sum of the first n terms of {b_n} is T_n = (-n^2 + n)/2.So, summarizing:(1) The general formula for the geometric sequence {a_n} is a_n = 3 * (1/2)^{n - 1}.(2) The sum of the first n terms of {b_n} is T_n = (-n^2 + n)/2.I think that's it. Let me just double-check my steps to make sure I didn't make any mistakes.Starting with part (1):- I used the given condition a_1 * a_6 = 32 * a_2 * a_{10} and expressed each term in terms of a_1 and q.- Simplified the equation to find q^5 = 1/32, so q = 1/2.- Then, used the sum of the first 3 terms to find a_1 = 3.- Plugged into the general formula for a geometric sequence.Looks good.For part (2):- Expressed b_n in terms of a_n, which gave me b_n = 1 - n.- Recognized it's an arithmetic sequence with first term 0 and common difference -1.- Applied the sum formula for an arithmetic sequence.- Verified with specific values, which worked.So, I think everything checks out.**Final Answer**(1) The general formula for the sequence is boxed{a_n = 3 cdot left( frac{1}{2} right)^{n-1}}.(2) The sum of the first ( n ) terms of ({b_n}) is boxed{frac{-n^2 + n}{2}}.

answer：Okay, so I have this problem here: The product of three consecutive integers, including a negative number, is -336. I need to find their sum. Hmm, let's see how to approach this.First, I know that consecutive integers mean numbers that come one after another without any gaps. So if I let the first integer be 'n', then the next two would be 'n+1' and 'n+2'. But wait, the problem mentions that one of them is negative. So, maybe 'n' is negative, or maybe one of the others is negative. I'll need to consider that.The product of these three numbers is -336. So, mathematically, that would be:n * (n + 1) * (n + 2) = -336Hmm, that's a cubic equation. Solving cubic equations can be tricky, but maybe I can find integer solutions by trial and error since the numbers aren't too big.Let me think about the cube root of 336 to get an approximate idea of where to start. The cube root of 336 is roughly 7, because 7 cubed is 343, which is close to 336. So, maybe the numbers are around 7 or -7.Let me test n = 7:7 * 8 * 9 = 504That's too big and positive, not -336.What about n = -7:-7 * -6 * -5 = -210Still not -336. Hmm.Wait, maybe I need to go a bit higher in the negative numbers. Let's try n = -8:-8 * -7 * -6 = -336Oh, that works! So the three consecutive integers are -8, -7, and -6.Now, to find their sum:-8 + (-7) + (-6) = -21So, the sum of these three consecutive integers is -21.But just to make sure I didn't miss any other possible sets, let me check n = -9:-9 * -8 * -7 = -504That's too low.And n = -6:-6 * -5 * -4 = -120Too high.So, it seems like -8, -7, and -6 are the only consecutive integers that multiply to -336. Therefore, their sum is definitely -21.

answer：First, identify the possible outcomes for both Paco and Manu. Paco spins a number from 1 to 7, and Manu spins a number from 1 to 11. The total number of possible outcomes is (7 times 11 = 77).Next, determine the conditions under which the product of the two numbers is a prime number. A prime number has exactly two distinct positive divisors: 1 and itself. Therefore, the product of Paco's and Manu's numbers can only be prime if one of the numbers is 1 and the other is a prime number.List the prime numbers within the range of each spinner:- Paco's primes: 2, 3, 5, 7- Manu's primes: 2, 3, 5, 7, 11Now, calculate the number of favorable outcomes:1. When Paco spins 1, Manu must spin a prime number: 2, 3, 5, 7, 11. This gives 5 favorable outcomes.2. When Manu spins 1, Paco must spin a prime number: 2, 3, 5, 7. This gives 4 favorable outcomes.However, the case where both Paco and Manu spin 1 results in the product 1, which is not prime. Therefore, we must subtract this overlapping case.Total favorable outcomes = 5 (from Paco spinning 1) + 4 (from Manu spinning 1) - 1 (overlapping case) = 8.Finally, calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:[text{Probability} = frac{8}{77}]

answer：First, I need to determine the time it takes me to paint the room alone, denoted as ( x ) hours.My painting rate is ( frac{1}{x} ) rooms per hour, and my brother's painting rate is ( frac{1}{4} ) rooms per hour. When we work together, our combined painting rate is ( frac{1}{x} + frac{1}{4} ) rooms per hour.We know that together we can paint the room in approximately 1.714 hours, so the combined rate is ( frac{1}{1.714} ) rooms per hour.Setting the combined rate equal to the sum of our individual rates, we get:[frac{1}{x} + frac{1}{4} = frac{1}{1.714}]To solve for ( x ), I'll first find a common denominator and then solve the equation:[frac{4 + x}{4x} = frac{1}{1.714}][4 + x = frac{4x}{1.714}][4 + x = 2.333x quad text{(rounded to three decimal places for simplicity)}][x - 2.333x = -4][-1.333x = -4][x = frac{-4}{-1.333}][x approx 3]So, it takes me approximately 3 hours to paint the room alone.