Algebra 2 Quiz: Rewrite Exponential Expressions Using Exponents
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Rewrite Exponential Expressions Using ExponentsQuestion 1 of 20

An account grows by 12% per year, so its value after tt years is multiplied by (1.12)t(1.12)^t. Rewrite (1.12)t(1.12)^t in an equivalent form that reveals the monthly growth factor (assume 12 months per year).​​

(1.1212)t\left(1.12^{12}\right)^t
(1.121/12)12t\left(1.12^{1/12}\right)^{12t}
(1.1212)12t\left(\dfrac{1.12}{12}\right)^{12t}
(1.12)t/12\left(1.12\right)^{t/12}
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Algebra 2 Quiz

Algebra 2 Quiz: Rewrite Exponential Expressions Using Exponents

Practice Rewrite Exponential Expressions Using Exponents in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rewrite Exponential Expressions Using Exponents, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An account grows by 12% per year, so its value after tt years is multiplied by (1.12)t(1.12)^t. Rewrite (1.12)t(1.12)^t in an equivalent form that reveals the monthly growth factor (assume 12 months per year).​​

  1. (1.1212)t\left(1.12^{12}\right)^t
  2. (1.121/12)12t\left(1.12^{1/12}\right)^{12t} (correct answer)
  3. (1.1212)12t\left(\dfrac{1.12}{12}\right)^{12t}
  4. (1.12)t/12\left(1.12\right)^{t/12}
Explanation: This question tests your ability to use exponent properties to transform exponential expressions into equivalent forms that reveal different information, like converting annual interest rates to monthly rates. The power-of-a-power property (b^a)^c = b^(ac) lets us rewrite expressions like (1.12)^t (12% annual growth) as ((1.12)^(1/12))^(12t) to reveal the monthly growth rate: we break each year into 12 months and find the factor that, when applied 12 times (raised to power 12), gives the yearly factor 1.12. The monthly factor is (1.12)^(1/12) ≈ 1.0095, meaning about 0.95% per month. Both forms equal the same value for any t, but one emphasizes annual compounding, the other monthly! To convert annual expression (1.12)^t to monthly form: (1) Recognize that t years equals 12t months, (2) Want form (something)^(12t), (3) That something is (1.12)^(1/12) because ((1.12)^(1/12))^(12t) = (1.12)^((1/12)×12t) = (1.12)^t by power-of-power property, (4) Calculate with calculator: (1.12)^(1/12) ≈ 1.0095, (5) So (1.12)^t = ((1.12)^(1/12))^(12t), revealing monthly rate of about 0.95% (since 1.0095 = 1 + 0.0095). The exponent properties make the conversion systematic! Choice B correctly applies the power-of-a-power property to transform the expression and identifies the equivalent monthly rate. Choice C divides the annual factor by 12 to get monthly (1.12/12 ≈ 0.093), but this gives SIMPLE interest division, not compound! For compound interest, the monthly factor isn't found by division—we need (1.12)^(1/12), which is the 12th root of 1.12 ≈ 1.0095. Division would work for simple interest, but compound interest requires the fractional exponent! The power-of-a-power property (b^a)^c = b^(ac) is your main tool for time-base conversion: to convert annual rate factor b^t to monthly, write as ((b)^(1/12))^(12t)—take 12th root of b for monthly factor, raise to 12t (12 months times t years). This property is foundation of all time-base transformations!

Question 2

A culture of bacteria triples every day. If tt is measured in weeks, its growth factor can be written as 37t3^{7t} (since 7 days per week). Rewrite 37t3^{7t} using exponent properties as a power of a power so it looks like a weekly growth factor raised to tt.

Which rewrite is correct?

  1. (31/7)t\left(3^{1/7}\right)^{t}
  2. (37)t\left(3^{7}\right)^{t} (correct answer)
  3. (3t)7\left(3^{t}\right)^{7}
  4. (37t)7\left(3^{7t}\right)^{7}
Explanation: This question tests your skill in rewriting exponential expressions using power-of-a-power to shift from daily to weekly growth factors while keeping the overall model equivalent. The property (b^a)^c = b^(a c) lets us transform 3^{7t} (tripling every day over 7t days) into (3^7)^t, where 3^7 is the weekly growth factor since seven daily triplings multiply to 3^7, or 2187 times per week. This form emphasizes growth per week raised to t weeks, making it easier to think in weekly terms. To rewrite: (1) Identify the weekly factor as (daily)^7 = 3^7, (2) Raise to t: (3^7)^t = 3^{7t}, confirming equivalence via the property. Choice B correctly applies this to show the weekly factor 3^7 raised to t, matching the requested form. Choice C gives (3^t)^7 = 3^{7t} too, but it's raised to 7 instead of t, so it doesn't present as weekly^t—note the order matters for the interpretation! Use this strategy for any period change: group the base by the new time unit using exponents, and verify by expanding—you're mastering these conversions beautifully!

