Solving & Explaining Steps - Math 3

1. A company's revenue R (in thousands) and cost C (in thousands) are given by R = 50x - x² and C = 20x + 100, where x is units produced (in hundreds). To find the break-even points where revenue equals cost, which solution process is correct?
   Answer: Set 50x - x² = 20x + 100, rearrange: -x² + 30x - 100 = 0, multiply by -1: x² - 30x + 100 = 0, factor: (x - 10)(x - 20) = 0, so x = 10 or x = 20 Choice A correctly sets revenue equal to cost, rearranges to standard form by moving all terms to one side, multiplies by -1 to make the leading coefficient positive, and factors correctly to find x = 10 or x = 20 (representing 1000 or 2000 units). Choice B makes an error in the discriminant calculation (should be 900 - 400 = 500, but then incorrectly calculates the roots). Choice C has correct algebra but fails to show the important step of multiplying by -1. Choice D shows the work clearly but doesn't explicitly show the multiplication by -1 step that makes factoring easier.
2. To find when the temperature reaches 90°F, Sarah sets up 75 + $45e^{-0.2t}$ = 90. Which solution method correctly handles the isolation and logarithmic steps?
   Answer: Subtract 75: $45e^{-0.2t}$ = 15, divide by 45: $e^{-0.2t}$ = 1/3, so -0.2t = -ln(3), and t = ln(3)/0.2 ≈ 5.49 minutes Choice D correctly isolates the exponential term first ($45e^{-0.2t}$ = 15), then divides by 45 to get $e^{-0.2t}$ = 1/3. Taking the natural logarithm: -0.2t = ln(1/3) = -ln(3), so t = ln(3)/0.2. Choice A shows correct steps but has an unnecessary complication. Choice B incorrectly takes the logarithm before isolating the exponential. Choice C has the right approach but shows ln(1/3) = -ln(3) without clear intermediate steps.
3. To determine when the concentration drops to 3 mg/L, which solution method correctly explains each step?
   Answer: Set $12e^{-0.2t}$ = 3, divide by 12: $e^{-0.2t}$ = 1/4, take natural log: -0.2t = ln(1/4) = -ln(4), so t = ln(4)/0.2 ≈ 6.93 hours When you encounter exponential decay problems like drug concentration, you're solving equations where the variable appears in an exponent. The key is systematically isolating the exponential term before applying logarithms. To find when the concentration drops to 3 mg/L, you start with $12e^{-0.2t} = 3$. First, isolate the exponential by dividing both sides by 12: $e^{-0.2t} = \frac{1}{4}$. Next, take the natural logarithm of both sides: $-0.2t = \ln(\frac{1}{4})$. Using the logarithm property $\ln(\frac{1}{a}) = -\ln(a)$, this becomes $-0.2t = -\ln(4)$. Finally, solve for t: $t = \frac{-\ln(4)}{-0.2} = \frac{\ln(4)}{0.2} \approx 6.93$ hours. This matches answer choice C exactly. Answer A makes an error by taking the natural logarithm before isolating the exponential term, incorrectly applying $\ln(12e^{-0.2t}) = \ln(12) - 0.2t$. Answer B correctly reaches $e^{-0.2t} = 0.25$ but writes this as a decimal instead of the fraction $\frac{1}{4}$, making the subsequent logarithm work less clear. Answer D makes a sign error when dividing by -0.2, incorrectly treating the operation as $\frac{\ln(\frac{1}{4})}{-0.2}$ instead of recognizing that $\ln(\frac{1}{4}) = -\ln(4)$. Always isolate the exponential term first in these problems, then apply natural logarithms. This systematic approach prevents the algebraic errors that create most wrong answer choices.
4. To find when the water height reaches 5 feet, Alex solves 2√(t + 4) - 3 = 5. Which step-by-step solution correctly identifies and addresses the key challenge in solving this radical equation?
   Answer: Isolate the radical term first: 2√(t + 4) = 8, then √(t + 4) = 4, square: t + 4 = 16, so t = 12, then verify solution Choice D correctly identifies that the key challenge is properly isolating the radical before squaring, and includes the crucial verification step. The solution gives t = 12, and checking: h = 2√(12 + 4) - 3 = 2√16 - 3 = 8 - 3 = 5 ✓. Choice A follows correct steps but omits verification. Choice B squares incorrectly without first dividing by 2. Choice C squares both sides prematurely, leading to a more complex equation.
