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3.3 — More Interesting
You already know that exponential growth multiplies an amount by the same factor over equal time intervals. In the previous lessons, half-year and other fractional time intervals introduced roots and fractional exponents. Today we explain why the exponent rules still work for those smaller intervals.
1. In class: A Solidify Understanding Task
One year's growth can happen in twelve equal steps
Suppose an account starts with $300 and grows by 5% over one year. A 5% increase means multiplying the current balance by 1.05. The annual model is
The balance after one year is \(300(1.05)=315\) dollars. How can we describe the same growth after one month, assuming growth follows the same exponential pattern throughout the year?
Let \(m\) be the exact monthly multiplication factor. Twelve months must reproduce one annual factor:
\[\underbrace{m\cdot m\cdots m}_{12\text{ factors}}=m^{12}=1.05.\]Because \(m\) is positive, the unique positive solution is the positive twelfth root:
\[m=\sqrt[12]{1.05}=(1.05)^{1/12}\approx1.00407412.\]The monthly increase is about \(0.4074\%\), not exactly \(5\%/12\) under this model. A monthly factor is approximately 1.004074; the corresponding percentage increase is approximately 0.4074%.
Let \(n\) count months since the deposit. Each month multiplies by \((1.05)^{1/12}\). Thus
\[A(n)=300\left((1.05)^{1/12}\right)^n=300(1.05)^{n/12}.\]At six months: \(A(6)=300(1.05)^{6/12}=300\sqrt{1.05}\approx\$307.41.\) At twelve months: \(A(12)=300(1.05)^1=\$315.\) These values match the stated annual growth.
In the packet, time \(t\) remains measured in years, so the number of elapsed months is \(12t\). Substituting \(n=12t\) into our monthly model yields
\[A(t)=300\left((1.05)^{1/12}\right)^{12t}.\]Using the power-of-a-power rule, the exponents multiply: \((1/12)(12t)=t\). Therefore these two exact equations predict precisely the same balance for every time \(t\) in their domain, not just after full years.
A rounded monthly multiplier is only an approximation
The packet also gives \(1.004\) as a convenient rounded monthly factor. Because \(1.004\neq(1.05)^{1/12}\), the expression \(300(1.004)^{12t}\) is not algebraically identical to \(300(1.05)^t\).
Rounding the monthly multiplier before repeatedly multiplying gives a balance about 28 cents smaller after one year. Use the exact twelfth root in calculations when accuracy matters; round the final balance only at the end.
Why exponent rules extend to fractional powers
With whole-number exponents, \(a^2a^3=a^5\) because the combined product contains five factors of \(a\). For positive bases, we want fractional powers to follow this same pattern.
Multiplying powers: add their exponents
Multiplying two square roots of a positive number must give that number:
\[a^{1/2}a^{1/2}=\sqrt a\,\sqrt a=a=a^1.\]The rule \(a^r a^s=a^{r+s}\) agrees because \(\tfrac12+\tfrac12=1\). Thus a fractional exponent measures a proportional part of a whole exponent.
Combine three one-month growth factors:
\[(1.05)^{1/12}(1.05)^{1/12}(1.05)^{1/12}=(1.05)^{3/12}=(1.05)^{1/4}.\]The resulting factor describes three months, or one-quarter of a year.
Raising a power to a power: multiply the exponents
Six monthly steps correspond to half a year:
\[\left((1.05)^{1/12}\right)^6=(1.05)^{6/12}=(1.05)^{1/2}.\]In general, \((a^r)^s=a^{rs}\) for positive \(a\) and real exponents \(r,s\). It works here because taking six equal monthly steps is the same as taking half of the annual exponent.
Indeed, \(16^{1/4}=2\), and \(2^2=4\). The two calculations confirm the same result.
Dividing powers: subtract the exponents
For a positive base \(a\), division reverses multiplication:
\[\frac{a^{3/4}}{a^{1/4}}=a^{3/4-1/4}=a^{1/2}.\]A negative exponent expresses a reciprocal, not a negative output: \(a^{-r}=\frac{1}{a^r}\) for \(a>0\).
This multiplier reverses one month of growth. It is not a factor for an ordinary month of forward growth.
For positive bases, these work for fractional exponents just as they do for integer exponents. Expressions with negative bases or even roots require additional care; today's account model uses positive bases throughout.