Question 3

An investment grows according to A(t)=P (1.15)tA(t)=P\,(1.15)^t (with tt in years). Which statement correctly interprets the transformation (1.15)t=(1.151/12)12t?(1.15)^t = \left(1.15^{1/12}\right)^{12t}?

  1. It shows the account earns 15% each month for 12t12t months.
  2. It shows the same growth written using a monthly growth factor applied over 12t12t months. (correct answer)
  3. It changes the annual rate to a simple monthly rate of 0.15/120.15/12.
  4. It shows the account earns 1.15121.15^{12} per year instead of 1.151.15 per year.
Explanation: This question tests your ability to interpret the meaning of exponential transformations when converting between different time periods in growth models. The transformation (1.15)t=((1.15)1/12)12t(1.15)^t = \left( (1.15)^{1/12} \right)^{12t} uses the power-of-a-power property to rewrite annual growth in terms of monthly growth: the annual factor 1.15 (representing 15% annual growth) is converted to its monthly equivalent by taking the 12th root, giving (1.15)1/12≈1.0117(1.15)^{1/12} \approx 1.0117 (about 1.17% monthly growth). This monthly factor is then applied 12t times (for 12t months in t years), maintaining the same overall growth but revealing the monthly compounding structure. The key insight is that both expressions produce identical values for any t, but the right side makes the monthly growth rate explicit! To understand why this works: (1) The original (1.15)t(1.15)^t means "multiply by 1.15 for each of t years", (2) The transformed ((1.15)1/12)12t\left( (1.15)^{1/12} \right)^{12t} means "multiply by (1.15)1/12(1.15)^{1/12} for each of 12t months", (3) By the power-of-a-power property, ((1.15)1/12)12t=(1.15)(1/12)×12t=(1.15)t\left( (1.15)^{1/12} \right)^{12t} = (1.15)^{ (1/12) \times 12t } = (1.15)^t, (4) So we get the same result but expressed differently. The transformation doesn't change the growth—it reveals it at a different time scale! Choice B correctly states that the transformation shows the same growth written using a monthly growth factor applied over 12t months. Choice A incorrectly claims the account earns 15% each month, but (1.15)1/12≈1.0117(1.15)^{1/12} \approx 1.0117 represents about 1.17% monthly growth, not 15%! The 15% is the annual rate, and when compounded monthly, you need the 12th root to find the equivalent monthly rate. This is a critical distinction in compound interest calculations! When interpreting exponential transformations: (1) The base raised to a fractional power gives the rate for that fraction of the time period, (2) (Annual factor)1/n^{1/n} gives the factor for 1/n of a year, (3) The exponent in the outer parentheses (like 12t) counts how many of these smaller periods occur. Always verify that the transformation maintains equality—the power-of-a-power property ensures it does!

Question 4

A population is modeled by P(t)=P0(1.08)tP(t)=P_0(1.08)^t, where tt is in years. Rewrite the growth part (1.08)t(1.08)^t to show an equivalent form with a quarterly growth factor raised to the number of quarters.

(Use the power-of-power property.)

  1. (1.08)t=((1.08)4)t(1.08)^t = \left((1.08)^{4}\right)^{t}
  2. (1.08)t=((1.08)1/4)4t(1.08)^t = \left((1.08)^{1/4}\right)^{4t} (correct answer)
  3. (1.08)t=((1.08)1/t)4(1.08)^t = \left((1.08)^{1/t}\right)^{4}
  4. (1.08)t=((1.08)1/4)t/4(1.08)^t = \left((1.08)^{1/4}\right)^{t/4}
Explanation: This question tests your ability to apply exponent properties to rewrite growth expressions in terms of different time intervals, such as quarterly from annual. The power-of-a-power property (b^a)^c = b^(a*c) enables us to transform (1.08)^t into ((1.08)^{1/4})^{4t}, where (1.08)^{1/4} is the quarterly factor—applied 4 times per year over t years, equaling 4t quarters, and both expressions match since (1/4)*4t = t. For the conversion: note t years = 4t quarters, seek (quarterly factor)^{4t}, set quarterly factor to (1.08)^{1/4} so the property multiplies exponents back to t— you can verify with a calculator if needed, but the focus is on the algebraic equivalence. Choice A correctly implements the power-of-a-power property to produce the quarterly form with the exponent 4t for total quarters. Choice B reverses the process by raising to the 4th power inside, leading to (1.08)^{4t}, which overstates growth—a frequent error is confusing roots with powers, but remember to take the root (divide exponent) when breaking into more periods! To master this, use the formula: for m periods per year, rewrite b^t as (b^{1/m})^{m t}—the exponents multiply to t, preserving value. Keep up the excellent work; experiment with m=2 for semi-annual to see how this strategy transfers seamlessly!

Question 5

A quantity is multiplied by (1.05)t(1.05)^t after tt years. Which expression is equivalent and reveals the daily growth factor (assume 365 days per year)?

