Advanced Math⚡ High Yield (1-3 Questions per Test)
Exponential Functions
Digital SAT Math Preparation & Desmos Strategies
4 Concepts19 Practice Qs30 Mock Qs⚡ Desmos Speed Hacks
Key Concepts & Worked Archetypes
Concept 1
Concept 1: Standard Exponential Form
An exponential function is represented as $f(x) = a \cdot b^x$, where $a$ represents the initial value and $b$ represents the constant multiplier or growth/decay factor per unit of $x$.
The initial value $a$ is always the y-intercept, found by evaluating $f(0)$.
If $b > 1$, the function models exponential growth.
If $0 < b < 1$, the function models exponential decay.
📘 Traditional Algebraic Method
Substitute two given coordinate points $(x_1, y_1)$ and $(x_2, y_2)$ into $y = a \cdot b^x$ to set up a system of equations, then divide them to solve for $b$ and subsequently $a$.
⚡ SAT Speed Trick & Desmos Hack
Type the table of values or given points directly into Desmos regression syntax: $y_1 \sim a \cdot b^{x_1}$ to immediately extract values for $a$ and $b$.
💡 Worked SAT Archetype Example
Problem: An exponential function passes through the points $(1, 6)$ and $(3, 54)$. What is the value of $f(2)$?
📘 Step-by-Step Textbook Solution:
Step 1
Write the general equation $f(x) = a \cdot b^x$
Step 2
Substitute point $(1, 6)$: $6 = a \cdot b^1$
Step 3
Substitute point $(3, 54)$: $54 = a \cdot b^3$
Step 4
Divide the second equation by the first: $\frac{54}{6} = \frac{a b^3}{a b}$
Step 5
Simplify to find $b^2$: $b^2 = 9$
Step 6
Solve for positive base $b$: $b = 3$
Step 7
Substitute $b = 3$ back into the first equation: $6 = a(3)$
Step 8
Solve for $a$: $a = 2$
Step 9
Construct the function: $f(x) = 2(3)^x$
Step 10
Evaluate for $f(2)$: $f(2) = 2(3)^2$
Step 11
Final calculation: $f(2) = 18$
⚡ Speed / Desmos Tactic:
Step 1
Open Desmos and create a table with columns $x_1$ and $y_1$
Step 2
Enter points $(1, 6)$ and $(3, 54)$ into the table
Step 3
Type the regression command: $y_1 \sim a \cdot b^{x_1}$
Step 4
Read off parameters $a = 2$ and $b = 3$
Step 5
Evaluate $2 \cdot 3^2 = 18$
Concept 2
Concept 2: Percentage Growth and Decay
Real-world contexts apply percentage changes over fixed intervals, transforming the base into $b = 1 \pm \frac{r}{100}$.
For annual percentage increase of $r\%$, use $b = 1 + \frac{r}{100}$.
For annual percentage decrease of $r\%$, use $b = 1 - \frac{r}{100}$.
If time units are scaled, adjust the exponent accordingly as $\frac{t}{k}$.
📘 Traditional Algebraic Method
Identify the initial amount, convert the percentage rate to a decimal multiplier, and construct the model $P(t) = P_0(1 \pm r)^t$.
⚡ SAT Speed Trick & Desmos Hack
Define $f(x) = P_0(1 \pm r)^x$ inside Desmos and use the evaluation feature or trace along the curve to find target values.
💡 Worked SAT Archetype Example
Problem: The value of a car depreciates by $15\%$ each year. If its initial purchase price was $\$25,000$, what is its value after 4 years?
📘 Step-by-Step Textbook Solution:
Step 1
Identify initial value $P_0 = 25000$
Step 2
Identify decay rate $r = 0.15$
Step 3
Formulate decay factor $b = 1 - 0.15 = 0.85$
Step 4
Write model equation: $V(t) = 25000(0.85)^t$
Step 5
Substitute $t = 4$: $V(4) = 25000(0.85)^4$
Step 6
Compute power: $(0.85)^4 = 0.52200625$
Step 7
Multiply by initial value: $V(4) = 25000 \cdot 0.52200625$
Step 8
Final calculation: $V(4) = 13050.16$
⚡ Speed / Desmos Tactic:
Step 1
Open Desmos and type: $V(t) = 25000(0.85)^t$
Step 2
Type on a new line: $V(4)$
Step 3
Read output directly: $13050.16$
Concept 3
Concept 3: Equivalent Exponential Expressions
Rewriting exponential expressions to reveal hidden features, such as changing monthly growth rates to annual growth rates using exponent rules.