Use the exponent to interpret time
For \(1.05^t\), the exponent \(t\) counts years. Thus the difference between two exponents represents elapsed time, and adding exponents combines time intervals.
| Expression | Simplified form | Meaning |
|---|---|---|
| \((1.05)^{1/12}\cdot(1.05)^{1/12}\cdot(1.05)^{1/12}\) | \((1.05)^{1/4}\) | Three months of growth |
| \(\big((1.05)^{1/12}\big)^6\) | \((1.05)^{1/2}\) | Six months of growth |
| \((1.05)^{-1/12}\) | \(1/(1.05)^{1/12}\) | Undo one month's growth |
| \((1.05)^{1/2}\cdot(1.05)^{1/4}\) | \((1.05)^{3/4}\) | Nine months of growth |
| \((1.05)^{1/2}/(1.05)^{1/2}\) | \(1\) | No net multiplicative change |
What does \((1.05)^{1/12}\) mean? Let \(q=(1.05)^{1/12}\). The power rule gives \(q^{12}=1.05\). Therefore \(q\) is the positive twelfth root of 1.05:
\[(1.05)^{1/12}=\sqrt[12]{1.05}.\]Similarly, \(7^{1/3}=\sqrt[3]{7}\): multiplying that positive number by itself three times gives 7. This is precisely why the denominator of a unit-fraction exponent specifies the root.
2. READY — Exponential forms and equivalent representations (1–11)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A positive whole-number exponent counts the number of identical factors. For example, \(4^3=4\cdot4\cdot4=64\). The exponential form, factor list, and standard numeral are three descriptions of the same value.
Write the product of six 2s in exponential and standard forms.
\[2\cdot2\cdot2\cdot2\cdot2\cdot2=2^6=64.\]Conversely, \(343=7\cdot7\cdot7=7^3\). For numbers such as 81 or 625, different bases may represent the same value: \(81=3^4=9^2\) and \(625=5^4=25^2\). The packet does not always specify a unique base, so more than one valid exponential representation can exist.
To write several equivalent expressions, use the product, quotient, and power rules. They are identities: each representation has the same value wherever it is defined.
For \(3^8\), three equivalent expressions are
\[3^5\cdot3^3,\qquad \frac{3^{11}}{3^3},\qquad (3^2)^4.\]The exponents check out: \(5+3=8\), \(11-3=8\), and \(2\cdot4=8\).
All three represent the same positive reciprocal. Similarly, \(7^{1/3}=\sqrt[3]7=(7^{1/6})^2=\frac{7^{4/3}}7\). Do not confuse \(7^{1/3}\) with \(\frac{7}{3}\).
Check your work — Ready 1–11 answer key
- a. \(7^7\); b. 823543.
- a. Ten factors of 2; b. 1024.
- One choice: a. \(3^4\); b. \(3\cdot3\cdot3\cdot3\). (Other valid representations include \(9^2\).)
- a. \(11\cdot11\cdot11\cdot11\cdot11\); b. 161051.
- a. \(3^{10}\); b. 59049.
- One choice: a. \(5^4\); b. \(5\cdot5\cdot5\cdot5\). (Also \(25^2\).)
- Examples: \(2^5\cdot2^5,\ (2^2)^5,\ \frac{2^{12}}{2^2}\).
- Examples: \(3^3\cdot3^4,\ \frac{3^9}{3^2},\ (3^{\frac72})^2\).
- Examples: \(\frac1{13^8},\ (13^{-4})^2,\ \frac{13^2}{13^{10}}\).
- Examples: \(\sqrt[3]7,\ (7^{\frac16})^2,\ \frac{7^{\frac43}}7\).
- Examples: \(5,\ 5^0\cdot5^1,\ \frac{5^3}{5^2}\).
3. SET — Decide whether expressions are equivalent (12–17)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Equivalent expressions give the same output for every allowed input. To decide, simplify each expression independently using distributive and exponent rules. One matching numerical test does not prove equivalence for all inputs, although one counterexample can prove they are not equivalent.
Compare \(6(2^{x+1})\), \(12(2^x)\), and \(3(2^{x+2})\).
\[6\cdot2^x\cdot2=12\cdot2^x,\qquad 3\cdot2^x\cdot4=12\cdot2^x.\]All three are equivalent. The key is that changing \(x+1\) to \(x\) removes a factor of 2, which the outside coefficient must compensate for.