  1. (1.051/365)365t\left(1.05^{1/365}\right)^{365t} (correct answer)
  2. (1.05365)t\left(1.05^{365}\right)^{t}
  3. (1.05)t/365\left(1.05\right)^{t/365}
  4. (1.05365)365t\left(\dfrac{1.05}{365}\right)^{365t}
Explanation: This question tests your ability to use exponent properties to transform exponential expressions into equivalent forms that reveal different information, like converting annual growth rates to daily rates. The power-of-a-power property (b^a)^c = b^(ac) lets us rewrite expressions like (1.05)^t (5% annual growth) as ((1.05)^(1/365))^(365t) to reveal the daily growth rate: we break each year into 365 days and find the factor that, when applied 365 times (raised to power 365), gives the yearly factor 1.05. The daily factor is (1.05)^(1/365) ≈ 1.000134, meaning about 0.0134% per day. Both forms equal the same value for any t, but one emphasizes annual compounding, the other daily! To convert annual expression (1.05)^t to daily form: (1) Recognize that t years equals 365t days, (2) Want form (something)^(365t), (3) That something is (1.05)^(1/365) because ((1.05)^(1/365))^(365t) = (1.05)^((1/365)×365t) = (1.05)^t by power-of-power property, (4) Calculate with calculator: (1.05)^(1/365) ≈ 1.000134, (5) So (1.05)^t = ((1.05)^(1/365))^(365t), revealing daily rate of about 0.0134%. The exponent properties make the conversion systematic! Choice A correctly applies the power-of-a-power property to transform the expression and identifies the equivalent daily rate. Choice D divides the annual factor by 365 to get daily (1.05/365 ≈ 0.00288), but this gives SIMPLE interest division, not compound! For compound interest, the daily factor isn't found by division—we need (1.05)^(1/365), which is the 365th root of 1.05 ≈ 1.000134. Division would give a daily simple interest rate, but compound interest requires the fractional exponent! The power-of-a-power property (b^a)^c = b^(ac) is your main tool for time-base conversion: to convert annual rate factor b^t to daily, write as ((b)^(1/365))^(365t)—take 365th root of b for daily factor, raise to 365t (365 days times t years). For very small time periods like days, the fractional power gives a factor very close to 1, but when compounded many times (365 days), it produces the full annual growth!

Question 6

A savings account is modeled by A(t)=A0(1.06)tA(t)=A_0(1.06)^t, where tt is in years. Which expression is equivalent and reveals the daily growth factor (assume 365 days per year) applied over 365t365t days?

  1. ((1.06)1/365)t\left(\left(1.06\right)^{1/365}\right)^{t}
  2. ((1.06)365)t\left(\left(1.06\right)^{365}\right)^{t}
  3. ((1.06)1/365)365t\left(\left(1.06\right)^{1/365}\right)^{365t} (correct answer)
  4. (1.06)365t\left(1.06\right)^{365t}
Explanation: This question tests your ability to apply exponent properties to reveal daily growth factors from an annual model, adapting for finer time scales. Using the power-of-a-power rule ((b)^a)^c = b^(a c), we can rewrite (1.06)^t as ((1.06)^{1/365})^{365t}, where (1.06)^{1/365} is the daily factor that compounds 365 times to match the 6% annual growth. This daily factor is tiny, about 1.000159 or 0.0159% per day, but over 365t days it equals the original. The steps: (1) t years = 365t days, (2) Seek (daily)^{365t}, (3) Daily = (1.06)^{1/365} since ((1.06)^{1/365})^{365t} = (1.06)^t. Choice C correctly transforms to show the daily factor raised to 365t, as requested. Choice B uses (1.06^{365})^t, which is like applying a huge yearly-overcompounded factor t times, not daily—remember to take the root for sub-periods! For daily or other frequent compounding, always use (annual^{1/n})^{n t} with n=365; approximate with a calculator if needed—this approach will serve you well in finance models, keep up the excellent work!

Question 7

An investment grows by 12% per year, so its value after tt years is multiplied by (1.12)t(1.12)^t. Rewrite (1.12)t(1.12)^t to show an equivalent form with a monthly growth factor raised to the number of months, and identify the correct expression.

(Use the power-of-power property.)