Product rule: $b^m \cdot b^n = b^{m+n}$
Power of a power rule: $(b^m)^n = b^{m \cdot n}$
Negative exponent rule: $b^{-n} = \frac{1}{b^n}$
📘 Traditional Algebraic Method
Apply algebraic exponent properties step-by-step to match the target form given in the multiple-choice options.
⚡ SAT Speed Trick & Desmos Hack
Graph the original expression in Desmos as $f(x)$, then test each option by graphing it; the correct option will perfectly overlap (cover) the original graph.
💡 Worked SAT Archetype Example
Problem: Which of the following is equivalent to $f(t) = 3^{2t+1}$?
📘 Step-by-Step Textbook Solution:
Step 1
Apply addition rule for exponents: $3^{2t+1} = 3^{2t} \cdot 3^1$
Step 2
Rewrite $3^{2t}$ using power property: $(3^2)^t$
Step 3
Simplify base: $3^2 = 9$
Step 4
Combine parts to form final expression: $3 \cdot 9^t$
⚡ Speed / Desmos Tactic:
Step 1
Enter original function into Desmos: $f(x) = 3^{2x+1}$
Step 2
Enter choice A: $9(3^x)$ (observe no overlap)
Step 3
Enter choice B: $3(9^x)$ (observe exact color overlap with $f(x)$)
Step 4
Select option matching choice B
Concept 4
Concept 4: Half-Life and Doubling Time
Special exponential applications where quantities halve or double over constant time intervals.
Half-life formula: $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$, where $h$ is the half-life period.
Doubling time formula: $N(t) = N_0 (2)^{\frac{t}{d}}$, where $d$ is the doubling period.
Ensure units of time for $t$ and $h$ (or $d$) match perfectly.
📘 Traditional Algebraic Method
Set up the structured half-life or doubling equation, isolate the exponential term, and use logarithms or logical exponent matching if powers are integers.
⚡ SAT Speed Trick & Desmos Hack
Graph the function in Desmos and use the intersection feature with a horizontal line equal to the target amount to find $t$.
💡 Worked SAT Archetype Example
Problem: A radioactive substance has a half-life of 3 years. If there are initially 80 grams, how many grams remain after 9 years?
📘 Step-by-Step Textbook Solution:
Step 1
Identify initial amount $N_0 = 80$
Step 2
Identify half-life $h = 3$
Step 3
Identify elapsed time $t = 9$
Step 4
Write the half-life equation: $N(9) = 80 \left(\frac{1}{2}\right)^{\frac{9}{3}}$
Open Desmos and type: $N(t) = 80(0.5)^{\frac{t}{3}}$
Step 2
Type on a new line: $N(9)$
Step 3
Read output directly: $10$
Practice Questions (19)
Question 1Evaluating Exponential Functions
Easy
A bacteria culture grows according to the function $f(t) = 300(1.5)^t$, where $t$ is the time in hours. What is the population of the bacteria after $2$ hours?
Hint: Substitute the given time value for $t$ directly into the exponential function.
📘 Step-by-Step Algebraic Solution
Step 1
Identify the given function $f(t) = 300(1.5)^t$ and the given time $t = 2$.
Step 2
Substitute $t = 2$ into the function: $f(2) = 300(1.5)^2$.
Step 3
Evaluate the exponent: $(1.5)^2 = 2.25$.
Step 4
Multiply by the initial amount: $300 \times 2.25 = 675$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Type 'f(t) = 300(1.5)^t' into Desmos.
Step 2
Type 'f(2)' on the next line to instantly see the output.
Question 2Evaluating Exponential Functions
Easy
The value of a car is modeled by the function $V(t) = 24,000(0.8)^t$, where $t$ is the age of the car in years. What is the value of the car when it is $1$ year old?
Hint: Plug $t = 1$ into the value function.
📘 Step-by-Step Algebraic Solution
Step 1
Identify the equation $V(t) = 24,000(0.8)^t$ and set $t = 1$.
Step 2
Substitute $t = 1$: $V(1) = 24,000(0.8)^1$.
Step 3
Calculate the product: $24,000 \times 0.8 = 19,200$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Open Desmos and type '24000(0.8)^1'.
Step 2
Read the result $19,200$ immediately.
