Understanding Unit 6: Exponents and Exponential Functions – Homework 10 Answer Key Explained
Exponents and exponential functions form the backbone of many high‑school mathematics curricula, and Unit 6 typically consolidates the rules, properties, and real‑world applications that students need to master. Homework 10 is often the culminating set of problems that tests both procedural fluency and conceptual insight. This article walks you through the complete answer key, explains why each step works, and highlights common pitfalls so you can verify your work with confidence and deepen your understanding of exponential mathematics And that's really what it comes down to. Simple as that..
1. Quick Review of Core Concepts
Before diving into the answer key, recall the essential ideas that appear throughout the homework:
| Concept | Key Rule | Example |
|---|---|---|
| Product of Powers | (a^{m}\cdot a^{n}=a^{m+n}) | (2^{3}\cdot2^{4}=2^{7}) |
| Quotient of Powers | (\dfrac{a^{m}}{a^{n}}=a^{m-n}) | (\dfrac{5^{6}}{5^{2}}=5^{4}) |
| Power of a Power | ((a^{m})^{n}=a^{mn}) | ((3^{2})^{3}=3^{6}) |
| Zero Exponent | (a^{0}=1) (for (a\neq0)) | (7^{0}=1) |
| Negative Exponent | (a^{-n}=\dfrac{1}{a^{n}}) | (4^{-2}=1/4^{2}=1/16) |
| Scientific Notation | (a\times10^{n}) where (1\le a<10) | (3.2\times10^{5}) |
| Exponential Function | (f(x)=a\cdot b^{x}) (base (b>0, b\neq1)) | (f(x)=2\cdot3^{x}) |
| Growth/Decay Rate | (P(t)=P_{0}e^{kt}) (continuous) | (P(t)=150e^{0.04t}) |
Understanding these rules will make each answer in Homework 10 feel intuitive rather than a series of memorized steps.
2. Homework 10 – Problem‑by‑Problem Answer Key
Below each problem statement you’ll find the final answer, a step‑by‑step solution, and a brief note on the underlying principle But it adds up..
Problem 1 – Simplify the expression
[ \frac{2^{5}\cdot 2^{-3}}{2^{2}} ]
Answer: (2^{0}=1)
Solution:
- Apply the product rule in the numerator: (2^{5}\cdot2^{-3}=2^{5+(-3)}=2^{2}).
- Divide by the denominator using the quotient rule: (\dfrac{2^{2}}{2^{2}}=2^{2-2}=2^{0}).
- Any non‑zero base raised to the zero power equals 1.
Why it matters: This problem reinforces combining positive and negative exponents before simplifying the fraction.
Problem 2 – Write the number in scientific notation
[ 0.000456 ]
Answer: (4.56\times10^{-4})
Solution:
- Move the decimal point four places to the right to obtain a coefficient between 1 and 10: (0.000456\rightarrow4.56).
- Because the decimal moved right, the exponent is negative: (10^{-4}).
Why it matters: Scientific notation is essential for handling very small or very large quantities, especially in physics and chemistry.
Problem 3 – Evaluate the exponential function (f(x)=3\cdot2^{x}) at (x=-2).
Answer: (\displaystyle \frac{3}{4}=0.75)
Solution:
- Substitute (x=-2): (f(-2)=3\cdot2^{-2}).
- Convert the negative exponent: (2^{-2}=1/2^{2}=1/4).
- Multiply: (3\cdot\frac{1}{4}= \frac{3}{4}=0.75).
Why it matters: Negative exponents produce fractional values, showing how exponential functions can model decay as well as growth No workaround needed..
Problem 4 – Solve for (x): (\displaystyle 5^{2x}=125)
Answer: (x= \dfrac{3}{2})
Solution:
- Recognize that (125=5^{3}).
- Rewrite the equation: (5^{2x}=5^{3}).
- Since the bases are equal, set the exponents equal: (2x=3).
- Solve: (x=3/2).
Why it matters: This demonstrates equating exponents when bases match, a common technique for logarithmic‑free solving That's the whole idea..
Problem 5 – Determine the continuous growth rate (k) if a population grows from 200 to 800 in 5 years, using (P(t)=P_{0}e^{kt}).
