🧠:<think>Alright, so I've got this problem about a British football club's fan base growing exponentially. It's modeled by the function ( F(t) = F_0 times e^{kt} ), where ( F_0 = 10,000 ) is the initial number of fans, ( t ) is time in years, and ( k ) is the growth constant. The club doesn't care about politics or economics, just the growth of their fan base. There are two parts to this problem. Let me tackle them one by one.Problem 1: Finding the growth constant ( k ) when the fan base doubles every 3 years.Okay, so the fan base doubles every 3 years. That means if I start with ( F_0 ) fans, after 3 years, I'll have ( 2F_0 ) fans. Let me write that down:( F(3) = 2F_0 )But according to the model, ( F(t) = F_0 e^{kt} ). So, plugging in ( t = 3 ):( F_0 e^{k times 3} = 2F_0 )Hmm, I can divide both sides by ( F_0 ) to simplify:( e^{3k} = 2 )Now, to solve for ( k ), I need to take the natural logarithm of both sides. Remember, the natural logarithm is the inverse function of the exponential function with base ( e ). So,( ln(e^{3k}) = ln(2) )Simplify the left side:( 3k = ln(2) )So, solving for ( k ):( k = frac{ln(2)}{3} )Let me calculate that value numerically to check if it makes sense. I know that ( ln(2) ) is approximately 0.6931, so:( k approx frac{0.6931}{3} approx 0.2310 ) per year.That seems reasonable. So, the growth constant ( k ) is ( ln(2)/3 ).Problem 2: Finding the time ( t ) when the revenue reaches 500 times its initial value ( R_0 ), given that revenue ( R ) is directly proportional to the square root of the fan base.Alright, so revenue ( R ) is directly proportional to the square root of the fan base. That means:( R = c times sqrt{F(t)} )where ( c ) is the constant of proportionality.Given that the initial revenue ( R_0 = 100 ), we can find ( c ) by plugging in ( t = 0 ):( R_0 = c times sqrt{F(0)} )We know ( F(0) = F_0 = 10,000 ), so:( 100 = c times sqrt{10,000} )Calculating ( sqrt{10,000} ):( sqrt{10,000} = 100 )So,( 100 = c times 100 )Divide both sides by 100:( c = 1 )So, the revenue function simplifies to:( R(t) = sqrt{F(t)} )But ( F(t) = 10,000 e^{kt} ), so:( R(t) = sqrt{10,000 e^{kt}} )Simplify the square root:( R(t) = sqrt{10,000} times sqrt{e^{kt}} )( R(t) = 100 times e^{k t / 2} )Because ( sqrt{e^{kt}} = e^{kt/2} ).We need to find the time ( t ) when the revenue reaches 500 times its initial value. The initial revenue ( R_0 = 100 ), so 500 times that is ( 500 times 100 = 50,000 ).So, set ( R(t) = 50,000 ):( 100 times e^{k t / 2} = 50,000 )Divide both sides by 100:( e^{k t / 2} = 500 )Take the natural logarithm of both sides:( ln(e^{k t / 2}) = ln(500) )Simplify the left side:( frac{k t}{2} = ln(500) )Solve for ( t ):( t = frac{2 ln(500)}{k} )But from Problem 1, we know that ( k = frac{ln(2)}{3} ). So, substitute that in:( t = frac{2 ln(500)}{ln(2)/3} )Simplify the division:( t = 2 ln(500) times frac{3}{ln(2)} )( t = frac{6 ln(500)}{ln(2)} )Hmm, that's the exact value. But maybe I can simplify ( ln(500) ) further. Let me see:500 is 5 x 100, which is 5 x 10^2. So,( ln(500) = ln(5 times 10^2) = ln(5) + ln(10^2) = ln(5) + 2 ln(10) )But I don't know if that helps. Alternatively, I can express 500 as 5^3 x 2^2 x 5, but that might complicate things. Maybe it's better to just leave it as ( ln(500) ).Alternatively, I can compute the numerical value to check. Let me compute ( ln(500) ) and ( ln(2) ):( ln(500) approx 6.2146 )( ln(2) approx 0.6931 )So,( t approx frac{6 times 6.2146}{0.6931} )First, compute 6 x 6.2146:6 x 6 = 366 x 0.2146 ≈ 1.2876So total ≈ 36 + 1.2876 ≈ 37.2876Now, divide by 0.6931:37.2876 / 0.6931 ≈ 53.78 yearsSo, approximately 53.78 years. But since the question asks for the exact value, I should keep it in terms of logarithms.Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from ( R(t) = sqrt{F(t)} ), which is ( sqrt{10,000 e^{kt}} = 100 e^{kt/2} ). That seems correct.Set ( 100 e^{kt/2} = 50,000 ), so ( e^{kt/2} = 500 ). Taking natural logs: ( kt/2 = ln(500) ), so ( t = (2 ln(500))/k ). Since ( k = ln(2)/3 ), substituting gives ( t = (2 ln(500)) / (ln(2)/3) = 6 ln(500)/ln(2) ). That seems right.Alternatively, I can express this as ( t = 6 log_2(500) ), since ( ln(500)/ln(2) = log_2(500) ). So, ( t = 6 log_2(500) ). That might be a cleaner way to write it.Let me check that:( log_b(a) = ln(a)/ln(b) ), so yes, ( ln(500)/ln(2) = log_2(500) ). Therefore, ( t = 6 log_2(500) ).But 500 is 5^3 x 2^2 x 5, but that might not help much. Alternatively, 500 = 5^3 x 2^2 x 5, but that's 5^4 x 2^2, which is 625 x 4 = 2500, which is not 500. Wait, no, 500 is 5^3 x 2^2, because 5^3 is 125, and 2^2 is 4, and 125 x 4 is 500. So,( log_2(500) = log_2(5^3 times 2^2) = log_2(5^3) + log_2(2^2) = 3 log_2(5) + 2 )So,( t = 6 (3 log_2(5) + 2) = 18 log_2(5) + 12 )But that might not necessarily be simpler. So, perhaps leaving it as ( 6 log_2(500) ) is the most straightforward exact form.Alternatively, since ( log_2(500) = log_2(5^3 times 2^2) = 3 log_2(5) + 2 ), as above, but unless the problem expects a specific form, either should be acceptable.But let me see if I can express 500 in terms of powers of 2 or 5 to make it simpler.Wait, 500 is 5^3 * 2^2, as I thought earlier. So, ( log_2(500) = log_2(5^3 times 2^2) = 3 log_2(5) + 2 ). So, plugging that back into ( t ):( t = 6 (3 log_2(5) + 2) = 18 log_2(5) + 12 )But I don't know if that's any better. Maybe not. So, perhaps the simplest exact form is ( t = frac{6 ln(500)}{ln(2)} ).Alternatively, since ( ln(500) = ln(5 times 100) = ln(5) + ln(100) = ln(5) + 2 ln(10) ), and ( ln(10) = ln(2 times 5) = ln(2) + ln(5) ), so:( ln(500) = ln(5) + 2(ln(2) + ln(5)) = 3 ln(5) + 2 ln(2) )Thus,( t = frac{6 (3 ln(5) + 2 ln(2))}{ln(2)} = frac{18 ln(5) + 12 ln(2)}{ln(2)} = 18 frac{ln(5)}{ln(2)} + 12 )Which is the same as:( t = 18 log_2(5) + 12 )So, that's another way to express it. But again, unless the problem expects a specific form, either is fine.But let me check if I made any mistakes in the earlier steps. Let me go through the steps again.1. Revenue is proportional to the square root of the fan base: ( R = c sqrt{F(t)} ).2. At ( t = 0 ), ( R_0 = 100 = c sqrt{F_0} = c sqrt{10,000} = c times 100 ). So, ( c = 1 ). Therefore, ( R(t) = sqrt{F(t)} ).3. ( F(t) = 10,000 e^{kt} ), so ( R(t) = sqrt{10,000 e^{kt}} = 100 e^{kt/2} ).4. We need ( R(t) = 500 R_0 = 500 times 100 = 50,000 ).5. So, ( 100 e^{kt/2} = 50,000 ) => ( e^{kt/2} = 500 ).6. Taking natural logs: ( kt/2 = ln(500) ) => ( t = (2 ln(500))/k ).7. From Problem 1, ( k = ln(2)/3 ), so ( t = (2 ln(500)) / (ln(2)/3) = 6 ln(500)/ln(2) ).Yes, that seems correct.Alternatively, I can write ( t = 6 log_2(500) ), which is the same thing.So, the exact value is ( t = frac{6 ln(500)}{ln(2)} ) years, or ( t = 6 log_2(500) ) years.I think that's the answer they're looking for.Summary of Thoughts:1. For the first part, using the doubling time formula, I found ( k = ln(2)/3 ).2. For the second part, by expressing revenue in terms of the fan base and setting it to 500 times the initial revenue, I derived the time ( t ) as ( 6 ln(500)/ln(2) ) or ( 6 log_2(500) ).I think that's solid. I didn't see any mistakes in my calculations, and the steps make sense. The key was recognizing that revenue is proportional to the square root of the fan base, which led to expressing ( R(t) ) in terms of ( F(t) ), and then solving for ( t ) when ( R(t) ) reaches 50,000.Final Answer1. The growth constant ( k ) is boxed{dfrac{ln 2}{3}}.2. The time ( t ) when the revenue reaches 500 times its initial value is boxed{dfrac{6 ln 500}{ln 2}} years.</think>

🧠:<think>Okay, so I have this problem about a counselor who uses differential equations and probability models to help families affected by Alzheimer's. There are two sub-problems here. Let me try to tackle them one by one.Starting with Sub-problem 1: Solving the differential equation for the progression of Alzheimer's modeled by a logistic growth equation. The equation is given as:[ frac{dP}{dt} = rPleft(1 - frac{P}{K}right) ]where ( r ) is the growth rate constant, ( K ) is the carrying capacity, and ( P(t) ) is the progression level at time ( t ). The initial condition is ( P(0) = P_0 ).Hmm, I remember that the logistic differential equation is a standard one in calculus. It's a separable equation, so I should be able to separate the variables ( P ) and ( t ) and integrate both sides. Let me write that out.First, rewrite the equation:[ frac{dP}{dt} = rPleft(1 - frac{P}{K}right) ]To separate variables, I can divide both sides by ( Pleft(1 - frac{P}{K}right) ) and multiply both sides by ( dt ):[ frac{dP}{Pleft(1 - frac{P}{K}right)} = r dt ]Now, I need to integrate both sides. The left side looks a bit tricky, but I think partial fractions can be used here. Let me set up the integral:[ int frac{1}{Pleft(1 - frac{P}{K}right)} dP = int r dt ]Let me simplify the denominator on the left side. Let me write ( 1 - frac{P}{K} ) as ( frac{K - P}{K} ). So the denominator becomes ( P cdot frac{K - P}{K} ) which is ( frac{P(K - P)}{K} ). Therefore, the integrand becomes:[ frac{K}{P(K - P)} ]So, the integral becomes:[ int frac{K}{P(K - P)} dP = int r dt ]Now, let's perform partial fraction decomposition on ( frac{K}{P(K - P)} ). Let me set:[ frac{K}{P(K - P)} = frac{A}{P} + frac{B}{K - P} ]Multiplying both sides by ( P(K - P) ):[ K = A(K - P) + BP ]Expanding the right side:[ K = AK - AP + BP ]Grouping like terms:[ K = AK + (B - A)P ]Since this must hold for all ( P ), the coefficients of like terms must be equal. So, for the constant term:[ K = AK implies A = 1 ]And for the coefficient of ( P ):[ 0 = (B - A) implies B = A = 1 ]So, the partial fractions decomposition is:[ frac{K}{P(K - P)} = frac{1}{P} + frac{1}{K - P} ]Therefore, the integral becomes:[ int left( frac{1}{P} + frac{1}{K - P} right) dP = int r dt ]Integrating term by term:Left side:[ int frac{1}{P} dP + int frac{1}{K - P} dP = ln|P| - ln|K - P| + C ]Right side:[ int r dt = rt + C ]So, combining both sides:[ ln|P| - ln|K - P| = rt + C ]Simplify the left side using logarithm properties:[ lnleft| frac{P}{K - P} right| = rt + C ]Exponentiating both sides to eliminate the natural log:[ frac{P}{K - P} = e^{rt + C} = e^{rt} cdot e^C ]Let me denote ( e^C ) as a constant ( C' ) for simplicity:[ frac{P}{K - P} = C' e^{rt} ]Now, solve for ( P ). Multiply both sides by ( K - P ):[ P = C' e^{rt} (K - P) ]Expand the right side:[ P = C' K e^{rt} - C' P e^{rt} ]Bring all terms involving ( P ) to the left side:[ P + C' P e^{rt} = C' K e^{rt} ]Factor out ( P ):[ P (1 + C' e^{rt}) = C' K e^{rt} ]Solve for ( P ):[ P = frac{C' K e^{rt}}{1 + C' e^{rt}} ]Now, apply the initial condition ( P(0) = P_0 ). Let's plug ( t = 0 ) into the equation:[ P_0 = frac{C' K e^{0}}{1 + C' e^{0}} = frac{C' K}{1 + C'} ]Solve for ( C' ):Multiply both sides by ( 1 + C' ):[ P_0 (1 + C') = C' K ]Expand:[ P_0 + P_0 C' = C' K ]Bring terms with ( C' ) to one side:[ P_0 = C' K - P_0 C' ]Factor out ( C' ):[ P_0 = C' (K - P_0) ]Solve for ( C' ):[ C' = frac{P_0}{K - P_0} ]So, substitute ( C' ) back into the expression for ( P(t) ):[ P(t) = frac{left( frac{P_0}{K - P_0} right) K e^{rt}}{1 + left( frac{P_0}{K - P_0} right) e^{rt}} ]Simplify numerator and denominator:Numerator:[ frac{P_0 K e^{rt}}{K - P_0} ]Denominator:[ 1 + frac{P_0 e^{rt}}{K - P_0} = frac{K - P_0 + P_0 e^{rt}}{K - P_0} ]So, ( P(t) ) becomes:[ P(t) = frac{ frac{P_0 K e^{rt}}{K - P_0} }{ frac{K - P_0 + P_0 e^{rt}}{K - P_0} } = frac{P_0 K e^{rt}}{K - P_0 + P_0 e^{rt}} ]Factor ( K ) in the denominator:Wait, actually, let me factor ( e^{rt} ) in the denominator:[ K - P_0 + P_0 e^{rt} = K - P_0 (1 - e^{rt}) ]But maybe it's better to write it as:[ P(t) = frac{P_0 K e^{rt}}{K - P_0 + P_0 e^{rt}} ]Alternatively, factor ( P_0 ) in the denominator:[ P(t) = frac{P_0 K e^{rt}}{K - P_0 (1 - e^{rt})} ]But perhaps the standard form is more straightforward. Let me see.Alternatively, we can factor out ( e^{rt} ) in the denominator:[ K - P_0 + P_0 e^{rt} = e^{rt} (P_0) + (K - P_0) ]But maybe that's not necessary. Let me check if this is the standard logistic growth solution.Yes, the standard solution is:[ P(t) = frac{K P_0 e^{rt}}{K + P_0 (e^{rt} - 1)} ]Wait, let me see if my expression can be rewritten to match that.Starting from my expression:[ P(t) = frac{P_0 K e^{rt}}{K - P_0 + P_0 e^{rt}} ]Factor ( K ) in the denominator:[ K - P_0 + P_0 e^{rt} = K + P_0 (e^{rt} - 1) ]So, substituting back:[ P(t) = frac{P_0 K e^{rt}}{K + P_0 (e^{rt} - 1)} ]Which is the standard form. So, that's correct.Alternatively, sometimes it's written as:[ P(t) = frac{K}{1 + left( frac{K - P_0}{P_0} right) e^{-rt}} ]Let me verify that.Starting from my expression:[ P(t) = frac{P_0 K e^{rt}}{K - P_0 + P_0 e^{rt}} ]Divide numerator and denominator by ( e^{rt} ):[ P(t) = frac{P_0 K}{(K - P_0) e^{-rt} + P_0} ]Factor ( P_0 ) in the denominator:[ P(t) = frac{P_0 K}{P_0 + (K - P_0) e^{-rt}} ]Which can be written as:[ P(t) = frac{K}{1 + left( frac{K - P_0}{P_0} right) e^{-rt}} ]Yes, that's another standard form. So, both forms are correct. Depending on how you want to present it.So, I think either form is acceptable, but perhaps the first form is more straightforward.Therefore, the solution to the differential equation is:[ P(t) = frac{K P_0 e^{rt}}{K + P_0 (e^{rt} - 1)} ]Alternatively,[ P(t) = frac{K}{1 + left( frac{K - P_0}{P_0} right) e^{-rt}} ]Either way, it's correct.Moving on to Sub-problem 2: The counselor uses a probability model to estimate the likelihood that a patient will require full-time care within the next year. The probability ( P(T) ) is modeled by:[ P(T) = 1 - e^{-lambda T} ]Given that 70% of patients require full-time care within one year (( T = 1 )), find ( lambda ).So, we have ( P(1) = 0.7 ). Plugging into the equation:[ 0.7 = 1 - e^{-lambda cdot 1} ]Simplify:[ e^{-lambda} = 1 - 0.7 = 0.3 ]Take the natural logarithm of both sides:[ -lambda = ln(0.3) ]Multiply both sides by -1:[ lambda = -ln(0.3) ]Compute the value:I know that ( ln(0.3) ) is negative, so ( -ln(0.3) ) is positive. Let me compute it numerically.First, ( ln(0.3) approx ln(3/10) = ln(3) - ln(10) approx 1.0986 - 2.3026 = -1.204 ). So, ( lambda approx 1.204 ).Alternatively, using a calculator, ( ln(0.3) approx -1.2039728043 ), so ( lambda approx 1.2039728043 ).Rounding to a reasonable number of decimal places, say four decimal places: ( lambda approx 1.2040 ).Alternatively, if we need an exact expression, it's ( lambda = ln(10/3) ), since ( e^{-lambda} = 0.3 = 3/10 ), so ( lambda = ln(10/3) ).Let me verify:( e^{-ln(10/3)} = e^{ln(3/10)} = 3/10 = 0.3 ). Yes, that's correct.So, ( lambda = ln(10/3) ) exactly, or approximately 1.2040.Therefore, the value of ( lambda ) is ( ln(10/3) ) or approximately 1.204.Let me recap:Sub-problem 1: Solved the logistic differential equation and found the expression for ( P(t) ).Sub-problem 2: Used the given probability model and solved for ( lambda ) when ( P(1) = 0.7 ), resulting in ( lambda = ln(10/3) ).I think that's all. Let me just double-check my steps.For Sub-problem 1, the separation of variables and partial fractions seemed correct. The integration steps looked right, and the final form matches the standard logistic growth solution. So, that should be correct.For Sub-problem 2, plugging in ( T = 1 ) and ( P(1) = 0.7 ), solving for ( lambda ) by taking natural logs. That seems straightforward. The exact value is ( ln(10/3) ), which is approximately 1.204. That makes sense because ( e^{-1.204} approx 0.3 ), so 1 - 0.3 = 0.7. Correct.So, I think both solutions are correct.Final AnswerSub-problem 1: The progression level as a function of time is (boxed{P(t) = dfrac{K P_0 e^{rt}}{K + P_0 (e^{rt} - 1)}}).Sub-problem 2: The value of (lambda) is (boxed{lnleft(dfrac{10}{3}right)}).</think>

🧠:<think>Okay, so I have this problem about Alex's jewelry collection. It's divided into necklaces, bracelets, and rings, each with an equal number of items. The total value of the collection is a quadratic function of the number of items in each category. The function is given as V(n) = an² + bn + c. Part 1 asks me to find the coefficients a, b, and c. They've given me three data points: when n=10, V=900; when n=20, V=1600; and when n=30, V=2500. So, I can set up three equations based on these points and solve for a, b, c.Let me write down the equations:1. When n=10: a*(10)² + b*(10) + c = 900 So, 100a + 10b + c = 9002. When n=20: a*(20)² + b*(20) + c = 1600 So, 400a + 20b + c = 16003. When n=30: a*(30)² + b*(30) + c = 2500 So, 900a + 30b + c = 2500Now, I have a system of three equations:1. 100a + 10b + c = 9002. 400a + 20b + c = 16003. 900a + 30b + c = 2500I need to solve for a, b, c. Let me subtract the first equation from the second to eliminate c.Equation 2 - Equation 1:(400a - 100a) + (20b - 10b) + (c - c) = 1600 - 900300a + 10b = 700Let me call this Equation 4: 300a + 10b = 700Similarly, subtract Equation 2 from Equation 3:(900a - 400a) + (30b - 20b) + (c - c) = 2500 - 1600500a + 10b = 900Let me call this Equation 5: 500a + 10b = 900Now, I have two equations:Equation 4: 300a + 10b = 700Equation 5: 500a + 10b = 900Subtract Equation 4 from Equation 5:(500a - 300a) + (10b - 10b) = 900 - 700200a = 200So, a = 200 / 200 = 1Now that I have a=1, plug