Question 3Evaluating Exponential Functions
Medium
An investment of $\$5,000$ grows according to the function $A(t) = 5000(1.06)^{t/2}$, where $t$ is the time in years. What is the value of the investment after $4$ years?
Hint: Substitute $t = 4$ into the exponent expression $t/2$ first.
📘 Step-by-Step Algebraic Solution
Step 1
Set $t = 4$ in the function $A(t) = 5000(1.06)^{t/2}$.
Step 2
Simplify the exponent: $4 / 2 = 2$.
Step 3
Compute the base power: $(1.06)^2 = 1.1236$.
Step 4
Multiply by $5,000$: $5000 \times 1.1236 = 5,618.00$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Enter '5000(1.06)^(4/2)' directly into Desmos.
Step 2
Read off the final value $5618$.
Question 4Evaluating Exponential Functions
Medium
The population of a town is given by $P(t) = 12,000(2)^{t/10}$, where $t$ is the number of years from the present. How many years will it take for the population to reach $24,000$?
Hint: Set $P(t) = 24,000$ and solve the exponential equation for $t$.
📘 Step-by-Step Algebraic Solution
Step 1
Set up the equation $24000 = 12000(2)^{t/10}$.
Step 2
Divide both sides by $12,000$: $2 = 2^{t/10}$.
Step 3
Equate the exponents since bases are equal: $1 = t/10$.
Step 4
Solve for $t$: $t = 10$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Type 'y = 12000(2)^(x/10)' and 'y = 24000' in Desmos.
Step 2
Click the intersection point to find $x = 10$.
Question 5Evaluating Exponential Functions
Hard
The mass $M$ (in grams) of a radioactive isotope remaining after $t$ days is given by $M(t) = 800(e)^{-0.05t}$. If the mass remaining is $400$ grams, which of the following represents the exact number of days $t$ that have passed?
Hint: Set $M(t) = 400$, isolate the exponential term, and take the natural logarithm of both sides.
📘 Step-by-Step Algebraic Solution
Step 1
Set $400 = 800e^{-0.05t}$.
Step 2
Divide by $800$: $0.5 = e^{-0.05t}$.
Step 3
Take natural log of both sides: $\ln(0.5) = -0.05t$.
Step 4
Rewrite $0.5$ as $1/2$: $\ln(1/2) = -\ln(2) = -0.05t$.
Step 5
Solve for $t$: $t = \frac{-\ln(2)}{-0.05} = 20\ln(2)$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Define expressions in Desmos as functions of $x$, e.g., 'f(x) = 800e^(-0.05x)'.
Step 2
Evaluate the given options numerically (e.g., $20\ln(2) \approx 13.86$) and check which one satisfies $f(x) = 400$.
Question 6Interpreting Exponential Expressions
Easy
The function $f(x) = 450(1.08)^x$ models the number of social media followers a page has, where $x$ represents the number of months since the page was created. What does the number $1.08$ represent in this context?
Hint: In an exponential function $y = a(b)^x$, $b$ is the growth factor per unit of $x$.
📘 Step-by-Step Algebraic Solution
Step 1
Identify the standard exponential growth model $f(x) = a(b)^x$.
Step 2
Recognize that $a = 450$ is the initial value (starting followers).
Step 3
Recognize that $b = 1.08$ is the base, representing the multiplier for each one-unit increase in $x$ (each month).
⚡ Desmos Shortcut / Speed Hack
Step 1
Recall the general form $y = a(1 + r)^x$ or $y = a(b)^x$.
Step 2
Match $b$ directly to 'growth factor / multiplier per time period'.
Question 7Interpreting Exponential Expressions
Easy
A certain compound decays according to the function $Q(t) = 1,200(0.85)^t$, where $t$ is measured in hours. Which of the following best describes the hourly change in the compound's quantity?
Hint: Relate the decay factor $0.85$ to $1 - r$, where $r$ is the percentage decrease.
📘 Step-by-Step Algebraic Solution
Step 1
The decay factor is $b = 0.85$.
Step 2
Set up the relation $b = 1 - r$, where $r$ is the rate of decrease.
Step 3
$0.85 = 1 - r \implies r = 0.15$.
Step 4
Convert to percentage: $0.15 = 15\%$ decrease.
⚡ Desmos Shortcut / Speed Hack
Step 1
Subtract the base from 1: $1 - 0.85 = 0.15$.
Step 2
Convert to a percentage: $15\%$ reduction.
