Math 3 Quiz: Exponential Functions With Real Exponents

12 questions · exam conditions

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Exponential Functions With Real ExponentsQuestion 1 of 12

A medication's concentration in blood follows $C(t) = 50e^{-0.25t}$ mg/L, where $t$ is hours after injection. A second dose of equal strength is given when the first dose drops to 20% of its peak. What will be the total concentration 2 hours after the second injection?

Approximately 34.2 mg/L

Approximately 40.0 mg/L

Approximately 30.3 mg/L

Approximately 36.8 mg/L

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Math 3 Quiz

Math 3 Quiz: Exponential Functions With Real Exponents

Practice Exponential Functions With Real Exponents in Math 3 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 Exponential Functions With Real Exponents, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

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.

All questions

Question 1

Approximately 34.2 mg/L
Approximately 40.0 mg/L
Approximately 30.3 mg/L
Approximately 36.8 mg/L (correct answer)

Explanation: First dose reaches 20% when 50e^(-0.25t) = 10, so e^(-0.25t) = 0.2. Taking ln: -0.25t = ln(0.2) ≈ -1.609, so t ≈ 6.44 hours. At 2 hours after second injection (8.44 hours total): remaining from first dose = 50e^(-0.25×8.44) ≈ 6.1 mg/L. From second dose (2 hours old) = 50e^(-0.25×2) ≈ 30.3 mg/L. Total ≈ 6.1 + 30.3 = 36.4 mg/L. Choice A miscalculates timing. Choice B ignores exponential decay. Choice C only counts the second dose.

Question 2

Two competing technologies have market shares described by $S_1(t) = \frac{60}{1 + 2e^{-0.3t}}$ and $S_2(t) = 40e^{-0.1t}$ , where $t$ is years. At what time will Technology 1's growth rate equal Technology 2's decay rate in magnitude?

When $t$ satisfies $\frac{36e^{-0.3t}}{(1+2e^{-0.3t})^2} = 4e^{-0.1t}$ , requiring numerical solution methods (correct answer)
At $t = \ln(9) \approx 2.20$ years, where the logistic and exponential derivatives balance
When $t = 10\ln(2) \approx 6.93$ years, using the characteristic time scales of both functions
At $t = 5\ln(3) \approx 5.49$ years, found by equating the instantaneous percentage changes

Explanation:

S_1'(t) = \frac{60 \cdot 0.6e^{-0.3t}}{(1+2e^{-0.3t})^2} = \frac{36e^{-0.3t}}{(1+2e^{-0.3t})^2}

and

S_2'(t) = -4e^{-0.1t}

. For equal magnitudes:

\frac{36e^{-0.3t}}{(1+2e^{-0.3t})^2} = 4e^{-0.1t}

. This transcendental equation cannot be solved analytically and requires numerical methods. Substituting

u = e^{-0.1t}

gives

u^3 = e^{-0.3t}

, but the equation still doesn't simplify to a closed form. Choice B assumes a specific solution without justification. Choice C uses irrelevant time scales. Choice D confuses percentage changes with absolute rate magnitudes.

Question 3

An exponential function $f(x) = ab^x$ passes through points $(2, 12)$ and $(5, 96)$ . If this function models temperature change, and the temperature must remain below 200°C for safety, what is the maximum safe value of $x$ ?

$x = 7$ , since this keeps temperature safely below the limit
$x = 6 + \log_2(25/12)$ , where temperature first reaches exactly 200°C
$x = 2 + \log_2(200/12)$ , solving $ab^x = 200$ directly (correct answer)
$x = 5 + \log_2(25/12)$ , using the second point as reference

Explanation: From the points:

12 = ab^2

and

96 = ab^5

. Dividing gives

8 = b^3

, so

b = 2

. Then

a = 12/4 = 3

. So

f(x) = 3 \cdot 2^x

. For the safety limit:

3 \cdot 2^x = 200

, so

2^x = 200/3

. Thus

x = \log_2(200/3) = \log_2(200/12) + \log_2(4) = 2 + \log_2(200/12)

. For safety,

x

must be less than this value.

Question 4

The concentration of a drug in blood follows $C(t) = 20e^{-0.4t} + 5e^{-0.1t}$ mg/L, where $t$ is hours after injection. At what time is the rate of concentration decrease exactly half of the initial rate?

When $t = \ln(2)/0.4 \approx 1.73$ hours, corresponding to the dominant exponential term's half-life
When $t = \ln(2)/0.25 \approx 2.77$ hours, using the average of the two decay constants
When $t$ satisfies $8e^{-0.4t} + 0.5e^{-0.1t} = 4.25$ , accounting for both exponential components (correct answer)
When $t = 2\ln(2) \approx 1.39$ hours, since the rate halving occurs before the half-life

Explanation:

C'(t) = -8e^{-0.4t} - 0.5e^{-0.1t}

. Initial rate:

C'(0) = -8 - 0.5 = -8.5

. We need

C'(t) = -4.25

, so

-8e^{-0.4t} - 0.5e^{-0.1t} = -4.25

, which gives

8e^{-0.4t} + 0.5e^{-0.1t} = 4.25

. This equation cannot be solved in closed form and requires numerical methods. Choice A only considers the first term. Choice B incorrectly averages decay constants. Choice D uses a simple doubling relationship that doesn't apply to the sum of exponentials.

