Math II · A-SSE.3.c

Using Exponent Properties to Transform Exponential Expressions

This objective teaches students how to translate exponential models into the time scale people actually care about. Annual growth can be rewritten as monthly growth, hourly decay can be rewritten as daily decay, and a confusing formula can be turned into a readable story about repeated multiplication.

Concept Algebra
Domain Seeing Structure in Expressions
Read time 9 minutes

What this learning objective is really asking you to learn

An exponential expression describes repeated multiplication. In a model like \(A(t) = 500(1.08)^t\), the starting amount is 500, the growth factor is 1.08, and the exponent tells how many times the growth factor is applied. If \(t\) is measured in years, the expression says the quantity is multiplied by 1.08 once per year. That means 8 percent growth per year. But real questions often use different time units. A family might want to know monthly growth. A scientist might want daily decay. A business might want quarterly compounding. This objective teaches students how to rewrite exponential expressions so the model speaks the right time language.

The key is that exponent rules are not just symbolic tricks. They encode the logic of repeated multiplication. The rule \((a^m)^n = a^{mn}\) says that if you repeat a repeated multiplication, the total number of repetitions multiplies. The rule \(a^{m+n} = a^m a^n\) says that growth over a combined time interval equals growth over the first interval multiplied by growth over the second. The rule \(a^{m-n} = a^m / a^n\) says that reversing growth means dividing by the factor. These rules are the machinery that lets one exponential expression be translated into another.

For example, the expression \(1.15^t\) represents a 15 percent growth factor per year if \(t\) is in years. To express the same growth using monthly steps, we need a monthly factor that, when applied 12 times, gives the annual factor 1.15. That monthly factor is \(1.15^{1/12}\). Therefore:

\[1.15^t = (1.15^{1/12})^{12t}\].

This is not a new model. It is the same model written with a different clock. Since \(1.15^{1/12}\) is about 1.0117, the model has an equivalent monthly growth factor of about 1.0117, or about 1.17 percent per month. A student who reads only the original expression sees annual growth. A student who rewrites it sees monthly growth. Same mathematics, clearer interpretation.

The objective also applies to decay. Suppose a medicine level is modeled by \(M(t) = 80(0.5)^t\), where \(t\) is measured in half-lives. If one half-life is 6 hours, then after \(h\) hours the number of half-lives is \(h/6\), so the model can be written as \(M(h) = 80(0.5)^{h/6}\). If students want a one-hour decay factor, they can rewrite this as \(80(0.5^{1/6})^h\). The quantity \(0.5^{1/6}\) is the factor that remains each hour.

This kind of rewriting answers the question, “What does the base mean?” The base of an exponential expression is not meaningful by itself unless we know the exponent unit. A base of 1.02 might mean 2 percent growth per month, per day, per year, or per generation. The same number can tell a completely different story depending on the time unit. This objective trains students to connect base and exponent instead of reading either one alone.

Why students should learn this math

Exponential growth and decay are everywhere in adult life. Savings accounts, credit cards, loans, inflation, population growth, depreciation, medicine dosage, radioactive decay, cooling, infection spread, social media growth, and technology adoption all involve repeated percentage change. People who cannot interpret exponential expressions are easy to mislead. A small percent per month can become a large percent per year. A small daily decay can create a large long-term drop. A financial advertisement may quote one rate while compounding happens on another schedule. Understanding exponent transformations gives students a practical defense against confusion.

Consider debt. A monthly interest rate of 2 percent may sound small. But the annual growth factor is \(1.02^12\), which is about 1.268. That means about 26.8 percent annual growth if unpaid. The expression \(1.02^12\) is a warning. It turns a harmless-looking monthly number into a yearly consequence. Conversely, an annual growth rate can be converted into a monthly factor so students can understand what happens between yearly statements.

The same issue appears in health and science. A medicine that decays by a fixed percentage every hour may stay in the body much longer than a simple subtraction model would suggest. A population that grows by a fixed percentage each year may accelerate dramatically. A radioactive substance described by a half-life uses exponential decay, not linear decline. If students understand exponent properties, they can translate these descriptions into useful formulas.

This objective also matters because exponential expressions can look different while describing the same relationship. \(100(1.12)^t\), \(100(1.0095)^{12t}\), and \(100(1.12^{1/365})^{365t}\) can represent the same annual growth viewed yearly, monthly, or daily. Students who only match surface appearance may think these are different models. Students who understand exponent laws can recognize equivalence and interpret each form.