  1. (1.12)t=((1.12)1/12)t/12(1.12)^t = \left((1.12)^{1/12}\right)^{t/12}
  2. (1.12)t=(1.12)t/12(1.12)^t = \left(1.12\right)^{t/12}
  3. (1.12)t=((1.12)12)t(1.12)^t = \left((1.12)^{12}\right)^{t}
  4. (1.12)t=((1.12)1/12)12t(1.12)^t = \left((1.12)^{1/12}\right)^{12t} (correct answer)
Explanation: This question tests your ability to use exponent properties to transform exponential expressions into equivalent forms that reveal different time scales, like converting an annual growth factor to a monthly one. The power-of-a-power property states that (b^a)^c = b^(a*c), which allows us to rewrite expressions like (1.12)^t as ((1.12)^{1/12})^{12t} to show the monthly growth factor raised to the number of months in t years. To convert the annual expression (1.12)^t to a monthly form: recognize that t years equal 12t months, aim for the form (monthly factor)^{12t}, where the monthly factor is (1.12)^{1/12} because ((1.12)^{1/12})^{12t} = (1.12)^{(1/12)*12t} = (1.12)^t by the power-of-a-power property—both forms are equivalent for any t, but the monthly version highlights the per-month multiplier. Choice B correctly applies the power-of-a-power property to transform the expression and identifies the equivalent monthly form with the exponent 12t representing the total months. A tempting distractor like Choice A raises the base to the 12th power instead of taking the 12th root, which would multiply the exponent by 12 instead of dividing, resulting in (1.12)^{12t}—that's a common mix-up between raising the base versus adjusting the exponent properly! Remember, the power-of-a-power property multiplies the exponents, so to keep the overall expression equivalent, if you multiply the outer exponent by 12 (for months), you must divide the inner exponent by 12 (take the root)—this balances out to the original. Practice this by verifying with a small t, like t=1: calculate both sides and see they match, building your confidence in transforming rates systematically!

Question 8

A population is modeled by P(t)=P0(1.08)tP(t)=P_0(1.08)^t, where tt is in years. To express the same growth using quarters, rewrite (1.08)t(1.08)^t as a quarterly factor raised to the number of quarters 4t4t.

Which rewrite is correct?

  1. ((1.08)4)t/4\left(\left(1.08\right)^{4}\right)^{t/4}
  2. ((1.08)1/4)4t\left(\left(1.08\right)^{1/4}\right)^{4t} (correct answer)
  3. ((1.08)4t)1/4\left(\left(1.08\right)^{4t}\right)^{1/4}
  4. (1.08)t/4\left(1.08\right)^{t/4}
Explanation: This question tests your ability to use exponent properties to rewrite exponential growth models for different time periods, such as converting annual to quarterly compounding. The power-of-a-power property ((b)^a)^c = b^(a c) enables us to express (1.08)^t as ((1.08)^{1/4})^{4t}, where (1.08)^{1/4} is the quarterly growth factor that, when applied four times per year, matches the annual 8% growth. The quarterly factor is approximately 1.0194, or about 1.94% per quarter, and this form highlights the growth over 4t quarters while remaining equivalent to the original expression. To make the conversion: (1) Note that t years equal 4t quarters, (2) Aim for (quarterly factor)^{4t}, (3) Set the quarterly factor to (1.08)^{1/4} since ((1.08)^{1/4})^{4t} = (1.08)^((1/4)*4t) = (1.08)^t, (4) Use a calculator to approximate if needed, (5) This systematically reveals the per-quarter rate. Choice A correctly uses the power-of-power property to show the quarterly factor raised to 4t, matching the requested form. A common mistake in Choice B is using (1.08^4)^{t/4}, which is mathematically equal but raises to t/4 instead of 4t, so it doesn't show the factor applied over the number of quarters—pay attention to the exponent matching the total periods! Remember the strategy: for n periods per year, rewrite b^t as (b^{1/n})^{n t}; this multiplies exponents to preserve equivalence and adapts to any time scale—great job exploring these transformations!

Question 9

Simplify the expression using properties of exponents (quotient of powers and power of a power): (5t53)2\left(\frac{5^{t}}{5^{3}}\right)^{2}

  1. 52t−65^{2t-6} (correct answer)
  2. 52t+65^{2t+6}
  3. 5t−15^{t-1}
  4. 5t−65^{t-6}
Explanation: This question tests your skills in simplifying expressions using quotient and power-of-a-power properties together. The quotient property gives 5t/53=5t−35^t / 5^3 = 5^{t-3}. Then power-of-a-power (ba)c=bac(b^a)^c = b^{a c} yields (5t−3)2=52(t−3)=52t−6(5^{t-3})^2 = 5^{2(t-3)} = 5^{2t-6}— distributing the 2 simplifies it neatly. Step-by-step: first simplify inside the parentheses by subtracting exponents (t−3t-3), then apply the outer power by multiplying the exponent by 2— check with t=3t=3: original ((53)/53)2=12=1((5^3)/5^3)^2 = 1^2 = 1, and 56−6=50=15^{6-6} = 5^0 = 1. Choice A correctly combines these properties to reach 52t−65^{2t-6}. Choice B adds 6 instead of subtracting, possibly from mishandling the quotient as addition—a common pitfall is forgetting subtraction for division! Use the order: simplify quotients first (subtract), then powers (multiply exponents)—try with variables like (xa/xb)c=xc(a−b)(x^a / x^b)^c = x^{c(a-b)}. Excellent effort; this combo will simplify many algebraic expressions, keep practicing!