Question 5

A bacterial culture starts with 500 bacteria and follows the growth model $P(t) = 500 \cdot 2^{0.3t}$ , where $t$ is time in hours. After how many hours will the culture first contain more than 8 times its initial population?

10 hours (correct answer)
8 hours
12 hours
15 hours

Explanation: We need to find when P(t) > 8 × 500 = 4000. Setting up: 500 × 2^(0.3t) > 4000, so 2^(0.3t) > 8. Since 8 = 2³, we have 2^(0.3t) > 2³, which means 0.3t > 3, so t > 10. The culture first exceeds 8 times its initial size after 10 hours. Choice B uses 8 (the multiplier) instead of solving the inequality. Choice C assumes linear growth. Choice D results from incorrectly solving 0.3t = 3 as t = 15.

Question 6

A city's population follows $P(t) = 50000 \cdot 1.02^t$ where $t$ is years since 2020. Climate change projections suggest the growth rate will decrease by 0.3 percentage points starting in 2030. Under this scenario, what will be the population in 2035?

Approximately 58,425 people
Approximately 60,949 people
Approximately 57,152 people
Approximately 59,687 people (correct answer)

Explanation: When you encounter exponential growth problems with changing rates, you need to break the timeline into segments and apply each rate to its respective period. Initially, the population grows at 2% annually with

P(t) = 50000 \cdot 1.02^t

. From 2020 to 2030 (10 years), the population becomes:

P(10) = 50000 \cdot 1.02^{10} = 50000 \cdot 1.2190 = 60,949

people. Starting in 2030, the growth rate decreases by 0.3 percentage points, changing from 2.0% to 1.7% annually. For the final 5 years (2030 to 2035), you apply the new rate to the 2030 population:

P_{2035} = 60949 \cdot 1.017^5 = 60949 \cdot 0.9792 = 59,687

people. Answer A (58,425) likely applies the reduced growth rate for the entire 15-year period, ignoring that the rate change doesn't begin until 2030. Answer B (60,949) represents the population in 2030, not 2035—this is the intermediate calculation, not the final answer. Answer C (57,152) probably compounds the rate reduction incorrectly, perhaps subtracting 0.3% from the population rather than from the growth rate. The correct answer is D (59,687 people). Remember: When growth rates change mid-problem, calculate each time period separately using its corresponding rate. The population at the end of one period becomes the starting value for the next period. Always verify which year the question asks for—intermediate calculations often appear as distractors.

Question 7

A viral video's view count grows according to $V(t) = 1000 \cdot 3^{t/2}$ for the first week, where $t$ is in days. If the growth rate slows to half its original percentage rate after one week, what will be the view count after 2 weeks?

$1000 \cdot 3^{7/2} \cdot (3^{1/4})^7 = 1000 \cdot 3^{14/4} = 1000 \cdot 3^{3.5} \approx 140,296$ views
$1000 \cdot 3^{7/2} \cdot 1.5^7 \approx 46,766 \cdot 17.086 \approx 799,203$ views using modified base
$1000 \cdot 3^{7/2} \cdot 3^{7/4} = 1000 \cdot 3^{21/4} \approx 1000 \cdot 3^{5.25} \approx 243,000$ views (correct answer)
$1000 \cdot 3^7 = 1000 \cdot 2187 = 2,187,000$ views, maintaining original exponential growth

Explanation: After 1 week:

V(7) = 1000 \cdot 3^{7/2} = 1000 \cdot 3^{3.5} \approx 46,766

views. The original growth rate is

\ln(3)/2

per day (since

V(t) = 1000e^{(\ln 3/2)t}

). Half this rate is

\ln(3)/4

. For the second week:

V(t) = V(7) \cdot e^{(\ln 3/4)(t-7)} = 46,766 \cdot 3^{(t-7)/4}

for

7 \leq t \leq 14

. At

t = 14

V(14) = 46,766 \cdot 3^{7/4} = 1000 \cdot 3^{7/2} \cdot 3^{7/4} = 1000 \cdot 3^{21/4}

. Choice A uses wrong exponent arithmetic. Choice B incorrectly changes the base. Choice D ignores the rate change after week 1.

Question 8

A population model shows $P(t) = 1200 \cdot 3^{t/5}$ where $t$ is years since 2020. In what year will the population growth rate (in people per year) first exceed 1000 people per year?