There is also a deeper reason. Algebra is often taught as if rewriting is about simplification. But in modeling, rewriting is about revelation. The “simplest” expression is not always the most useful expression. A yearly expression may be simple, but a monthly expression may answer the real question. A model written with a half-life may be scientifically natural, while a model written with an hourly factor may be easier for scheduling. Students need to learn that mathematical form is chosen for purpose.

The historical machinery behind exponential rewriting

Exponential thinking grew from repeated multiplication, but its full power developed slowly. Ancient mathematicians understood powers in geometric and arithmetic contexts. Squares and cubes had strong geometric meaning. Higher powers became easier to use as notation improved. The modern exponent notation we use today was refined over centuries, with important contributions from mathematicians such as René Descartes and later algebraists who standardized symbolic rules.

The need for exponential models became especially urgent in finance and astronomy. Compound interest forced people to understand repeated percentage growth. If money grows by a percentage each period, the amount after many periods is not found by adding the percentage repeatedly. It is found by multiplying repeatedly. This distinction is the heart of exponential growth.

Logarithms, developed in the early seventeenth century by John Napier and others, were partly created to make multiplication and powers easier to compute. Before calculators, logarithms transformed multiplication into addition and powers into multiplication. This was revolutionary for navigation, astronomy, engineering, and finance. The same relationship underlies this objective: exponents control repeated multiplication, and changing the exponent scale changes how we understand the model.

Scientific growth and decay models expanded the importance of exponentials. Radioactive decay, population models, cooling laws, and later electrical circuits and differential equations all used exponential functions. In each case, a model could be written in different but equivalent forms depending on whether the natural unit was seconds, years, half-lives, doubling times, or continuous rates. Modern technology hides many calculations, but it does not remove the need for interpretation. A scientist or analyst still must know what the base and exponent mean.

The historical lesson is clear: exponent notation became powerful because it compressed repeated multiplication. Exponent laws became powerful because they let people translate that compression across different scales. This objective gives students access to that machinery.

Technical execution: how to transform exponential expressions

The first rule is to identify the whole model. In \(A(t) = a(b)^t\), \(a\) is the initial value when \(t = 0\), and \(b\) is the growth or decay factor per one unit of \(t\). If \(b > 1\), the model grows. The percent growth rate per time unit is \(b - 1\), expressed as a percent. If \(0 < b < 1\), the model decays. The percent decay rate per time unit is \(1 - b\), expressed as a percent.

The second rule is to identify the time unit. A base without a time unit is incomplete. The expression \(1.03^t\) means 3 percent growth per unit of \(t\), but the unit might be a month, year, week, or generation. Students should write the unit in words before interpreting the rate.

The third rule is to use exponent laws to change time scales. If a yearly factor is \(B\), then the monthly factor is \(B^{1/12}\) because twelve monthly steps must equal one yearly step. The daily factor is approximately \(B^{1/365}\) if using 365 days. In general, if one large time unit contains \(n\) smaller time units, then the smaller-unit factor is \(B^{1/n}\).

For example, rewrite \(A(t) = 2000(1.09)^t\), where \(t\) is in years, in terms of months \(m\). Since \(m = 12t\), we have \(t = m/12\). Substitute:

\[A(m) = 2000(1.09)^{m/12}\].

This can also be written as:

\[A(m) = 2000(1.09^{1/12})^m\].

The monthly factor is \(1.09^{1/12}\), about 1.0072, so the monthly growth rate is about 0.72 percent. The original form revealed annual growth. The transformed form reveals monthly growth.

For decay, suppose \(D(t) = 500(0.80)^t\), where \(t\) is in days. To find the weekly factor, use 7 days per week. After one week, the factor is \(0.80^7\), about 0.2097. That means about 20.97 percent remains after a week, so the weekly decay is about 79.03 percent. Again, the transformed expression shows something not obvious from the original daily factor.

Students should also learn to rewrite expressions by decomposing exponents. For \(3(2)^{x+4}\), use \(2^{x+4} = 2^x 2^4\), so the expression becomes \(48(2)^x\). This reveals that shifting the exponent by 4 multiplies the starting coefficient by 16. For \(5(9)^x\), recognizing that \(9 = 3^2\) gives \(5(3)^{2x}\). This may help compare with another expression using base 3.