back into Equation 4:300*(1) + 10b = 700300 + 10b = 70010b = 700 - 300 = 400So, b = 400 / 10 = 40Now, with a=1 and b=40, plug into Equation 1 to find c:100*(1) + 10*(40) + c = 900100 + 400 + c = 900500 + c = 900c = 900 - 500 = 400So, the coefficients are a=1, b=40, c=400.Let me double-check with the third equation to make sure:900a + 30b + c = 900*1 + 30*40 + 400 = 900 + 1200 + 400 = 2500, which matches. So, that seems correct.Part 2: Alex wants the collection to be valued at 4900. So, we need to find n such that V(n) = 4900.Given V(n) = n² + 40n + 400 = 4900So, set up the equation:n² + 40n + 400 = 4900Subtract 4900:n² + 40n + 400 - 4900 = 0n² + 40n - 4500 = 0Now, solve this quadratic equation for n.Quadratic equation: n² + 40n - 4500 = 0We can use the quadratic formula: n = [-b ± sqrt(b² - 4ac)] / (2a)Here, a=1, b=40, c=-4500Discriminant D = 40² - 4*1*(-4500) = 1600 + 18000 = 19600sqrt(19600) = 140So, n = [-40 ± 140]/2We have two solutions:n = (-40 + 140)/2 = 100/2 = 50n = (-40 - 140)/2 = -180/2 = -90Since n cannot be negative, we discard -90.Thus, n=50.So, Alex should include 50 necklaces, 50 bracelets, and 50 rings.Let me verify this:V(50) = 50² + 40*50 + 400 = 2500 + 2000 + 400 = 4900, which is correct.So, the answers are a=1, b=40, c=400 and n=50.Final Answer1. The coefficients are (a = boxed{1}), (b = boxed{40}), and (c = boxed{400}).2. Alex should include (boxed{50}) items of each category.</think>

🧠:<think>Alright, so I've got these two math problems to solve, both inspired by Kathryn Kuhlman's events. Let me take them one at a time.Starting with the first problem: modeling attendance using the logistic growth model. Hmm, okay, I remember the logistic growth equation from my ecology class. It's a differential equation that models population growth considering a carrying capacity. The equation is:[frac{dA}{dt} = rAleft(1 - frac{A}{K}right)]Where:- ( A(t) ) is the attendance at time ( t ),- ( r ) is the intrinsic growth rate,- ( K ) is the carrying capacity.Given:- At ( t = 0 ), ( A(0) = 100 ),- After 5 years, ( A(5) = 500 ),- The carrying capacity ( K = 1000 ).We need to find the intrinsic growth rate ( r ).Okay, so I think the solution to the logistic differential equation is:[A(t) = frac{K}{1 + left(frac{K - A(0)}{A(0)}right)e^{-rt}}]Let me verify that. Yes, that's the standard solution. So plugging in the known values:At ( t = 0 ), ( A(0) = 100 ), so:[100 = frac{1000}{1 + left(frac{1000 - 100}{100}right)e^{0}} = frac{1000}{1 + 9} = frac{1000}{10} = 100]Good, that checks out.Now, at ( t = 5 ), ( A(5) = 500 ). Let's plug that into the equation:[500 = frac{1000}{1 + left(frac{1000 - 100}{100}right)e^{-5r}} = frac{1000}{1 + 9e^{-5r}}]So, let's solve for ( r ).First, write the equation:[500 = frac{1000}{1 + 9e^{-5r}}]Multiply both sides by ( 1 + 9e^{-5r} ):[500(1 + 9e^{-5r}) = 1000]Divide both sides by 500:[1 + 9e^{-5r} = 2]Subtract 1 from both sides:[9e^{-5r} = 1]Divide both sides by 9:[e^{-5r} = frac{1}{9}]Take the natural logarithm of both sides:[-5r = lnleft(frac{1}{9}right) = -ln(9)]So,[r = frac{ln(9)}{5}]Calculating that, ( ln(9) ) is approximately 2.1972, so:[r approx frac{2.1972}{5} approx 0.4394]So, the intrinsic growth rate ( r ) is approximately 0.4394 per year. Let me double-check my steps.1. Plugged in the known values into the logistic growth solution.2. Solved for ( e^{-5r} ) correctly.3. Took the natural log and solved for ( r ).4. Calculated the numerical