Answer: (k\approx0.2773\ \text{yr}^{-1})
Solution:
- Plug values: (800 = 200e^{5k}).
- Divide by 200: (4 = e^{5k}).
- Take natural logarithm: (\ln 4 = 5k).
- Solve: (k = \frac{\ln 4}{5}\approx\frac{1.3863}{5}=0.2773).
Why it matters: Continuous growth models are vital in biology, finance, and physics. The natural log appears because the base of the exponential function is e And it works..
Problem 6 – Simplify ((3^{2} \cdot 3^{-5})^{4})
Answer: (3^{-12}) or (\displaystyle \frac{1}{3^{12}})
Solution:
- Inside the parentheses, combine exponents: (3^{2-5}=3^{-3}).
- Apply the power‑of‑a‑power rule: ((3^{-3})^{4}=3^{-3\cdot4}=3^{-12}).
- Optionally rewrite as a positive exponent: (1/3^{12}).
Why it matters: Nested exponent rules often cause errors; handling them step‑by‑step avoids sign mistakes Took long enough..
Problem 7 – Convert ((7\times10^{3})^{2}) to standard form.
Answer: (4.9\times10^{7})
Solution:
- Square the coefficient: (7^{2}=49).
- Square the power of ten: ((10^{3})^{2}=10^{6}).
- Multiply: (49\times10^{6}=4.9\times10^{7}) (shift decimal one place left, increase exponent by 1).
Why it matters: This reinforces the power of a product rule and the need to keep the coefficient between 1 and 10 Worth keeping that in mind. Simple as that..
Problem 8 – Find the inverse function of (g(x)=4\cdot5^{x}).
Answer: (g^{-1}(x)=\log_{5}!\left(\dfrac{x}{4}\right))
Solution:
- Replace (g(x)) with (y): (y=4\cdot5^{x}).
- Swap (x) and (y): (x=4\cdot5^{y}).
- Isolate the exponential term: (\dfrac{x}{4}=5^{y}).
- Apply log base 5 to both sides: (\log_{5}!\left(\dfrac{x}{4}\right)=y).
- Hence, (g^{-1}(x)=\log_{5}!\left(\dfrac{x}{4}\right)).
Why it matters: Inverse exponential functions are logarithms; recognizing this relationship is crucial for solving real‑world problems involving unknown time periods or rates.
Problem 9 – If (f(x)=2^{x}+3^{x}), compute (f(2)).
Answer: (13)
Solution:
- Evaluate each term: (2^{2}=4) and (3^{2}=9).
- Add: (4+9=13).
Why it matters: Mixed‑base exponential expressions appear in combinatorial contexts; evaluating them is straightforward once each base is handled separately.
Problem 10 – Determine the half‑life (t_{1/2}) of a substance that follows (A(t)=A_{0}e^{-0.06t}).
Answer: (\displaystyle t_{1/2}\approx11.55\ \text{units of time})
Solution:
- Half‑life definition: (A(t_{1/2})=\dfrac{A_{0}}{2}).
- Set up equation: (\dfrac{A_{0}}{2}=A_{0}e^{-0.06t_{1/2}}).
- Cancel (A_{0}): (\dfrac{1}{2}=e^{-0.06t_{1/2}}).
- Take natural log: (\ln!\left(\dfrac{1}{2}\right)=-0.06t_{1/2}).
- Solve: (t_{1/2}= \dfrac{\ln 2}{0.06}\approx\frac{0.6931}{0.06}=11.55).
Why it matters: Half‑life calculations are a classic application of exponential decay, linking the constant (k) to a tangible time measure But it adds up..