2028
2025
2027 (correct answer)
2030

Explanation: The growth rate is P'(t) = 1200 × 3^(t/5) × ln(3) × (1/5) = 240 ln(3) × 3^(t/5). We need 240 ln(3) × 3^(t/5) > 1000. Since ln(3) ≈ 1.099, this becomes 263.8 × 3^(t/5) > 1000, so 3^(t/5) > 3.79. Taking log: t/5 > log₃(3.79) ≈ 1.24, so t > 6.2. This corresponds to year 2026.2, so 2027. Choice B uses when population reaches 1000. Choice A miscalculates the derivative. Choice D uses the wrong logarithm base.

Question 9

A research team models the spread of an invasive species using the function $S(t) = \frac{1000}{1 + 49e^{-0.4t}}$ , where $S(t)$ represents the number of organisms and $t$ is time in months. Initially, there are 20 organisms present.

The research team also considers an alternative exponential model $E(t) = 20 \cdot r^t$ that matches the logistic model's behavior during the early growth phase. What value of $r$ best approximates the initial exponential growth?

$r \approx 1.33$
$r \approx 1.49$ (correct answer)
$r \approx 1.40$
$r \approx 1.25$

Explanation: When you encounter problems asking you to match exponential and logistic models during their early growth phases, you need to find where their growth rates are equivalent. This requires understanding how both models behave when the population is still small. To find the exponential growth rate

r

, you need to match the derivative (growth rate) of the logistic model at

t = 0

. First, let's verify the initial condition:

S(0) = \frac{1000}{1 + 49e^0} = \frac{1000}{50} = 20

✓ Next, find the derivative of the logistic function:

S'(t) = \frac{1000 \cdot 49 \cdot 0.4 \cdot e^{-0.4t}}{(1 + 49e^{-0.4t})^2}

t = 0

S'(0) = \frac{1000 \cdot 49 \cdot 0.4}{(1 + 49)^2} = \frac{19600}{2500} = 7.84

For the exponential model

E(t) = 20r^t

, the derivative is

E'(t) = 20r^t \ln(r)

, so

E'(0) = 20\ln(r)

. Setting them equal:

20\ln(r) = 7.84

, which gives

\ln(r) = 0.392

, so

r = e^{0.392} \approx 1.48

. Choice B (

r \approx 1.49

) is correct. Choice A (1.33) underestimates the growth rate, likely from calculation errors. Choice C (1.40) is close but represents rounding the natural log incorrectly. Choice D (1.25) significantly underestimates the exponential base, possibly from misunderstanding the relationship between continuous and discrete growth rates. Strategy tip: When matching exponential and logistic models, always equate their derivatives at the starting point—this captures the instantaneous growth rate that defines early exponential behavior.

Question 10

An investment account follows the model $A(t) = 5000(1.08)^t$ for annual compounding. Compare this to continuous compounding at rate $r$ that produces the same value after exactly 2 years. What is the approximate value of $r$ ?

0.077 or about 7.7% (correct answer)
0.080 or exactly 8.0%
0.083 or about 8.3%
0.076 or about 7.6%

Explanation: After 2 years, annual compounding gives 5000(1.08)² = 5000(1.1664) = 5832. For continuous compounding: 5000e^(2r) = 5832, so e^(2r) = 1.1664. Taking ln: 2r = ln(1.1664) ≈ 0.1542, so r ≈ 0.077. Choice B assumes the rates are equal, ignoring the compounding difference. Choice C results from incorrectly using ln(1.08). Choice D comes from computational errors in the logarithm calculation.

Question 11

The radioactive decay of a substance follows $N(t) = N_0 e^{-0.1t}$ , where $t$ is in years. If the half-life is approximately 6.93 years, what does the coefficient -0.1 represent in this context?

The annual percentage rate of decay expressed as a decimal fraction
The fraction of the substance that remains after each time period
The instantaneous rate of decay per unit of original substance per year (correct answer)
The time constant representing the reciprocal of the average lifetime

Explanation: The coefficient -0.1 is the decay constant, representing the instantaneous rate of change per unit of current substance per year. This is the continuous decay rate parameter. Choice A incorrectly treats it as a simple percentage rate. Choice B confuses it with the base of exponential decay. Choice D describes a time constant concept but not specifically what -0.1 represents in this decay model.

Question 12

The function $h(x) = 3 \cdot 4^{x-2} + 5$ represents a transformed exponential. If this function has the same percentage rate of growth as $k(x) = 7 \cdot 2^{bx}$ , what is the value of $b$ ?

$b = 1$
$b = 2$ (correct answer)
$b = \frac{1}{2}$
$b = 4$

Explanation: The percentage growth rate depends on the base of the exponential. For h(x) = 3 × 4^(x-2) + 5, ignoring the vertical shift and horizontal translation, the growth factor per unit is 4. For k(x) = 7 × 2^(bx), the growth factor per unit is 2^b. Setting equal: 2^b = 4 = 2², so b = 2. Choice A gives growth factor 2, not 4. Choice C gives growth factor √2. Choice D gives growth factor 16, much too large.