Question 8Interpreting Exponential Expressions
Medium
The population of a species of bird is modeled by $P(t) = 500(1.04)^{4t}$, where $t$ is the time in years. Which of the following expressions is equivalent to $P(t)$ highlighting the *quarterly* growth factor?
Hint: Use exponent rules like $(a^b)^c = a^{bc}$ to rewrite the base and exponent.
📘 Step-by-Step Algebraic Solution
Step 1
Start with the given function $P(t) = 500(1.04)^{4t}$.
Step 2
Apply the power rule of exponents: $(x^a)^b = x^{ab}$.
Step 3
Rewrite $(1.04)^{4t}$ as $((1.04)^4)^t$.
Step 4
Conclude that the annual growth factor is $(1.04)^4 \approx 1.1699$, leading to $500(1.04^4)^t$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Check the structure of options. We want an expression in terms of $t$ in the exponent.
Step 2
$(1.04)^{4t} = ((1.04)^4)^t$, which directly matches option C.
Question 9Interpreting Exponential Expressions
Medium
The value of an account is represented by $A(m) = 4,000(1.005)^{12m}$, where $m$ is the number of years. Which of the following interpretations is correct regarding the expression?
Hint: The exponent is $12m$. If $m$ is years, $12m$ represents months.
📘 Step-by-Step Algebraic Solution
Step 1
Identify that $m$ represents years, so $12m$ represents the total number of months.
Step 2
The base inside is $1.005$, which equals $1 + 0.005$.
Step 3
$0.005$ as a percentage is $0.5\%$.
Step 4
Therefore, it grows by $0.5\%$ each month ($12$ times a year).
⚡ Desmos Shortcut / Speed Hack
Step 1
Note that $12m$ suggests monthly compounding over $m$ years.
The population of a city doubles every $15$ years. If the initial population is $P_0$, the population $t$ years from now can be modeled by $P(t) = P_0(2)^{kt}$. What is the value of $k$?
Hint: Use the condition that when $t = 15$, the population is $2P_0$.
📘 Step-by-Step Algebraic Solution
Step 1
Set up the doubling condition: $P(15) = 2P_0$.
Step 2
Substitute into the model: $2P_0 = P_0(2)^{k(15)}$.
Step 3
Divide both sides by $P_0$: $2 = 2^{15k}$.
Step 4
Equate exponents: $1 = 15k \implies k = \frac{1}{15}$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Doubling time models always take the form $2^{t / \text{doubling time}}$.
Step 2
Here the exponent is $t/15$, which is equivalent to $\frac{1}{15}t$. Thus $k = 1/15$.
Question 11Constructing Exponential Equations
Easy
A digital asset starts with $500$ users and triples every year. Which function $f(t)$ models the number of users after $t$ years?
Hint: The general form is $f(t) = \text{initial} \times (\text{growth factor})^t$.
📘 Step-by-Step Algebraic Solution
Step 1
Identify the initial value (starting users) $a = 500$.
Step 2
Identify the growth factor (triples means multiply by $3$) $b = 3$.
Step 3
Substitute into $f(t) = a(b)^t$ to get $f(t) = 500(3)^t$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Initial value is $500$, base is $3$.
Step 2
Immediately pick $500(3)^t$.
Question 12Constructing Exponential Equations
Easy
An office printer loses $20\%$ of its resale value each year. If it was purchased for $\$1,200$, which equation represents its value $V$ after $t$ years?
Hint: A loss of $20\%$ means retaining $80\%$ of the value each year.
📘 Step-by-Step Algebraic Solution
Step 1
Initial value $a = 1,200$.
Step 2
Decay rate $r = 20\% = 0.20$.
Step 3
Decay factor $b = 1 - r = 1 - 0.20 = 0.80$.
Step 4
Construct the function: $V = 1200(0.8)^t$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Retained value is $100\% - 20\% = 80\% = 0.8$.
Step 2
Match with initial value $1200$ to get $1200(0.8)^t$.
Question 13Constructing Exponential Equations
Medium
A culture of mold increases by $75\%$ every $3$ hours. If the initial count is $200$ spores, which function models the number of spores $N$ after $t$ hours?
Hint: The growth factor occurs every $3$ hours, so the exponent should be $t/3$.
📘 Step-by-Step Algebraic Solution
Step 1
Initial count $a = 200$.
Step 2
Growth factor for a $75\%$ increase is $1 + 0.75 = 1.75$.