Another common transformation uses negative exponents. Since \(b^{-x} = (1/b)^x\), an expression like \(100(2)^{-t}\) can be rewritten as \(100(1/2)^t\). The second form makes decay clearer. It says the quantity is halved each time unit. Students should not treat negative exponents as mysterious; they usually mean reciprocal growth factors.

What this math represents in real life

This math represents repeated proportional change across time scales. A price that increases by a fixed percent each year, a bacteria population that doubles every few hours, a medicine concentration that decays by a fixed percentage, a car value that depreciates, a subscription base that grows, and a radioactive sample that halves all use the same machinery. The world often changes multiplicatively, not additively.

In finance, exponent transformations reveal the real cost or value of rates. A monthly rate can be annualized. An annual rate can be converted to a monthly rate. A nominal rate can be compared with an effective rate. Students do not need to become financial experts in Math II, but they should understand that the time unit of compounding matters.

In science, exponent transformations help match the model to measurement. A lab might collect data every hour while a textbook gives a half-life in days. A climate model might use annual factors while a simulation steps through months. A medication schedule might require hour-by-hour estimates. Exponent laws let the same relationship be expressed at the right resolution.

In technology, exponential growth and decay appear in algorithms, networks, sound, images, machine learning, and data storage. A quantity that is multiplied repeatedly can explode or vanish quickly. Understanding exponentials gives students a mental radar for nonlinear change.

Where this fits in the big map of mathematics

This objective is a bridge between basic exponent laws and full exponential modeling. In Math I, students learned to distinguish linear from exponential relationships. In Math II, they learn to read and rewrite exponential expressions more flexibly. Later, they will use rational exponents, radicals, logarithms, and perhaps continuous exponential functions. All of that depends on understanding that exponent rules preserve the meaning of repeated multiplication while changing the form of the expression.

It also prepares students for logarithms. Logarithms answer questions like, “How long does it take to reach a certain amount?” But before students solve exponential equations, they need to understand what the exponential expression is saying. Rewriting \(1.15^t\) in monthly form is a conceptual step toward understanding bases, exponents, and time scales deeply enough to use logarithms responsibly.

In the full map, exponentials are one of the major function families. Linear functions model constant addition. Quadratic functions model curved change with constant second differences. Exponential functions model constant multiplication. This objective gives students the tools to manipulate the language of constant multiplication.

Common student traps and how to avoid them

One trap is confusing growth factor with growth rate. A factor of 1.15 means 15 percent growth, not 115 percent growth. A factor of 0.85 means 15 percent decay, not 85 percent decay.

A second trap is ignoring the time unit. The base only has meaning relative to the exponent. A factor of 1.01 per day is very different from a factor of 1.01 per year.

A third trap is dividing percent rates when converting time scales. A 12 percent annual growth rate is not exactly 1 percent per month under multiplicative compounding. The monthly factor is \(1.12^{1/12}\), not \(1 + 0.12/12\) unless using an approximation in a specific simple-interest context.

A fourth trap is thinking equivalent expressions must look similar. Exponential expressions can look very different while producing the same outputs. Students should test equivalence using exponent laws, not surface appearance.

Problem Library

Problems in the App From This Objective

168 problems across 12 archetypes in the app.

use `(b^k)^t = b^{kt}` structure.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Rewrite P(m)=100(1.02)^m, m is months to show growth over a larger interval.

Problem 2

Rewrite A(d)=50(1.01)^d, d is days to show growth over a larger interval.

Problem 3

Rewrite N(q)=200(1.05)^q, q is quarters to show growth over a larger interval.

Problem 4

Rewrite B(h)=80(0.99)^h, h is hours to show growth over a larger interval.

Problem 5

Rewrite C(s)=10(1.1)^s, s is seconds to show growth over a larger interval.

Problem 6

Rewrite V(m)=1000(0.95)^m, m is minutes to show growth over a larger interval.

Open in simulator
Problem 7

Rewrite P(w)=300(1.03)^w, w is weeks to show growth over a larger interval.

Problem 8

Rewrite M(y)=500(1.08)^y, y is years to show growth over a larger interval.

Problem 9

Rewrite R(d)=25(1.005)^d, d is days to show growth over a larger interval.