They are equivalent for every real \(x\), because the base 3 is positive and nonzero.
Compare \(40(2^{x+2})\), \(20(2^{x+3})\), and \(40(4^x)\).
\[40(2^{x+2})=160\cdot2^x,\qquad 20(2^{x+3})=160\cdot2^x.\]The first two agree. At \(x=0\), the third is \(40(4^0)=40\), while each of the first two is 160. Thus not all three are equivalent. Be especially careful: \(4^x=(2^2)^x=2^{2x}\), not \(2^{x+2}\).
Compare \(10(1.04^x)\), \(10((1.04^{1/4})^{4x})\), and \(10((1.04^{x/2})^2)\).
\[\left(1.04^{1/4}\right)^{4x}=1.04^{(1/4)(4x)}=1.04^x,\] \[\left(1.04^{x/2}\right)^2=1.04^{(x/2)2}=1.04^x.\]Therefore all three are equivalent. An important difference is where the exponent is placed: \((1.04^{1/4})^{4x}\) does not mean \(1.04^{1/(4\cdot4x)}\).
Because the constant terms differ, these are not equivalent. Distribute the number outside the parentheses to every term inside.
Check your work — Set 12–17 answer key
- Yes. Each equals \(\frac53\,3^x\), by separating the constant powers of 3.
- Yes. \(2^{-x}=\frac1{2^x}=(\frac12)^x\).
- No. The first and third equal \(3x+1\); the middle is \(3x-1\).
- No. The second and third equal \(50(4^x)\); the first equals \(200(2^x)\). For example, at \(x=0\) the first gives 200 and the other two give 50.
- Yes. The power-of-a-power rule makes every exponent \(x\).
- Yes. The combined exponents are \(x,(-1)(-x)=x\), and \((\frac15)(5x)=x\).
4. GO — Fluency with exponent rules (18–29)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Today's Go exercises ask you to leave your answer in exponential form. For products with the same base, add the exponents; for quotients with the same nonzero base, subtract; for a power raised to another power, multiply. You may have to do several steps.
| Operation | Rule | Quick example |
|---|---|---|
| Same-base product | \(a^ma^n=a^{m+n}\) | \(z^3z^5=z^8\) |
| Power of a power | \((a^m)^n=a^{mn}\) | \((p^4)^3=p^{12}\) |
| Same-base quotient | \(a^m/a^n=a^{m-n}\) | \(q^8/q^3=q^5\) |
| Zero and negative powers | \(a^0=1,\ a^{-m}=1/a^m\) | \(r^0=1\), \(r^{-2}=1/r^2\) |
Additional fully worked Go practice
Simplify \(4^2\cdot4^6\cdot4^3\), then simplify \((m^5)^3\).
\[4^2\cdot4^6\cdot4^3=4^{2+6+3}=\boxed{4^{11}}.\] \[(m^5)^3=m^{5\cdot3}=\boxed{m^{15}}.\]All factors in the first expression use the same base. For the second, multiplication of exponents is required because there is a power outside parentheses.
Assuming nonzero denominators, simplify \(\frac{x^7y^2z^4\cdot x^3y^5}{x^4y^3z^2}\).
\[\begin{aligned}\frac{x^7y^2z^4\cdot x^3y^5}{x^4y^3z^2}&=x^{7+3-4}y^{2+5-3}z^{4-2}\\&=\boxed{x^6y^4z^2}.\end{aligned}\]Work with each variable separately. You may subtract exponents only between powers of the same base.
Simplify \(\frac{(a^4)^5a^2}{a^9}\), assuming \(a\neq0\).
\[\frac{(a^4)^5a^2}{a^9}=\frac{a^{20}a^2}{a^9}=a^{20+2-9}=\boxed{a^{13}}.\]First multiply the exponents inside and outside the parentheses, then add and subtract as appropriate. Do not distribute the outside exponent by adding.
Simplify \(\frac{u^{r+5}u^{2r-1}}{u^{r+2}}\), for \(u\neq0\).
\[u^{(r+5)+(2r-1)-(r+2)}=\boxed{u^{2r+2}}.\]Place the entire denominator exponent in parentheses before subtracting to avoid a sign error.