Step 3
Since this growth happens every $3$ hours, divide time $t$ by $3$.
Step 4
Combine into $N(t) = 200(1.75)^{t/3}$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Look for time divided by the period of growth ($t/3$).
Step 2
Confirm base is $1.75$ and initial value is $200$.
Question 14Constructing Exponential Equations
Medium
A table passes through the points $(0, 3)$ and $(2, 48)$ and represents an exponential function of the form $f(x) = a(b)^x$. What is the function?
Hint: Use the point $(0, 3)$ to find $a$, then use $(2, 48)$ to find $b$.
📘 Step-by-Step Algebraic Solution
Step 1
Substitute $(0,3)$ into $f(x) = a(b)^x$: $3 = a(b)^0 \implies a = 3$.
Step 2
Substitute $(2,48)$ and $a = 3$: $48 = 3(b)^2$.
Step 3
Divide by $3$: $16 = b^2$.
Step 4
Solve for $b$ (growth factor must be positive): $b = 4$.
Step 5
Write the function: $f(x) = 3(4)^x$.
⚡ Desmos Shortcut / Speed Hack
Step 1
Test options with $(0,3)$: all give $3$.
Step 2
Test options with $(2,48)$: $3(4)^2 = 3(16) = 48$ matches.
Question 15Constructing Exponential Equations
Hard
An exponential function $f(x) = ab^x$ passes through the points $(1, 6)$ and $(3, 54)$. What is the value of $f(4)$?
Hint: Find the ratio $f(3)/f(1)$ to determine $b^2$, then find $b$ and $a$.
📘 Step-by-Step Algebraic Solution
Step 1
Set up equations: $ab^1 = 6$ and $ab^3 = 54$.
Step 2
Divide the second equation by the first: $\frac{ab^3}{ab^1} = \frac{54}{6}$.
Function $F$ is defined by $F(x) = 3(2)^x$ and function $G$ is defined by $G(x) = 2(3)^x$. For what value of $x$ do $F(x)$ and $G(x)$ have the same value?
Hint: Set $3(2)^x = 2(3)^x$ and take logarithms of both sides to solve for $x$.
📘 Step-by-Step Algebraic Solution
Step 1
Set $3(2)^x = 2(3)^x$.
Step 2
Divide both sides by $2 \cdot 2^x$: $\frac{3}{2} = \frac{3^x}{2^x} = \left(\frac{3}{2}\right)^x$.
Step 3
Wait, let's rewrite carefully: divide by $3^x$ and $3$: $\frac{2^x}{3^x} = \frac{2}{3} \implies \left(\frac{2}{3}\right)^x = \left(\frac{2}{3}\right)^1$.
Analysis of College Board question patterns and recent exam distributions.
🏛️
Official SAT PYQ Drill Bank (2023–2026)
We are compiling real verified College Board exam patterns with step-by-step Desmos shortcuts for Exponential Functions.
Updated for 2026 Testing Season
Revision, Traps & Desmos Syntax
General Exponential Growth/Decay
$f(x) = a \cdot b^x$
Used for standard point-based or table-based exponential modeling.
Percentage Growth/Decay Model
$P(t) = P_0 (1 \pm \frac{r}{100})^t$
Used for word problems involving annual or periodic percentage change.
Half-Life and Doubling Time
$A(t) = A_0 (2)^{\pm \frac{t}{k}}$
Used when quantities scale by factors of 2 over fixed duration $k$.
🚨 Top SAT Traps & Misconceptions
⚠️ SAT Trap: Confusing Growth Rate with Growth Factor
If a problem states a value increases by $8\%$, students often mistakenly use $b = 0.08$ instead of the correct growth factor $b = 1.08$.
⚠️ SAT Trap: Mismatched Time Units
Failing to convert rates when a model gives an annual rate but asks for monthly accumulation, leading to incorrect exponent scaling.
⚡ Essential Desmos Cheatsheet
🎯 Exponential Regression
y1 ~ a * b^x1
Instantly find exponential functions from a table of values or two coordinate points.
🎯 Graph Overlap Verification
f(x) = [original], g(x) = [option]
Verify equivalent expressions by checking curve superimposition in Desmos.
3-Level Mock Test (30 Questions)
🟢 Level 1: Foundation
10 Qs · Sub-600 Score
🟡 Level 2: Target 700+
10 Qs · 600–740 Score
🔴 Level 3: 800-Mastery
10 Qs · 750–800 Score
Question 1Level 1: Foundation
What is the initial value of the exponential function represented by $f(x) = 3(2)^x$?