5. When solving the exponential equation $3^{2x-1}$ = $27^{x-2}$ for a population growth model, what is the most appropriate first step and why?
   Answer: Rewrite 27 as 3³ to create a common base, since $27^{x-2}$ = $(3³)^{x-2}$ = $3^{3x-6}$ Choice B is correct because recognizing that 27 = 3³ allows us to rewrite the equation as $3^{2x-1}$ = $3^{3x-6}$, making it possible to equate exponents directly: 2x-1 = 3x-6, which gives x = 5. Choice A would work but is unnecessarily complex. Choice C is algebraically valid but more complicated than needed. Choice D is inappropriate since this equation doesn't fit the substitution pattern for polynomial transformation.
6. To find when the concentration equals 3 mg/L, Ben sets up $5te^{-0.4t}$ = 3. This equation cannot be solved algebraically using elementary functions. What is the most appropriate method to explain the solution process?
   Answer: Use iterative approximation or graphing technology to find that t ≈ 0.73 hours and t ≈ 6.59 hours are the two solutions When you encounter transcendental equations like $5te^{-0.4t} = 3$ that mix polynomial and exponential terms, they typically cannot be solved using standard algebraic methods. The question tests your understanding of appropriate solution strategies for such equations. The most appropriate method combines accessibility with completeness. Using iterative approximation or graphing technology provides a practical approach that reveals both solutions: $t ≈ 0.73$ hours and $t ≈ 6.59$ hours. This method is accessible to most students and clearly shows that the concentration reaches 3 mg/L at two different times—once as the drug concentration increases initially, and again as it decreases later. This makes biological sense for drug concentration models. Looking at the other options: Option B suggests Newton's method, which is mathematically sophisticated but requires calculus knowledge and doesn't emphasize finding all solutions. Option C mentions the Lambert W function, which is the exact analytical solution but involves advanced mathematics beyond typical coursework and only gives the principal solution initially. Option D proposes graphical analysis, which is helpful for visualization but less precise than iterative methods for determining actual numerical values. The key insight is that this concentration function has a maximum somewhere between the two solutions, creating the two intersection points with the horizontal line $y = 3$. Remember: when facing transcendental equations in applied contexts, combining graphical visualization with numerical approximation methods gives you both conceptual understanding and practical solutions. Always consider whether multiple solutions might exist by thinking about the function's behavior.
7. The temperature T (in °F) of a cooling object follows Newton's law of cooling: T = 75 + $125e^{-0.1t}$, where t is time in minutes. To find when the temperature reaches 100°F, which solution approach correctly shows all steps?
   Answer: Set T = 100: 75 + $125e^{-0.1t}$ = 100, rearrange: $125e^{-0.1t}$ = 25, divide: $e^{-0.1t}$ = 0.2, take ln: -0.1t = ln(0.2), solve: t = -ln(0.2)/0.1 = ln(5)/0.1 Choice D correctly isolates the exponential term by subtracting 75 and dividing by 125, takes the natural logarithm properly, and correctly handles the negative signs to get t = ln(5)/0.1 ≈ 16.1 minutes. Choice A makes an error in the final step by not properly handling the negative sign. Choice B has the wrong final expression, showing ln(5)/0.1 but deriving it from ln(0.2). Choice C correctly identifies that ln(1/5) = -ln(5) but then makes an error in solving for t, getting 10ln(5) instead of ln(5)/0.1.
8. To find when the population reaches 800, Jamie sets up 1200/(1 + $5e^{-0.3t}$) = 800. Which sequence of algebraic steps correctly isolates the exponential term?
   Answer: Cross multiply: 1200 = 800(1 + $5e^{-0.3t}$) = 800 + $4000e^{-0.3t}$, so 400 = $4000e^{-0.3t}$, giving $e^{-0.3t}$ = 0.1 Choice B correctly shows the complete isolation process: cross multiply to get 1200 = 800 + $4000e^{-0.3t}$, subtract 800 to get 400 = $4000e^{-0.3t}$, then divide by 4000 to get $e^{-0.3t}$ = 0.1. From here, taking ln gives -0.3t = ln(0.1), so t = -ln(0.1)/0.3. Choice A stops before the final division. Choice C uses an unnecessarily complex approach with reciprocals. Choice D incorrectly inverts the equation setup.