Check your work — Go 18–29 answer key
- \(7^{10}\).
- \(3^{20}\).
- \(5^{19}\).
- \(x^8\).
- \(\frac1{x^b}\).
- \(x^{a+b}\).
- \(x^{ab}\).
- \(y^{a-b}\).
- \(y^{ac-b}\).
- \(3^{17}\).
- \(r^4s\).
- \(\frac{y^3}{x^3}\).
Packet clarifications
The account is described as earning 5% annually and the task asks for equal monthly steps. The intended model assumes a fixed monthly growth factor whose twelfth power is 1.05. This is an effective annual 5% model. A bank quoting a 5% nominal annual rate compounded monthly would instead use a factor of \(1+0.05/12\), giving a different effective yearly rate. The packet is asking for the former.
In problem 5, \(1.004\) is a rounded approximation, not an exact replacement for \(1.05^{1/12}\).
Given only the standard values 81 and 625, the exponential form is not unique unless a base is specified. For example, \(81=3^4=9^2\) and \(625=5^4=25^2\). Choose a correct repeated-factor form; either is mathematically valid.
Additional learning resources
These optional materials are selected to reinforce Section 3.3 rather than to introduce the later lesson's extended radical manipulations. They require internet access.
| Resource and direct link | What it covers / why use it | Scope / What to skip |
|---|---|---|
| Paul's Online Math Notes — Rational Exponents | Written explanation of rational exponents and the familiar exponent properties, useful for the monthly-root argument. | Some extra material: focus on definitions and exponent laws. Save lengthy radical simplification for Section 3.4. |
| Paul's Online Math Notes — Rational Exponents Practice | Practice questions with step-by-step worked solutions, including products and quotients of fractional powers. | Some extra material: focus on exponent-rule exercises; postpone multi-variable fractional-power manipulations that are more advanced than today's Go problems. |
| Khan Academy — Rational-exponent standards and lessons | Links to short instructional and practice activities on fractional powers and equivalent exponent forms. | Some extra material: use unit-fraction exponents and equivalent forms. Skip later radical-expression challenges for now. |
| Professor Leonard — Intermediate Algebra Lecture 10.2: Rational Exponents | Optional extended classroom-style explanation of the connection between roots and fractional exponents. | Significantly broader: use the introductory explanation of fractional exponents as a review; skip extended conversion/simplification procedures reserved for 3.4. Exact relevant timestamps could not be independently verified. |
Teacher's review — Mathematics and wording
Section 3.3 main task, problems 1–5: The phrase “earns 5% annually” is potentially ambiguous without specifying whether 5% is an effective annual yield or a nominal annual rate. The task expects twelve monthly growth factors to multiply to exactly 1.05, which corresponds to an effective annual rate of 5%. A clearer statement would be: “The balance grows by exactly 5% over one year, and we model growth with a constant monthly multiplier.”
Section 3.3 main task, problem 5: It is important to describe \(1.004\) as an approximation, not an equivalent multiplier. The true monthly multiplier is about \(1.00407412\). After one year, use of \(1.004\) produces roughly $314.72 rather than $315. A clearer instruction would explicitly ask students to interpret this rounded factor.
Section 3.3 Ready, problems 3 and 6: Specifying standard form alone permits multiple exponent bases (for example \(81=3^4=9^2\)); accept mathematically correct alternatives or add the intended base.
Curriculum coverage map
| Packet assignment | Targeted skill | Where taught |
|---|---|---|
| Main task 1–2 | Exact monthly factor from annual growth; monthly balances | In class and Example 1 |
| Main task 3–5 | Equivalent year/month models; meaning of \(12t\); interpret rounded factor | In class, Examples 2–3 |
| Main task 6–11 | Product, power, quotient, reciprocal, and root with rational exponents | In class, Examples 4–7 |
| Ready 1–6 | Exponential form, factors, numerical value | Section 2, Example 8 |
| Ready 7–11 | Three equivalent exponential representations | Section 2, Examples 9–10 |
| Set 12–17 | Prove or refute equivalence in linear and exponential expressions | Section 3, Examples 11–15 |
| Go 18–29 | Simplify exponent products, quotients, nested powers, variables and symbolic exponents | Section 4, Problems A–D |
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