Problem 10

Rewrite S(h)=150(1.015)^h, h is hours to show growth over a larger interval.

Problem 11

Rewrite G(ms)=75(1.001)^ms, ms is milliseconds to show growth over a larger interval.

Problem 12

Rewrite D(s)=2000(0.999)^s, s is seconds to show growth over a larger interval.

Problem 13

Rewrite F(q)=120(1.06)^q, q is quarters to show growth over a larger interval.

Problem 14

Rewrite H(d)=40(1.002)^d, d is days to show growth over a larger interval.

Problem 15

Rewrite J(h)=600(0.98)^h, h is hours to show growth over a larger interval.

use roots or fractional exponents.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Rewrite P(y)=100(1.12)^y, y is years to show growth over a smaller interval.

Problem 17

Rewrite A(w)=80(0.7)^w, w is weeks to show growth over a smaller interval.

Problem 18

Rewrite N(d)=500(2)^d, d is days to show growth over a smaller interval.

Problem 19

Rewrite B(q)=1000(1.08)^q, q is quarters to show growth over a smaller interval.

Problem 20

Rewrite C(y)=200(1.05)^y, y is years to show growth over a smaller interval.

Problem 21

Rewrite M(d)=750(0.95)^d, d is decades to show growth over a smaller interval.

Problem 22

Rewrite R(h)=150(1.2)^h, h is hours to show growth over a smaller interval.

Problem 23

Rewrite G(m)=50(0.8)^m, m is minutes to show growth over a smaller interval.

Open in simulator
Problem 24

Rewrite P(f)=300(1.15)^f, f is fortnights to show growth over a smaller interval.

Problem 25

Rewrite E(y)=600(0.98)^y, y is years to show growth over a smaller interval.

Problem 26

Rewrite L(d)=250(1.01)^d, d is days to show growth over a smaller interval.

Problem 27

Rewrite J(y)=400(1.07)^y, y is years to show growth over a smaller interval.

Problem 28

Rewrite T(b)=5000(1.06)^b, b is bi-annual periods to show growth over a smaller interval.

Problem 29

Rewrite U(s)=10000(0.9)^s, s is semi-annual periods to show growth over a smaller interval.

Problem 30

Rewrite W(t)=2500(1.02)^t, t is tri-annual periods to show growth over a smaller interval.

use power relationships between bases.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Convert 5(4)^t to an equivalent exponential expression with base 2.

Problem 32

Convert 7(27)^x to an equivalent exponential expression with base 3.

Problem 33

Convert 9(1/8)^n to an equivalent exponential expression with base 1/2.

Problem 34

Convert 2(100)^t to an equivalent exponential expression with base 10.

Problem 35

Convert 3(9)^y to an equivalent exponential expression with base 3.

Open in simulator
Problem 36

Convert (64)^m to an equivalent exponential expression with base 4.

Problem 37

Convert 12(1/25)^k to an equivalent exponential expression with base 1/5.

Problem 38

Convert 6(16)^p to an equivalent exponential expression with base 2.

Problem 39

Convert 4(1/5)^t to an equivalent exponential expression with base 5.

Problem 40

Convert 8(1/9)^x to an equivalent exponential expression with base 3.

Problem 41

Convert 10(1/64)^y to an equivalent exponential expression with base 4.

Problem 42

Convert 25(625)^a to an equivalent exponential expression with base 5.

Problem 43

Convert 3(81)^z to an equivalent exponential expression with base 9.

Problem 44

Convert 5(16)^b to an equivalent exponential expression with base 1/4.

Problem 45

Convert 7(32)^w to an equivalent exponential expression with base 2.

convert multiplier to percent increase.
15 problems Warmup Practice Mixed Review Assessment
Problem 46

Interpret the percent growth in exponential model P(t)=400(1.07)^t.

Problem 47

Interpret the percent growth in exponential model A(t)=1200(1.25)^t.

Open in simulator
Problem 48

Interpret the percent growth in exponential model N(t)=60(2)^t.

Problem 49

Interpret the percent growth in exponential model B(t)=90(1.005)^t.

Problem 50

Interpret the percent growth in exponential model Y(x)=500(1.10)^x.

Problem 51

Interpret the percent growth in exponential model C(m)=2000(1.03)^m.

Problem 52

Interpret the percent growth in exponential model V(d)=100(1.5)^d.