9. To find when the tank is completely empty, which step-by-step solution correctly shows the process?
   Answer: Set V = 0: 500 - 25t - 0.5t² = 0, multiply by -2: t² + 50t - 1000 = 0, factor: (t + 70)(t - 20) = 0, so t = 20 minutes (reject t = -70) Choice B correctly sets V = 0, multiplies by -2 to eliminate decimals and make the leading coefficient positive, then factors the quadratic correctly as (t + 70)(t - 20) = 0, giving t = 20 minutes as the physical solution. Choice A sets up correctly but doesn't complete the calculation or factoring. Choice C rearranges unnecessarily and uses completing the square when factoring is simpler. Choice D makes an error in the factorization, getting (t - 20)(t + 50) instead of (t - 20)(t + 70).
10. To find when the company first breaks even (profit equals zero), which sequence of steps correctly solves the equation -2t² + 16t - 24 = 0?
    Answer: Factor out -2: -2(t² - 8t + 12) = 0, then factor the quadratic: -2(t - 2)(t - 6) = 0, giving t = 2 or t = 6 Choice A correctly factors out the GCF of -2, then factors the resulting quadratic t² - 8t + 12 as (t - 2)(t - 6). Setting each factor equal to zero gives t = 2 and t = 6. Choice B has an error in the discriminant calculation (should be 256 - 192, not 256 + 192). Choice C makes an error in completing the square by not properly handling the constant term. Choice D correctly factors out -2 but then incorrectly applies the quadratic formula.
11. A population grows according to P(t) = 1200$(1.05)^t$, where t is years after 2020. To find when the population doubles from its 2020 value, which approach correctly shows the solution steps?
    Answer: Set 1200$(1.05)^t$ = 2(1200), simplify: $(1.05)^t$ = 2, take natural log: t·ln(1.05) = ln(2), so t = ln(2)/ln(1.05) ≈ 14.2 years Choice B correctly identifies that doubling means P(t) = 2400, isolates the exponential expression to get $(1.05)^t$ = 2, then properly applies the natural logarithm and logarithm properties to solve for t. Choice A uses common log instead of natural log, which gives the same numerical result but natural log is more standard for exponential growth problems. Choice C overcomplicates the setup and uses unfamiliar logarithm notation. Choice D is mathematically correct but uses a more complex approach by taking log of both sides before simplification.
12. To find when the voltage reaches 8 volts, Lisa sets up 12(1 - $e^{-t/2}$) = 8. Which solution process correctly handles the exponential isolation and logarithmic conversion?
    Answer: After getting $e^{-t/2}$ = 1/3, apply ln: -t/2 = ln(1/3), so t = -2ln(1/3) = 2ln(3), since ln(1/3) = -ln(3), giving t ≈ 2.20 seconds When solving exponential equations like this capacitor voltage problem, you need to systematically isolate the exponential term, then apply logarithms to "undo" the exponential function. Starting with $12(1 - e^{-t/2}) = 8$, you first divide both sides by 12 to get $1 - e^{-t/2} = \frac{2}{3}$. Then subtract 1 from both sides (or equivalently, subtract $\frac{2}{3}$ from 1) to isolate the exponential: $e^{-t/2} = \frac{1}{3}$. Now apply the natural logarithm to both sides: $-\frac{t}{2} = \ln(\frac{1}{3})$. Using the logarithm property that $\ln(\frac{1}{3}) = -\ln(3)$, you get $-\frac{t}{2} = -\ln(3)$. Multiplying both sides by -2 gives $t = 2\ln(3) \approx 2.20$ seconds. Choice A correctly shows this complete process and explicitly demonstrates the key logarithm property that converts $\ln(\frac{1}{3})$ to $-\ln(3)$. Choice B stops at $t = -2\ln(\frac{1}{3})$ without showing the simplification to the cleaner form $2\ln(3)$. Choice C has the right steps but doesn't show the logarithm property conversion. Choice D makes an error by writing $-\frac{t}{2} = \ln(\frac{1}{3}) = -\ln(3)$ and then incorrectly jumping to $t = 2\ln(3)$ without showing the multiplication by -2. Remember: when solving exponential equations, always work methodically—isolate the exponential term first, then apply logarithms, and finally use logarithm properties to simplify your answer to its cleanest form.