Question 10

A population of bacteria triples every 4 hours. If the initial population is P0P_0, which expression is equivalent to 3t/43^{t/4} and reveals the approximate growth factor per hour?

  1. 30.25t≈(1.316)t3^{0.25t} \approx (1.316)^t
  2. 34t≈(81)t3^{4t} \approx (81)^t
  3. (31/4)t≈(1.316)t(3^{1/4})^t \approx (1.316)^t (correct answer)
  4. (0.75)t≈(0.422)t(0.75)^t \approx (0.422)^t
Explanation: To find the hourly growth factor, we rewrite 3t/43^{t/4} as (31/4)t(3^{1/4})^t. Since 31/4≈1.3163^{1/4} \approx 1.316, this shows the population multiplies by approximately 1.316 each hour. Choice A incorrectly writes the exponent as 0.25t0.25t instead of using the base transformation. Choice B uses the wrong transformation 34t3^{4t}. Choice D incorrectly assumes decay rather than growth.

Question 11

The brightness of a star as observed from Earth follows B(d)=B0⋅2−d/5B(d) = B_0 \cdot 2^{-d/5}, where dd is distance in parsecs. Which transformation shows how the brightness changes when the distance is measured in light-years (1 parsec ≈ 3.26 light-years)?

  1. B0⋅2−3.26L/5B_0 \cdot 2^{-3.26L/5}, where LL is distance in light-years
  2. B0⋅2−L/(5⋅3.26)B_0 \cdot 2^{-L/(5 \cdot 3.26)}, where LL is distance in light-years
  3. B0⋅2−L/16.3B_0 \cdot 2^{-L/16.3}, where LL is distance in light-years (correct answer)
  4. B0⋅2−5L/3.26B_0 \cdot 2^{-5L/3.26}, where LL is distance in light-years
Explanation: If dd parsecs = LL light-years, then d=L/3.26d = L/3.26. Substituting: B(L)=B0⋅2−(L/3.26)/5=B0⋅2−L/(5⋅3.26)=B0⋅2−L/16.3B(L) = B_0 \cdot 2^{-(L/3.26)/5} = B_0 \cdot 2^{-L/(5 \cdot 3.26)} = B_0 \cdot 2^{-L/16.3}. Choice A incorrectly multiplies the light-year distance by the conversion factor. Choice B shows the intermediate step but doesn't simplify. Choice D inverts the conversion relationship.

Question 12

A savings account earns 12% annual interest. To compare with a credit card that charges 1.2% monthly interest, which form of (1.12)t(1.12)^t reveals the equivalent monthly rate?

  1. (1.12)t=(1.121/12)12t≈(1.0095)12t(1.12)^t = (1.12^{1/12})^{12t} \approx (1.0095)^{12t} (correct answer)
  2. (1.12)t=(1.121/12)12t≈(1.0949)12t(1.12)^t = (1.12^{1/12})^{12t} \approx (1.0949)^{12t}
  3. (1.12)t=(1.01)12t(1.12)^t = (1.01)^{12t}, assuming simple division
  4. (1.12)t=(1.1212)t/12≈(3.896)t/12(1.12)^t = (1.12^{12})^{t/12} \approx (3.896)^{t/12}
Explanation: When you encounter compound interest problems with different compounding periods, you need to find equivalent rates that produce the same growth over the same time period. The key insight is using exponent properties to rewrite expressions in forms that reveal monthly rates. To find the equivalent monthly rate from an annual rate, you need to determine what monthly multiplier, when compounded 12 times, gives the same result as the annual multiplier. Since (1.12)t(1.12)^t represents annual compounding, you can rewrite this using the property amn=(am)na^{mn} = (a^m)^n. Working backwards: (1.12)t=(1.121/12)12t(1.12)^t = (1.12^{1/12})^{12t}. This form shows that 1.121/121.12^{1/12} is the monthly growth factor. Calculating this gives approximately 1.0095, meaning the equivalent monthly rate is 0.95%. Option A correctly applies this transformation and provides the accurate calculation, revealing that 12% annual interest is equivalent to about 0.95% monthly interest - slightly less than the credit card's 1.2% monthly rate. Option B uses the correct algebraic form but miscalculates 1.121/121.12^{1/12}, giving 1.0949 instead of 1.0095. Option C incorrectly assumes you can simply divide the annual rate by 12, which ignores the compounding effect - this would give 1% monthly instead of the true 0.95%. Option D flips the exponent relationship, calculating 1.12121.12^{12} instead of 1.121/121.12^{1/12}, which would represent converting from monthly to annual rather than annual to monthly. Remember: when converting between different compounding periods, use fractional exponents to find equivalent rates, and always double-check your calculations since small errors lead to very different financial outcomes.

Question 13

A chemical reaction rate is modeled by R(T)=k⋅3(T−298)/10R(T) = k \cdot 3^{(T-298)/10}, where TT is temperature in Kelvin and kk is a constant. To convert this to Celsius (C=T−273C = T - 273), which form correctly expresses the rate in terms of Celsius temperature?