Problem 53

Interpret the percent growth in exponential model F(h)=75(1.01)^h.

Problem 54

Interpret the percent growth in exponential model G(y)=10(3)^y.

Problem 55

Interpret the percent growth in exponential model H(k)=15000(1.001)^k.

Problem 56

Interpret the percent growth in exponential model J(s)=250(1.75)^s.

Problem 57

Interpret the percent growth in exponential model K(w)=800(1.06)^w.

Problem 58

Interpret the percent growth in exponential model L(z)=300(1.2)^z.

Problem 59

Interpret the percent growth in exponential model M(q)=50(4)^q.

Problem 60

Interpret the percent growth in exponential model Q(r)=1000(1.025)^r.

convert multiplier to percent decrease.
15 problems Warmup Practice Mixed Review Assessment
Problem 61

Interpret the percent decay in exponential model A(t)=800(0.9)^t.

Problem 62

Interpret the percent decay in exponential model M(t)=50(0.75)^t.

Problem 63

Interpret the percent decay in exponential model V(t)=12000(0.6)^t.

Problem 64

Interpret the percent decay in exponential model B(t)=300(0.995)^t.

Problem 65

Interpret the percent decay in exponential model P(t)=100(0.8)^t.

Problem 66

Interpret the percent decay in exponential model C(t)=2000(0.5)^t.

Problem 67

Interpret the percent decay in exponential model D(t)=500(0.95)^t.

Problem 68

Interpret the percent decay in exponential model E(t)=150(0.2)^t.

Problem 69

Interpret the percent decay in exponential model F(t)=10000(0.99)^t.

Problem 70

Interpret the percent decay in exponential model G(t)=750(0.7)^t.

Problem 71

Interpret the percent decay in exponential model H(t)=250(0.1)^t.

Problem 72

Interpret the percent decay in exponential model I(t)=5000(0.999)^t.

Problem 73

Interpret the percent decay in exponential model J(t)=600(0.45)^t.

Open in simulator
Problem 74

Interpret the percent decay in exponential model K(t)=1200(0.88)^t.

Problem 75

Interpret the percent decay in exponential model L(t)=900(0.33)^t.

use negative exponents or reciprocal bases.
12 problems Warmup Practice Mixed Review Assessment
Problem 76

Rewrite decay expression 100(1/2)^t using reciprocal growth or negative exponent form.

Problem 77

Rewrite decay expression 80(1/3)^n using reciprocal growth or negative exponent form.

Problem 78

Rewrite decay expression 500(0.25)^t using reciprocal growth or negative exponent form.

Open in simulator
Problem 79

Rewrite decay expression 60(2)^(-d) using reciprocal growth or negative exponent form.

Problem 80

Rewrite decay expression 200(1/5)^x using reciprocal growth or negative exponent form.

Problem 81

Rewrite decay expression 10(1/10)^y using reciprocal growth or negative exponent form.

Problem 82

Rewrite decay expression 75(3)^(-m) using reciprocal growth or negative exponent form.

Problem 83

Rewrite decay expression 120(5)^(-p) using reciprocal growth or negative exponent form.

Problem 84

Rewrite decay expression 25(0.5)^t using reciprocal growth or negative exponent form.

Problem 85

Rewrite decay expression 300(0.1)^k using reciprocal growth or negative exponent form.

Problem 86

Rewrite decay expression 15(2/3)^t using reciprocal growth or negative exponent form.

Problem 87

Rewrite decay expression 50(4/5)^(-r) using reciprocal growth or negative exponent form.

factor powers to isolate coefficient.
15 problems Warmup Practice Mixed Review Assessment
Problem 88

Transform 5(2)^(t+3) to reveal the initial value at input 0.

Problem 89

Transform 12(3)^(t-1) to reveal the initial value at input 0.

Problem 90

Transform 100(1.1)^(t+2) to reveal the initial value at input 0.

Open in simulator
Problem 91

Transform 80(0.5)^(t-2) to reveal the initial value at input 0.

Problem 92

Transform 3(4)^(t+2) to reveal the initial value at input 0.

Problem 93

Transform 20(5)^(t-2) to reveal the initial value at input 0.

Problem 94

Transform 50(1.2)^(t+1) to reveal the initial value at input 0.

Problem 95

Transform 100(2.5)^(t-1) to reveal the initial value at input 0.