3. Common Mistakes & How to Avoid Them
| Mistake | Typical Symptom | Correct Approach |
|---|---|---|
| Treating negative exponents as “subtracting” the base | Writing (2^{-3}=2-3) | Remember a negative exponent inverts the base: (2^{-3}=1/2^{3}). , divide by the coefficient), then apply the logarithm. 56\times10^{3}) for a small number |
| Skipping the step of isolating the exponential term before taking logs | Directly applying (\log) to an expression that still has a coefficient attached | Isolate the term with the exponent first (e. Plus, |
| Forgetting to adjust the exponent when converting to scientific notation | Writing (4560=4. g. | |
| Equating bases without checking the base is the same | Solving (3^{x}=9^{2}) by setting (x=2) | Rewrite both sides with a common base: (9^{2}=(3^{2})^{2}=3^{4}) → then (3^{x}=3^{4}) → (x=4). |
| Misapplying the power‑of‑a‑product rule | Squaring ((2\times10^{3})) as (2^{2}\times10^{3}) | Square both the coefficient and the power of ten: ((2\times10^{3})^{2}=2^{2}\times10^{6}). |
Being aware of these pitfalls helps you double‑check each answer in the key and develop a more reliable problem‑solving routine.
4. Extending the Learning – Real‑World Applications
- Population Modeling – The continuous growth formula used in Problem 5 accurately predicts bacterial cultures, human populations, and investment growth when compounded continuously.
- Radioactive Decay – Problem 10’s half‑life computation mirrors how scientists calculate the age of archaeological samples or the safety timeline for nuclear waste.
- Finance – The inverse function in Problem 8 is the backbone of logarithmic return calculations, essential for understanding interest rates and stock performance.
By linking each abstract algebraic manipulation to a tangible scenario, you reinforce the why behind the how Not complicated — just consistent. That's the whole idea..
5. Frequently Asked Questions (FAQ)
Q1: When should I use the natural exponential base (e) versus a generic base like 2 or 5?
A: Use (e) whenever the problem involves continuous growth or decay, such as radioactive decay, continuously compounded interest, or differential equations. Generic bases appear in discrete contexts (e.g., doubling time, geometric sequences).
Q2: Can I combine exponents with different bases?
A: Only if you first rewrite them with a common base (e.g., (9=3^{2})). Otherwise, the product or quotient rules do not apply across different bases Practical, not theoretical..
Q3: How do I check my answer for a negative exponent problem?
A: Compute the reciprocal of the positive‑exponent form. Take this case: verify (4^{-2}=1/16) by calculating (1/4^{2}=1/16).
Q4: What is the quickest way to find the inverse of an exponential function?
A: Isolate the exponential term, then apply the logarithm of the same base. Remember to swap (x) and (y) before solving for the new (y).
Q5: Why does scientific notation require the coefficient to be between 1 and 10?
A: This convention ensures a unique representation for each number, eliminating ambiguity (e.g., (0.5\times10^{1}) would be rewritten as (5\times10^{0})) That's the part that actually makes a difference..
6. Final Thoughts – Turning the Answer Key into Mastery
The Homework 10 answer key is more than a list of final numbers; it is a roadmap that showcases the logical flow of exponent and exponential function concepts. By reading each solution actively, re‑deriving the steps on paper, and connecting the algebra to real‑world contexts, you transform passive correction into active mastery Nothing fancy..
Remember these three takeaways:
- Master the exponent rules – they are the grammar of exponential language.
- Translate between forms (standard, scientific, logarithmic) fluidly; each offers a different perspective on the same relationship.
- Apply the concepts to authentic problems (population, half‑life, finance) to cement understanding and see the relevance beyond the classroom.
Armed with the detailed answer key and the explanations above, you’re ready not only to score perfectly on Homework 10 but also to tackle any future challenge involving exponents and exponential functions with confidence. Keep practicing, stay curious, and let the power of exponents work for you!
ConclusionThe study of exponents and exponential functions is not merely a mathematical exercise; it is a lens through which we interpret and interact with the world. From the decay of radioactive materials to the growth of investments, these concepts underpin countless phenomena that shape our understanding of time, change, and scale. The Homework 10 answer key, when approached with intention, becomes a bridge between abstract theory and practical application, fostering a deeper appreciation for the elegance of mathematics. By embracing the challenge of mastering these principles, you cultivate a mindset of precision and adaptability—skills that are invaluable in both academic and real-world contexts. As you continue to explore, remember that every exponent, every logarithmic transformation, and every real-world problem solved is an opportunity to strengthen your analytical abilities. The power of exponents lies not just in their calculations, but in their capacity to reveal the hidden patterns that govern our universe. Keep questioning, keep calculating, and let the beauty of mathematics guide your journey.