  1. k⋅3(C+273−298)/10=k⋅3(C−25)/10k \cdot 3^{(C+273-298)/10} = k \cdot 3^{(C-25)/10} (correct answer)
  2. k⋅3(C−25)/10k \cdot 3^{(C-25)/10}, since 298−273=25298 - 273 = 25
  3. k⋅3(C−298)/10k \cdot 3^{(C-298)/10}, directly substituting CC for TT
  4. k⋅3(C+25)/10k \cdot 3^{(C+25)/10}, accounting for the temperature shift
Explanation: When you encounter function transformations involving unit conversions, you need to carefully substitute the conversion relationship and simplify step by step. Starting with R(T)=k⋅3(T−298)/10R(T) = k \cdot 3^{(T-298)/10} where TT is in Kelvin, you're given that C=T−273C = T - 273, which means T=C+273T = C + 273. To express the rate in terms of Celsius, substitute this relationship for TT: R(C)=k⋅3((C+273)−298)/10R(C) = k \cdot 3^{((C + 273) - 298)/10} Simplifying the exponent: (C+273)−298=C+273−298=C−25(C + 273) - 298 = C + 273 - 298 = C - 25 Therefore: R(C)=k⋅3(C−25)/10R(C) = k \cdot 3^{(C-25)/10} Choice A is correct because it shows both the substitution step and the final simplified form, demonstrating the complete mathematical reasoning. Choice B gives the right final answer but lacks the substitution work that proves why 298−273=25298 - 273 = 25 is relevant to the conversion. Choice C makes the error of directly replacing TT with CC without accounting for the conversion relationship. This ignores that Celsius and Kelvin have different zero points. Choice D incorrectly adds 25 instead of subtracting it, showing a sign error in the algebraic simplification of the exponent. Study tip: For unit conversion problems in exponential functions, always write out the substitution step explicitly before simplifying. This prevents direct replacement errors and helps you track how constants change during the conversion process.

Question 14

A certain investment is modeled by V(t)=V0(1.10)tV(t)=V_0(1.10)^t, where tt is in years. Which expression correctly rewrites the growth factor to show an equivalent monthly rate and uses the correct exponent properties?

(You do not need to compute the decimal approximation.)

  1. (1.10)t=(1.1012)12t(1.10)^t = \left(\frac{1.10}{12}\right)^{12t}
  2. (1.10)t=((1.10)1/12)12t(1.10)^t = \left((1.10)^{1/12}\right)^{12t} (correct answer)
  3. (1.10)t=((1.10)12)t/12(1.10)^t = \left((1.10)^{12}\right)^{t/12}
  4. (1.10)t=((1.10)1/12)t(1.10)^t = \left((1.10)^{1/12}\right)^{t}
Explanation: This question tests your ability to correctly apply exponent properties to rewrite annual growth in monthly terms without needing approximations. The power-of-a-power property (b^a)^c = b^(a*c) supports rewriting (1.10)^t as ((1.10)^{1/12})^{12t}, explicitly showing the monthly rate factor raised to 12t (total months)— this keeps the expression equivalent since (1/12)*12t = t. To achieve this: identify 12 months per year, take the 12th root for monthly factor, exponentiate by 12t— the property ensures it matches without changing the growth. Choice C properly uses the property to display the monthly rate with the correct 12t exponent. Choice B uses ((1.10)^{12})^{t/12}, which is equivalent algebraically but shows a yearly compounded factor over t/12 (fractions of years), not explicitly the monthly rate—a subtle distractor, but the question seeks the per-month factor over total months! For any n periods, use (b^{1/n})^{n t} to reveal the sub-period rate clearly—verify exponents multiply to t. You're doing fantastically; apply to bi-monthly (n=6) to solidify this useful technique!

Question 15

An investment grows according to A(t)=A0(1.09)tA(t)=A_0(1.09)^t, where tt is in years. Which expression is equivalent and uses exponent properties to show an approximate monthly growth model of the form (monthly factor)12t(\text{monthly factor})^{12t}?

(You may use 1.091/12≈1.00721.09^{1/12}\approx 1.0072.)

  1. A(t)=A0 (1.0072)tA(t)=A_0\,(1.0072)^{t}
  2. A(t)=A0 (1.0072)12tA(t)=A_0\,(1.0072)^{12t} (correct answer)
  3. A(t)=A0 (1.09)12tA(t)=A_0\,(1.09)^{12t}
  4. A(t)=A0 (0.09)12tA(t)=A_0\,(0.09)^{12t}
Explanation: This question tests your ability to rewrite annual growth using exponent properties to approximate a monthly model, incorporating given numerical values. The power-of-a-power property lets us express (1.09)^t as ((1.09)^{1/12})^{12t}, and with 1.09^{1/12} ≈ 1.0072, this becomes (1.0072)^{12t}, showing about 0.72% monthly growth over 12t months. This equivalent form reveals compounding at a monthly scale. Steps: (1) Use the rewrite ((1.09)^{1/12})^{12t} = (1.09)^t, (2) Substitute the approximation for the monthly factor. Choice B correctly applies this with the given ≈1.0072 raised to 12t, matching the monthly model. Choice A uses (1.0072)^t, which understates growth by applying monthly factor yearly—ensure the exponent is n t for n periods! When approximations are provided, plug them into the property-based rewrite; this is key for real-world finance—excellent work, you're ready for more!