Problem 96

Transform 1000(0.8)^(t+1) to reveal the initial value at input 0.

Problem 97

Transform 60(0.2)^(t-1) to reveal the initial value at input 0.

Problem 98

Transform 2(3)^(t+4) to reveal the initial value at input 0.

Problem 99

Transform 1000(10)^(t-3) to reveal the initial value at input 0.

Problem 100

Transform 16(1/2)^(t+2) to reveal the initial value at input 0.

Problem 101

Transform 9(1/3)^(t-2) to reveal the initial value at input 0.

Problem 102

Transform (7)^(t+2) to reveal the initial value at input 0.

explain same growth with different time units.
15 problems Warmup Practice Mixed Review Assessment
Problem 103

Compare equivalent exponential models P(m)=100(1.01)^m and P(y)=100(1.01^12)^y in context monthly versus yearly growth.

Problem 104

Compare equivalent exponential models A(d)=50(0.95)^d and A(w)=50(0.95^7)^w in context daily versus weekly decay.

Problem 105

Compare equivalent exponential models N(t)=200(2)^t and N(t)=25(2)^(t+3) in context same input unit.

Open in simulator
Problem 106

Compare equivalent exponential models B(q)=1000(1.04)^q and B(y)=1000(1.04^4)^y in context quarterly versus yearly compounding.

Problem 107

Compare equivalent exponential models C(h) = 500(1.005)^h and C(d) = 500(1.005^24)^d in context hourly versus daily compounding.

Problem 108

Compare equivalent exponential models M(d) = 10000(0.98)^d and M(b) = 10000(0.98^14)^b in context daily versus bi-weekly decay.

Problem 109

Compare equivalent exponential models S(w) = 2000(1.02)^w and S(m) = 2000(1.02^4)^m in context weekly versus monthly growth (assuming 4 weeks per month).

Problem 110

Compare equivalent exponential models Y(t) = 300(1.5)^t and Y(t) = 200(1.5)^(t+1) in context same input unit.

Problem 111

Compare equivalent exponential models Z(x) = 1000(0.8)^x and Z(x) = 800(0.8)^(x-1) in context same input unit.

Problem 112

Compare equivalent exponential models P(y) = 10000(1.03)^y and P(d) = 10000(1.03^10)^d in context yearly versus decadal growth.

Problem 113

Compare equivalent exponential models D(s) = 500(0.99)^s and D(m) = 500(0.99^60)^m in context seconds versus minutes decay.

Problem 114

Compare equivalent exponential models V(h) = 2000(1.025)^h and V(y) = 2000(1.025^2)^y in context bi-annual versus annual compounding.

Problem 115

Compare equivalent exponential models F(t) = 60(3)^t and F(t) = 20(3)^(t+1) in context same input unit.

Problem 116

Compare equivalent exponential models G(x) = 400(0.5)^x and G(x) = 100(0.5)^(x-2) in context same input unit.

Problem 117

Compare equivalent exponential models R(d) = 750(0.9)^d and R(h) = 750(0.9^(1/24))^h in context daily versus hourly decay.

combine powers before computing.
12 problems Warmup Practice Mixed Review Assessment
Problem 118

Use exponent properties to evaluate 2^3*2^5 efficiently.

Problem 119

Use exponent properties to evaluate 3^6/3^4 efficiently.

Problem 120

Use exponent properties to evaluate (5^2)^3 efficiently.

Problem 121

Use exponent properties to evaluate 4^3*4^-1 efficiently.

Problem 122

Use exponent properties to evaluate 7^2 * 7^1 efficiently.

Problem 123

Use exponent properties to evaluate 6^5 / 6^3 efficiently.

Problem 124

Use exponent properties to evaluate (3^2)^3 efficiently.

Problem 125

Use exponent properties to evaluate 5^2 / 5^-1 efficiently.

Problem 126

Use exponent properties to evaluate (3^-1)^3 efficiently.

Problem 127

Use exponent properties to evaluate 9^5 / 9^5 efficiently.

Problem 128

Use exponent properties to evaluate (2^3 * 2^4) / 2^5 efficiently.

Open in simulator
Problem 129

Use exponent properties to evaluate (10^2)^2 / 10^3 efficiently.

apply exponent rules and compare.
15 problems Warmup Practice Mixed Review Assessment
Problem 130

Determine whether exponential expressions 2^(t+3) and 8*2^t are equivalent.