Explanation:
Step 1
Recall the standard form of an exponential function, $f(x) = a(b)^x$, where $a$ represents the initial value and $b$ represents the base or growth factor.
Step 2
Identify the coefficient multiplying the exponential term in the given function $f(x) = 3(2)^x$.
Step 3
Conclude that the initial value $a$ is $3$, corresponding to option B.
Question 2Level 1: Foundation
An exponential function is given by $g(x) = 5(4)^x$. What is the value of $g(2)$?
Explanation:
Step 1
Substitute $x = 2$ into the given function $g(x) = 5(4)^x$.
Step 2
Evaluate the exponent: $4^2 = 16$.
Step 3
Multiply by the initial coefficient: $5(16) = 80$, which matches option B.
Question 3Level 1: Foundation
Which of the following functions represents exponential growth?
Explanation:
Step 1
Understand that an exponential function $f(x) = a(b)^x$ represents growth when the base $b$ is greater than $1$.
Step 2
Analyze the bases of the given choices: choice A has $b = 0.8$ (decay), choice B has $b = 1$ (neither), choice C has $b = 1.05$ (growth), and choice D is linear.
Step 3
Therefore, $f(x) = 2(1.05)^x$ represents exponential growth, matching option C.
Question 4Level 1: Foundation
The value of a car depreciates according to the function $V(t) = 25000(0.85)^t$, where $t$ is the time in years. What is the annual percent decrease?
Explanation:
Step 1
Recall that a decay function is written as $V(t) = a(1 - r)^t$, where $r$ is the rate of decrease.
Step 2
Set the decay factor equal to $1 - r$: $1 - r = 0.85$.
Step 3
Solve for $r$: $r = 1 - 0.85 = 0.15$, which is equivalent to a $15\%$ decrease, matching option B.
Question 5Level 1: Foundation
What is the $y$-intercept of the graph of $y = 7(3)^x + 2$ in the $xy$-plane?
Explanation:
Step 1
Find the $y$-intercept by setting $x = 0$ in the equation $y = 7(3)^x + 2$.
Express the $y$-intercept as a coordinate pair: $(0, 9)$, matching option B.
Question 6Level 1: Foundation
Which of the following equations represents a horizontal asymptote for the graph of $f(x) = 4(0.5)^x - 3$?
Explanation:
Step 1
Recognize that for an exponential function of the form $f(x) = a(b)^x + c$, the horizontal asymptote is determined by the vertical shift $c$.
Step 2
As $x$ approaches infinity or negative infinity, the term $4(0.5)^x$ approaches $0$.
Step 3
Thus, $f(x)$ approaches $-3$, making the horizontal asymptote $y = -3$, which corresponds to option B.
Question 7Level 1: Foundation
If $2^x = 32$, what is the value of $x$?
Explanation:
Step 1
Rewrite both sides of the equation with the same base: $2^x = 2^5$.
Step 2
Use the property of equality for exponential functions: if $b^x = b^y$, then $x = y$.
Step 3
Therefore, $x = 5$, matching option C.
Question 8Level 1: Foundation
An initial population of $500$ bacteria grows at a rate of doubling every hour. Which function models the population $P$ after $h$ hours?
Explanation:
Step 1
Identify the initial amount, which is $500$.
Step 2
Identify the growth factor for doubling, which is $2$ per hour $h$.
Step 3
Combine these into the exponential model $P(h) = 500(2)^h$, matching option A.
Question 9Level 1: Foundation
Which of the following points lies on the graph of the exponential function $f(x) = 3(2)^x$?
Explanation:
Step 1
Test the coordinates from option A in the function: $f(1) = 3(2)^1 = 6$.
Step 2
Since the output matches the $y$-coordinate ($6$), the point $(1, 6)$ is on the graph.
Step 3
Verify other options if necessary; for example, $f(2) = 3(4) = 12 \neq 18$, confirming option A is correct.
Question 10Level 1: Foundation
What is the multiplier for an exponential function representing a $12\%$ increase?
Explanation:
Step 1
Recall that growth factor is calculated as $1 + r$, where $r$ is the percentage written as a decimal.
Step 2
Convert $12\%$ to a decimal: $r = 0.12$.
Step 3
Add to $1$: $1 + 0.12 = 1.12$, matching option B.