value.Everything seems to add up. So, I think that's the answer for part 1.Moving on to part 2: Volunteering hours.The volunteer rate is given by ( V(t) = at + b ), where ( a ) and ( b ) are constants. We're told that in the first year, ( t = 1 ), the volunteer hours were 20, and in the third year, ( t = 3 ), it was 50 hours. We need to express ( V(t) ) in terms of ( t ) and then find the total hours volunteered by the end of the 10th year.Wait, hold on. Is ( t ) measured in years? The problem says "volunteer at a rate that increases linearly over time, defined by the function ( V(t) = at + b )", and "volunteered 20 hours in the first year" and "50 hours in the third year". So, does ( t ) represent the year number? So, ( t = 1 ) is the first year, ( t = 3 ) is the third year.So, we have two points: (1, 20) and (3, 50). We can use these to find ( a ) and ( b ).First, let's write the equations:At ( t = 1 ):[20 = a(1) + b implies 20 = a + b]At ( t = 3 ):[50 = a(3) + b implies 50 = 3a + b]Now, we have a system of two equations:1. ( a + b = 20 )2. ( 3a + b = 50 )Let's subtract equation 1 from equation 2:( (3a + b) - (a + b) = 50 - 20 )Simplify:( 2a = 30 implies a = 15 )Now, plug ( a = 15 ) into equation 1:( 15 + b = 20 implies b = 5 )So, the volunteer rate function is:[V(t) = 15t + 5]Now, the question is, how many total hours will be volunteered by the end of the 10th year.Wait, hold on. Is ( V(t) ) the rate, meaning the amount per year? Or is it the cumulative hours? The wording says "volunteer at a rate that increases linearly over time, defined by the function ( V(t) = at + b )". So, I think ( V(t) ) is the rate, i.e., the amount volunteered per year at time ( t ).Therefore, to find the total hours volunteered by the end of the 10th year, we need to sum ( V(t) ) from ( t = 1 ) to ( t = 10 ).Alternatively, since ( V(t) ) is a linear function, the total can be calculated as the sum of an arithmetic series.Wait, let me think. If ( V(t) ) is the amount volunteered in year ( t ), then the total is the sum from ( t = 1 ) to ( t = 10 ) of ( V(t) ).Given ( V(t) = 15t + 5 ), so each year's contribution is 15t + 5.So, the total hours ( T ) is:[T = sum_{t=1}^{10} (15t + 5)]We can split this sum into two parts:[T = 15 sum_{t=1}^{10} t + 5 sum_{t=1}^{10} 1]Compute each sum:First, ( sum_{t=1}^{10} t ) is the sum of the first 10 natural numbers:[sum_{t=1}^{10} t = frac{10(10 + 1)}{2} = 55]Second, ( sum_{t=1}^{10} 1 ) is just 10, since we're adding 1 ten times.So,[T = 15 times 55 + 5 times 10 = 825 + 50 = 875]Therefore, the total hours volunteered by the end of the 10th year is 875 hours.Wait, let me double-check. Alternatively, since ( V(t) ) is linear, the average rate over 10 years is the average of the first and last year's rates.First year: ( V(1) = 15(1) + 5 = 20 )Tenth year: ( V(10) = 15(10) + 5 = 155 )Average rate: ( frac{20 + 155}{2} = frac{175}{2} = 87.5 ) hours per year.Total over 10 years: ( 87.5 times 10 = 875 ). Yep, same result.So, that seems correct.Just to recap:1. Found ( a ) and ( b ) using the given points.2. Expressed ( V(t) ) as ( 15t + 5 ).3. Calculated the total hours by summing the series or using the arithmetic series average method.4. Both methods gave the same result, so confident it's 875.I think that's solid.Final Answer1. The intrinsic growth rate ( r ) is boxed{dfrac{ln(9)}{5}}.2. The total hours volunteered by the end of the 10th year is boxed{875}.</think>