Question 16

A bank advertises an annual growth factor of 1.151.15. A student rewrites (1.15)t=((1.15)1/12)12t(1.15)^t = \left((1.15)^{1/12}\right)^{12t}. What does this transformation reveal about the growth model (assume 12 months per year)?

  1. The account grows by 15% each month.
  2. The account grows by a constant monthly factor of (1.15)1/12(1.15)^{1/12}, applied over 12t12t months. (correct answer)
  3. The account grows by a constant monthly factor of 1.15/121.15/12, applied over 12t12t months.
  4. The account grows by 15% total over 12t12t months, not per year.
Explanation: This question tests your understanding of how exponent properties reveal underlying growth rates in rewritten exponential expressions, like interpreting monthly compounding from an annual factor. The power-of-a-power property shows that (1.15)t=((1.15)1/12)12t(1.15)^t = ((1.15)^{1/12})^{12t}, where (1.15)1/12(1.15)^{1/12} is the constant monthly growth factor applied over 12t12t months, equivalent to the annual 15% growth when compounded monthly. This monthly factor is approximately 1.0117, or about 1.17% per month, emphasizing compound growth rather than simple division. The transformation works because ((1.15)1/12)12t=(1.15)t((1.15)^{1/12})^{12t} = (1.15)^t via exponent multiplication, (1/12)×12t=t(1/12) \times 12t = t, and it reveals the per-month multiplier directly. Choice B correctly identifies that the rewrite shows a constant monthly factor of (1.15)1/12(1.15)^{1/12} over 12t12t months, capturing the essence of compound interest. A tempting distractor like Choice C uses 1.15/121.15/12, which is simple interest division (about 0.0958 or 9.58% monthly, way off), but compounding requires the root, not arithmetic division—watch for that mix-up! To interpret such rewrites, always check the inner exponent: 1/n1/n for n periods per year gives the per-period factor; practice calculating these with a calculator to see how compounding boosts effective rates—you're building strong skills here!

Question 17

A quantity decays by 10% per year, so its value after tt years is multiplied by (0.90)t(0.90)^t. Which expression correctly rewrites (0.90)t(0.90)^t to show an equivalent daily factor raised to the number of days?

(Assume 365 days in a year.)

  1. (0.90)t=((0.90)365)t(0.90)^t = \left((0.90)^{365}\right)^{t}
  2. (0.90)t=((0.90)1/365)365t(0.90)^t = \left((0.90)^{1/365}\right)^{365t} (correct answer)
  3. (0.90)t=((0.90)/365)365t(0.90)^t = \left((0.90)/365\right)^{365t}
  4. (0.90)t=((0.90)1/365)t/365(0.90)^t = \left((0.90)^{1/365}\right)^{t/365}
Explanation: This question tests your skills in using exponent properties to convert decay expressions to finer time scales, like daily from annual, even for decay factors less than 1. The power-of-a-power property (b^a)^c = b^(a*c) allows rewriting (0.90)^t as ((0.90)^{1/365})^{365t}, where (0.90)^{1/365} is the daily decay factor—raised to 365t for total days in t years, equating back since (1/365)*365t = t, maintaining the same overall decay. To perform the rewrite: recognize t years ≈ 365t days, form (daily factor)^{365t} with daily factor = (0.90)^{1/365} via the property— this works for decay just as for growth, as exponents handle fractions and decimals fine. Choice B accurately applies the power-of-a-power property to create the daily equivalent with the correct exponent 365t. Choice C divides by 365 instead of taking the 365th root, which might tempt if confusing with simple rates, but for compounding decay, we need the root—division would give a linear approximation, not exponential! A transferable approach is: for n periods per unit time, use (b^{1/n})^{n t}—verify by simplifying the exponent: (1/n)*n t = t. You're building strong skills; try this with weekly (n=52) to reinforce how the property systematically reveals per-period factors!

Question 18

A savings account has an annual growth factor of 1.151.15. Which expression is equivalent to (1.15)t(1.15)^t and makes the approximate monthly growth factor explicit?

(You may use a calculator for the approximation.)