Question 1Level 2: Target 700+
The population of a town is modeled by the function $P(t) = 12000(1.03)^{t/2}$, where $t$ is the number of years since 2010. Which of the following best describes the meaning of the constant $1.03$ in this context?
Explanation:
Step 1
Examine the exponent of the growth factor, which is given as $t/2$.
Step 2
Recognize that $t$ represents the number of years, so $t/2$ represents the number of 2-year periods.
Step 3
The base $1.03$ corresponds to a $3\%$ increase per unit of the exponent, which is every 2 years, matching option B.
Question 2Level 2: Target 700+
An exponential function $f(x) = ab^x$ passes through the points $(1, 6)$ and $(3, 54)$. What is the value of $b$, assuming $b > 0$?
Explanation:
Step 1
Set up a system of equations using the given points: $ab^1 = 6$ and $ab^3 = 54$.
Step 2
Divide the second equation by the first equation: $\frac{ab^3}{ab^1} = \frac{54}{6}$.
Step 3
Simplify to solve for $b$: $b^2 = 9 \implies b = 3$ since $b > 0$, matching option B.
Question 3Level 2: Target 700+
Which of the following functions is equivalent to $f(x) = 3^{2x + 1}$?
Explanation:
Step 1
Use exponent rules to split the addition in the exponent: $3^{2x + 1} = 3^{2x} \cdot 3^1$.
Step 2
Rewrite $3^{2x}$ using power of a power property: $(3^2)^x = 9^x$.
Step 3
Combine terms to get $3(9^x)$, which corresponds to option B.
Question 4Level 2: Target 700+
The half-life of a radioactive substance is $15$ years. If the initial mass is $200$ grams, which function models the remaining mass $M$ after $t$ years?
Explanation:
Step 1
Recall the half-life exponential decay model: $M(t) = a(0.5)^{t/h}$, where $a$ is the initial mass and $h$ is the half-life.
Step 2
Substitute the given values $a = 200$ and $h = 15$ into the formula.
Step 3
Obtain $M(t) = 200(0.5)^{t/15}$, matching option A.
Question 5Level 2: Target 700+
If $4^{x-1} = 8^{x+2}$, what is the value of $x$?
Explanation:
Step 1
Express both bases $4$ and $8$ as powers of $2$: $4 = 2^2$ and $8 = 2^3$.
Step 2
Substitute these into the equation: $(2^2)^{x-1} = (2^3)^{x+2} \implies 2^{2(x-1)} = 2^{3(x+2)}.$
Step 3
Equate the exponents and solve: $2x - 2 = 3x + 6 \implies x = -8$, matching option A.
Question 6Level 2: Target 700+
The function $f(t) = 450(0.82)^{t}$ models the number of active users on a platform, where $t$ is measured in months. By what percentage does the number of active users decrease each month?
Explanation:
Step 1
Identify the decay factor $b = 0.82$ from the exponential function.
Step 2
Set up the equation relating the decay factor to the rate of decrease: $1 - r = 0.82$.
Step 3
Solve for $r$: $r = 1 - 0.82 = 0.18$, which equals an $18\%$ decrease, matching option B.
Question 7Level 2: Target 700+
Which of the following describes the transformation from the graph of $f(x) = 2^x$ to the graph of $g(x) = 2^{x-3} + 4$?
Explanation:
Step 1
Analyze the replacement of $x$ with $x-3$, which causes a horizontal shift of $3$ units to the right.
Step 2
Analyze the addition of $4$ outside the function, which causes a vertical shift of $4$ units upward.
Step 3
Combine these transformations to conclude it is shifted right $3$ units and up $4$ units, matching option A.
Question 8Level 2: Target 700+
An investment grows according to the model $A(t) = 5000(1.06)^t$. Approximately how many years will it take for the investment to double?
Explanation:
Step 1
Set the investment amount to twice the initial value: $10000 = 5000(1.06)^t$.
Step 2
Simplify to $2 = (1.06)^t$.
Step 3
Test values or use logarithms: $(1.06)^{12} \approx 2.01$, so it takes approximately $12$ years, matching option C.
Question 9Level 2: Target 700+
If $3^{x} + 3^{x} + 3^{x} = 243$, what is the value of $x$?
Explanation:
Step 1
Combine like terms on the left side: $3(3^x) = 243$.
Step 2
Divide both sides by $3$: $3^x = \frac{243}{3} = 81$.