13. A rectangular garden has dimensions where the length is 3 feet more than twice the width. If the area is 90 square feet, which approach correctly identifies the steps to find the width w?
    Answer: Set up w(2w + 3) = 90, expand to get 2w² + 3w - 90 = 0, factor as (2w + 15)(w - 6) = 0, giving w = 6 since width must be positive Choice C correctly sets up the equation w(2w + 3) = 90 where w is width and (2w + 3) is length, expands correctly, and factors the quadratic properly. The factorization (2w + 15)(w - 6) = 0 gives w = -15/2 or w = 6, and since width must be positive, w = 6. Choice A stops before completing the solution. Choice B incorrectly interprets 'length is 3 more than twice the width' as w + 3 instead of 2w + 3. Choice D uses an incorrect setup by creating a rational equation instead of the direct area formula.
14. A student solving log₃(x + 6) - log₃(x - 2) = 2 makes the following steps: log₃((x + 6)/(x - 2)) = 2, then (x + 6)/(x - 2) = 9. What should be the next steps to complete the solution correctly?
    Answer: Solve: x + 6 = 9x - 18, so x = 3; check both x + 6 > 0 and x - 2 > 0, then verify in original equation Choice C provides the most complete solution process. After getting x = 3, we must verify the domain restrictions: x + 6 > 0 (satisfied since 9 > 0) and x - 2 > 0 (satisfied since 1 > 0), then check the original equation: log₃(9) - log₃(1) = 2 - 0 = 2 ✓. Choice A checks the equation but not domain restrictions. Choice B checks only one domain condition. Choice D is incomplete about what specifically to verify.
15. A student is solving the rational equation (x + 2)/(x - 3) - 1/(x + 1) = 2 for a circuit analysis problem. After finding a common denominator and solving, they get x = 1 and x = -5. What should be the next step and why?
    Answer: Check both solutions in the original equation to verify they don't make any denominator zero Choice B is correct because rational equations can introduce extraneous solutions during the solving process, especially when multiplying by expressions containing variables. Both x = 1 and x = -5 must be checked: neither makes x - 3 = 0 or x + 1 = 0, so both are mathematically valid. Choice A skips the essential verification step. Choice C incorrectly assumes negative values are automatically invalid. Choice D is unnecessary since the question states the solutions were already found algebraically.
16. When solving the absolute value equation |2x - 5| = |x + 3| for a distance optimization problem, what is the most systematic approach to ensure all solutions are found?
    Answer: Set up two equations: 2x - 5 = x + 3 and 2x - 5 = -(x + 3), then solve each separately Choice D is the most systematic algebraic approach. When |A| = |B|, either A = B or A = -B. This gives: 2x - 5 = x + 3 (so x = 8) and 2x - 5 = -(x + 3) (so 3x = 2, x = 2/3). Both solutions check in the original equation. Choice B works but is more complex. Choice C is theoretically complete but unnecessarily complicated. Choice A is graphical rather than algebraic and may lack precision.
17. A student is solving the system: 2x + 3y = 12 and x² + y² = 13 using substitution. They solve the first equation for x = (12 - 3y)/2 and substitute into the second. What should be their next steps to avoid common algebraic errors?