Problem 131

Determine whether exponential expressions 3^(2t) and 9^t are equivalent.

Problem 132

Determine whether exponential expressions 5^t+5^2 and 5^(t+2) are equivalent.

Problem 133

Determine whether exponential expressions 4^t and 2^(2t) are equivalent.

Problem 134

Determine whether exponential expressions 7^(x-2) and 7^x / 49 are equivalent.

Problem 135

Determine whether exponential expressions 16^(t/2) and 2^(2t) are equivalent.

Problem 136

Determine whether exponential expressions 3^(-x) and 1 / 3^x are equivalent.

Open in simulator
Problem 137

Determine whether exponential expressions (2*3)^t and 2^t + 3^t are equivalent.

Problem 138

Determine whether exponential expressions (2x)^3 and 8x^3 are equivalent.

Problem 139

Determine whether exponential expressions (10^x) / 100 and 10^(x-2) are equivalent.

Problem 140

Determine whether exponential expressions 27^t and 3^(t+3) are equivalent.

Problem 141

Determine whether exponential expressions (x+1)^0 and 1 are equivalent.

Problem 142

Determine whether exponential expressions (1/4)^(-t) and 4^t are equivalent.

Problem 143

Determine whether exponential expressions (2^x)^3 and 2^(x+3) are equivalent.

Problem 144

Determine whether exponential expressions sqrt(9^t) and 3^t are equivalent.

connect base to chosen time step.
12 problems Warmup Practice Mixed Review Assessment
Problem 145

Interpret the growth factor over the specified interval in P(t)=100(1.02)^(12t), where t is years.

Problem 146

Interpret the growth factor over the specified interval in A(t)=50(0.9)^(t/2), where t is days.

Problem 147

Interpret the growth factor over the specified interval in N(t)=200(3)^(t/4), where t is hours.

Open in simulator
Problem 148

Interpret the growth factor over the specified interval in B(t)=80(0.5)^(24t), where t is days.

Problem 149

Interpret the growth factor over the specified interval in M(t)=1000(1.05)^(t/3), where t is months.

Problem 150

Interpret the growth factor over the specified interval in C(t)=20(2)^(6t), where t is weeks.

Problem 151

Interpret the growth factor over the specified interval in V(t)=5000(0.8)^(t/10), where t is years.

Problem 152

Interpret the growth factor over the specified interval in R(t)=10(1.1)^(t/5), where t is minutes.

Problem 153

Interpret the growth factor over the specified interval in W(t)=300(0.95)^(2t), where t is days.

Problem 154

Interpret the growth factor over the specified interval in S(t)=75(4)^(t/6), where t is hours.

Problem 155

Interpret the growth factor over the specified interval in D(t)=150(0.7)^(3t), where t is months.

Problem 156

Interpret the growth factor over the specified interval in E(t)=2500(1.01)^(t/24), where t is hours.

catch additive exponent mistakes, percent/base confusion, and time-unit errors.
12 problems Warmup Practice Mixed Review Assessment
Problem 157

Correct the exponent-property or growth-rate error in 1.08 means 108% growth per year.

Problem 158

Correct the exponent-property or growth-rate error in 2^(t+3)=2^t+8.

Problem 159

Correct the exponent-property or growth-rate error in 0.7 means a 70% decrease.

Problem 160

Correct the exponent-property or growth-rate error in Monthly base for annual growth 1.12 is 1.12/12.

Problem 161

Correct the exponent-property or growth-rate error in A base of 1.25 means 25% decrease.

Problem 162

Correct the exponent-property or growth-rate error in (x^2)^3 = x^5.

Problem 163

Correct the exponent-property or growth-rate error in 3^t * 3^2 = 3^(2t).

Problem 164

Correct the exponent-property or growth-rate error in If a monthly growth rate is 1.015, the annual growth rate is 1.015 * 12.

Problem 165

Correct the exponent-property or growth-rate error in A decay factor of 0.95 means a 95% decrease.

Open in simulator
Problem 166

Correct the exponent-property or growth-rate error in (2x)^3 = 2x^3.

Problem 167

Correct the exponent-property or growth-rate error in Any number raised to the power of 0 is 0.

Problem 168

Correct the exponent-property or growth-rate error in To find the daily decay factor from an annual factor of 0.8, divide 0.8 by 365.