Step 3
Rewrite $81$ as a power of $3$: $3^x = 3^4$, so $x = 4$, matching option B.
Question 10Level 2: Target 700+
The population of a colony of insects triples every 4 days. If the initial population is $N_0$, which expression represents the population after $d$ days?
Explanation:
Step 1
Identify the initial value $N_0$ and the growth factor multiplier $3$ for every 4-day period.
Step 2
Construct the exponent by dividing the total time $d$ by the duration of the growth period, giving $d/4$.
Step 3
Combine into the function $N_0(3)^{d/4}$, matching option B.
Question 1Level 3: 800 Mastery
The value of an asset is given by $V(t) = P(1 + r)^t$. If the asset's value triples every $8$ years, which of the following represents the annual growth rate $r$?
Explanation:
Step 1
Set the condition that the value triples over 8 years: $P(1+r)^8 = 3P$.
Step 2
Divide both sides by $P$ to get $(1+r)^8 = 3$.
Step 3
Solve for $1+r$ and then $r$: $1+r = 3^{1/8} \implies r = 3^{1/8} - 1$, matching option A.
Question 2Level 3: 800 Mastery
Let $f(x) = 5^x$. If $f(a + b) = 40$ and $f(a - b) = 10$, what is the value of $5^{2a}$?
Explanation:
Step 1
Translate the given conditions into exponential equations: $5^{a+b} = 40$ and $5^{a-b} = 10$.
Step 2
Multiply the two equations together: $5^{a+b} \cdot 5^{a-b} = 40 \cdot 10$.
Step 3
Simplify using exponent laws: $5^{2a} = 400$, wait, $40 \times 10 = 400$, matching option C.
Question 3Level 3: 800 Mastery
If $9^x - 3^{x} - 6 = 0$, what is the value of $3^x$?
Explanation:
Step 1
Rewrite the equation in terms of $u = 3^x$, noting that $9^x = (3^2)^x = (3^x)^2 = u^2$.
Step 2
Substitute $u$ into the equation: $u^2 - u - 6 = 0$.
Step 3
Factor the quadratic equation: $(u - 3)(u + 2) = 0$, yielding $u = 3$ or $u = -2$. Since $3^x > 0$, $u = 3$, matching option B.
Question 4Level 3: 800 Mastery
An exponential function is defined by $f(x) = ab^x$. If $f(2) = 12$ and $f(5) = 96$, what is the value of $a + b$?
Explanation:
Step 1
Set up the equations based on the given points: $ab^2 = 12$ and $ab^5 = 96$.
Step 2
Divide the equations to find $b$: $\frac{ab^5}{ab^2} = \frac{96}{12} \implies b^3 = 8 \implies b = 2$.
Step 3
Substitute $b = 2$ back into $ab^2 = 12$ to find $a$: $a(4) = 12 \implies a = 3$. Thus, $a + b = 3 + 2 = 5$, matching option B.
Question 5Level 3: 800 Mastery
If $\left(\frac{1}{2}\right)^{3x-1} = 8^{x+2}$, what is the value of $x$?
Explanation:
Step 1
Express both sides with a base of $2$: $\left(\frac{1}{2}\right)^{3x-1} = (2^{-1})^{3x-1} = 2^{-3x+1}$ and $8^{x+2} = (2^3)^{x+2} = 2^{3x+6}$.
Step 2
Equate the exponents since the bases are equal: $-3x + 1 = 3x + 6$.
Step 3
Solve for $x$: $6x = -5 \implies x = -\frac{5}{6}$, wait, re-evaluate: $-3x - 3x = 6 - 1 \implies -6x = 5 \implies x = -\frac{5}{6}$ which is option A.
Question 6Level 3: 800 Mastery
The population of bacteria in a dish doubles every $3$ hours. If there are $N$ bacteria initially, after how many hours will the population be $32N$?
Explanation:
Step 1
Set up the population model: $P(t) = N(2)^{t/3}$.
Step 2
Set the population equal to $32N$: $N(2)^{t/3} = 32N \implies 2^{t/3} = 32$.
Step 3
Rewrite $32$ as $2^5$ and solve for $t$: $\frac{t}{3} = 5 \implies t = 15$, matching option B.
Question 7Level 3: 800 Mastery
If $2^{x+2} + 2^x = 320$, what is the value of $x$?
Explanation:
Step 1
Factor the expression on the left side: $2^x(2^2 + 1) = 320$.