    Answer: Substitute and simplify: (12 - 3y)²/4 + y² = 13, so (144 - 72y + 9y²)/4 + 4y²/4 = 13, giving (144 - 72y + 13y²)/4 = 13 When solving systems involving linear and quadratic equations, careful fraction handling is crucial to avoid algebraic mistakes that can derail your entire solution. After substituting $x = \frac{12 - 3y}{2}$ into $x^2 + y^2 = 13$, you get $\left(\frac{12 - 3y}{2}\right)^2 + y^2 = 13$. The key insight is recognizing that you need a common denominator before combining terms. Since the first term has denominator 4 when expanded, you should express $y^2$ as $\frac{4y^2}{4}$ to match. This gives you $\frac{144 - 72y + 9y^2}{4} + \frac{4y^2}{4} = 13$, which combines to $\frac{144 - 72y + 13y^2}{4} = 13$. Option C follows this correct approach. Option A makes the error of not converting $y^2$ to the common denominator before adding, leading to the incorrect expression $\frac{144 - 72y + 9y^2}{4} + y^2$. Option B suggests clearing fractions first, but this approach is more prone to errors when dealing with squared expressions in the denominator. Option D incorrectly describes dividing the expanded numerator by 4, which isn't the proper way to handle the fraction. Study tip: When substituting expressions involving fractions into equations, always establish a common denominator before combining terms. This systematic approach prevents the mixing of fractional and whole number terms that often leads to algebraic errors in system problems.
18. When solving the equation x⁴ - 13x² + 36 = 0 using substitution, a student lets u = x². After solving 4u² - 13u + 36 = 0, they get u = 4 and u = 9. What error did the student make, and what is the correct approach?
    Answer: The substitution equation should be u² - 13u + 36 = 0, giving u = 4 and u = 9, so x = ±2 and x = ±3 Choice A correctly identifies that if u = x², then x⁴ = u², so the equation becomes u² - 13u + 36 = 0, not 4u² - 13u + 36 = 0. Solving u² - 13u + 36 = 0 using factoring or the quadratic formula gives u = 4 and u = 9. Then x² = 4 yields x = ±2, and x² = 9 yields x = ±3. Choice B identifies the error but doesn't complete the solution. Choice C assumes the wrong quadratic was solved correctly. Choice D doesn't address the coefficient error.
19. Which sequence of steps correctly solves for the time when N = 2000?
    Answer: Set $500e^{0.3t}$ = 2000, divide by 500 to get $e^{0.3t}$ = 4, take natural log: 0.3t = ln(4), so t = ln(4)/0.3 ≈ 4.62 hours When you encounter exponential growth problems, you're solving equations where the variable appears in an exponent. The key is systematically isolating that variable using logarithms. To find when the bacteria population reaches 2000, you start by setting up the equation: $500e^{0.3t} = 2000$. The most efficient approach is to first isolate the exponential term by dividing both sides by 500, giving you $e^{0.3t} = 4$. Then apply the natural logarithm to both sides. Since $\ln(e^{0.3t}) = 0.3t$ (the natural log and exponential function are inverses), you get $0.3t = \ln(4)$. Finally, solve for t: $t = \frac{\ln(4)}{0.3} \approx 4.62$ hours. Option A makes a critical error by writing $0.3t \cdot \ln(e) = \ln(4)$. While $\ln(e) = 1$, this step is unnecessary and shows a misunderstanding of how logarithms work with exponents. Option B incorrectly uses log base 10 instead of the natural logarithm. When dealing with equations involving $e$, you must use $\ln$, not $\log_{10}$. This leads to a wrong numerical answer. Option C takes the natural log of both sides before isolating the exponential term. While this eventually works out to the same answer, it's unnecessarily complicated and more prone to errors. Remember: when solving exponential equations with base $e$, isolate the exponential term first, then apply the natural logarithm. This direct approach minimizes calculation errors and is faster on timed exams.
20. A ball is thrown upward from a building. Its height above ground is modeled by h = -5t² + 20t + 60, where h is height in meters and t is time in seconds. To find when the ball hits the ground, which step-by-step solution is correct?
    Answer: Set h = 0: -5t² + 20t + 60 = 0, divide by -5: t² - 4t - 12 = 0, factor: (t - 6)(t + 2) = 0, so t = 6 seconds (reject t = -2) Choice A correctly sets height equal to 0 for ground level, divides the entire equation by -5 to simplify, factors the resulting quadratic correctly as (t - 6)(t + 2) = 0, and properly rejects the negative time solution. Choice B incorrectly sets height equal to 60 (the initial height) instead of 0. Choice C uses the quadratic formula correctly but makes a sign error in the denominator. Choice D starts correctly but doesn't complete the solution process clearly.

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