  1. (1.15)t=(1.012)12t(1.15)^t = (1.012)^{12t} (monthly factor ≈1.012\approx 1.012)
  2. (1.15)t=(1.15)12t(1.15)^t = (1.15)^{12t}
  3. (1.15)t≈(1.0123)t(1.15)^t \approx (1.0123)^t (monthly factor ≈1.0123\approx 1.0123)
  4. (1.15)t=((1.15)1/12)12t≈(1.0117)12t(1.15)^t = \left((1.15)^{1/12}\right)^{12t} \approx (1.0117)^{12t} (correct answer)
Explanation: This question tests your ability to use exponent properties to rewrite exponential expressions in forms that make sub-period growth factors explicit, such as monthly from annual. The power-of-a-power property (b^a)^c = b^(a*c) lets us express (1.15)^t as ((1.15)^{1/12})^{12t}, where (1.15)^{1/12} is the monthly factor—approximately 1.0117, meaning about 1.17% growth per month when compounded 12 times to match the 15% annual. To make the monthly factor explicit: identify that t years are 12t months, compute the 12th root of 1.15 using a calculator for ≈1.0117, and confirm ((1.0117))^{12t} ≈ (1.15)^t since the exponents multiply: (1/12)12t = t— this reveals the compounding structure without changing the value. Choice D correctly uses the power-of-a-power property to show the equivalent form and provides the accurate approximation for the monthly factor. Choice A uses a slightly off approximation (1.012 instead of 1.0117) and omits the root notation, which might come from mistakenly dividing the rate by 12 (15%/12=1.25%, then 1.0125 rounded) instead of taking the proper compound root—remember, simple division works for simple interest, but compounding requires the fractional exponent! The key strategy is to always take the nth root of the annual factor for n periods per year, then raise to (nt) for total periods—this ensures equivalence via exponent multiplication. You're doing great; try calculating (1.15)^{1/12} yourself to see it's indeed ≈1.0117, and apply this to other periods like quarterly for more practice!

Question 19

A bacteria culture doubles every hour. One model is N(t)=N0 2tN(t)=N_0\,2^{t} where tt is in hours. Rewrite 2t2^{t} to show an equivalent form with a per-minute factor raised to the number of minutes.

(There are 60 minutes in an hour.)

  1. 2t=(260)60t2^{t} = \left(\frac{2}{60}\right)^{60t}
  2. 2t=(260)t2^{t} = \left(2^{60}\right)^{t}
  3. 2t=(21/60)60t2^{t} = \left(2^{1/60}\right)^{60t} (correct answer)
  4. 2t=(21/60)t/602^{t} = \left(2^{1/60}\right)^{t/60}
Explanation: This question tests your use of exponent properties to rescale time units in exponential models, like hourly doubling to per-minute factors. The power-of-a-power property (b^a)^c = b^(a*c) allows rewriting 2^t as (2^{1/60})^{60t}, where 2^{1/60} is the per-minute factor—raised to 60t minutes in t hours, simplifying back to 2^t since (1/60)*60t = t, showing equivalence. For the transformation: note t hours = 60t minutes, set up (minute factor)^{60t} with minute factor = 2^{1/60} using the property— this breaks the doubling into 60 equal minute-by-minute multipliers. Choice B properly applies the power-of-a-power property with the correct exponent 60t for total minutes. Choice C divides by 60 instead of taking the 60th root, which might appeal for linear thinking, but exponential growth requires the root—division underestimates the compound effect! Always use (b^{1/m})^{m t} for m sub-units per original unit, verifying exponent multiplication to t. You're progressing wonderfully; apply this to seconds (m=3600) to see the pattern and build even more confidence!

Question 20

A savings balance is modeled by A(t)=A0(1.05)tA(t)=A_0(1.05)^t, where tt is in years. Which statement best describes what the transformation (1.05)t=((1.05)1/12)12t(1.05)^t = \left((1.05)^{1/12}\right)^{12t} reveals about the growth?

  1. It changes the annual rate from 5% to 12%.
  2. It shows the same total growth, but expressed using a monthly growth factor applied over 12t12t months. (correct answer)
  3. It means the account grows by 5% each month.
  4. It shows the balance is constant because the exponents cancel.
Explanation: This question tests your understanding of what exponent transformations reveal about growth models, focusing on equivalence without changing rates. The power-of-a-power property shows (1.05)t=((1.05)1/12)12t(1.05)^t = \left( (1.05)^{1/12} \right)^{12t}, which expresses the same annual 5% growth as a monthly factor applied over 12t12t months— the total growth remains identical, but it highlights the compounding structure. In detail: the transformation takes the 12th root for the monthly equivalent, raises to 12t12t (total months), and by the property, exponents multiply to tt, preserving value— it doesn't alter the rate, just reframes it. Choice B accurately describes that it shows the same growth via a monthly lens over 12t12t periods. Choice C misinterprets as 5% per month, tempting if ignoring the root, but that would be (1.05)12t(1.05)^{12t}, massively overgrowing— the root adjusts to a smaller monthly rate like ≈1.0041\approx 1.0041! To interpret such rewrites, always note the inner exponent (1/12)(1/12) divides the rate for sub-periods, outer multiplies for total count. You're grasping this well; think about how this applies to continuous growth for deeper insight!