Unit 6Exponents and Exponential Functions Homework 10 Answer Key: Mastering Advanced Concepts
Navigating the complexities of Unit 6 Exponents and Exponential Functions can be challenging, especially when tackling Homework 10. This leads to this comprehensive answer key serves as a vital resource, not just for checking answers, but for deepening your understanding of these fundamental mathematical concepts. Exponential functions model phenomena where quantities grow or decay at a rate proportional to their current value, making them essential in fields ranging from finance and biology to physics and computer science. This guide provides detailed solutions and explanations for Homework 10, empowering you to grasp the material thoroughly and build a strong foundation for future mathematical endeavors.
Understanding the Core Concepts
Before diving into the specific solutions, it's crucial to recall the core principles covered in Unit 6. Now, , (a^n = a \times a \times \ldots \times a), n times). Exponents represent repeated multiplication (e.g.Key skills include evaluating exponential expressions, graphing exponential functions, solving exponential equations using logarithms, and interpreting real-world applications like compound interest and population dynamics. The base (b) determines the function's behavior: if (b > 1), it exhibits exponential growth; if (0 < b < 1), it exhibits exponential decay. Now, exponential functions are defined as (f(x) = a \cdot b^x), where (b > 0), (b \neq 1). Homework 10 likely assesses your ability to apply these skills to more complex problems, including multi-step scenarios and interpreting function behavior.
Detailed Solutions and Explanations for Homework 10
Below are the solutions for each problem in Unit 6 Exponents and Exponential Functions Homework 10, presented with step-by-step reasoning to illuminate the process.
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Problem: Solve for (x): (2^{x+1} = 8).
- Solution: Recognize that (8 = 2^3). Rewrite the equation as (2^{x+1} = 2^3). Since the bases are the same and greater than 0 and not equal to 1, set the exponents equal: (x + 1 = 3). Solve for (x): (x = 2).
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Problem: Evaluate (3^{-2} \times 3^{4}) Worth keeping that in mind..
- Solution: Apply the Product of Powers Property ((a^m \times a^n = a^{m+n})): (3^{-2} \times 3^{4} = 3^{-2+4} = 3^{2} = 9).
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Problem: Graph the function (f(x) = 2^x) and state its domain, range, and asymptote.
- Solution: The graph starts at (0,1) and increases rapidly to the right. As (x) approaches negative infinity, (f(x)) approaches 0 but never touches the x-axis. That's why, the domain is all real numbers, the range is (y > 0), and the horizontal asymptote is (y = 0).
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Problem: Solve the exponential equation: (5^{2x} = 125) Nothing fancy..
- Solution: Recognize that (125 = 5^3). Rewrite the equation as (5^{2x} = 5^3). Set the exponents equal: (2x = 3). Solve for (x): (x = \frac{3}{2} = 1.5).
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Problem: A population of bacteria doubles every hour. If there are initially 100 bacteria, write an exponential function for the population (P(t)) after (t) hours and find the population after 5 hours Which is the point..
- Solution: The general form is (P(t) = P_0 \cdot b^t), where (P_0) is the initial population and (b) is the growth factor. Here, (P_0 = 100) and (b = 2) (since it doubles). So, (P(t) = 100 \cdot 2^t). After 5 hours: (P(5) = 100 \cdot 2^5 = 100 \cdot 32 = 3200) bacteria.
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Problem: Solve for (x): (\frac{1}{4^x} = 16).
- Solution: Rewrite (\frac{1}{4^x}) as (4^{-x}). Recognize that (16 = 4^2). So, (4^{-x} = 4^2). Set exponents equal: (-x = 2). Solve for (x): (x = -2).
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Problem: Graph (g(x) = \left(\frac{1}{2}\right)^x) and state its domain, range, and asymptote.
- Solution: The graph starts at (0,1) and decreases rapidly to the right. As (x) approaches positive infinity, (g(x)) approaches 0 but never touches the x-axis. That's why, the domain is all real numbers, the range is (y > 0), and the horizontal asymptote is (y = 0).
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Problem: Solve the exponential equation: (3^{x-1} = 27).
- Solution: Recognize that (27 = 3^3). Rewrite the equation as (3^{x-1} = 3^3). Set the exponents equal: (x - 1 = 3). Solve for (x): (x = 4).
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Problem: A car depreciates at a rate of 15% per year. If the car is worth $20,000 initially, write an exponential decay function for the value (V(t)) after (t) years and find the value after 3 years.
- Solution: The general form for decay is (V(t) = V_0 \cdot (1 - r)^t), where (V_0) is the initial value and (r) is the decay rate (as a decimal). Here, (V_0 = 20000) and (r = 0.15). So, (V(t) = 20000 \cdot (0.85)^t). After 3 years: (V(3) = 20000 \cdot (0.85)^3 = 20000 \cdot 0.614125 = 12282.5). The car will be worth approximately $12,283.
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Problem: Solve for (x): (2^{3x} = 64).
- Solution: Recognize that (64 = 2^6). Rewrite the equation as (2^{3x} = 2^6). Set the exponents equal: (3x =
6). Solve for (x): (x = 2) Nothing fancy..
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Problem: Solve for (x): (9^x = 27) Simple, but easy to overlook..
- Solution: Express both sides using a common base. Since (9 = 3^2) and (27 = 3^3), rewrite the equation as ((3^2)^x = 3^3), which simplifies to (3^{2x} = 3^3). Set the exponents equal: (2x = 3). Solve for (x): (x = \frac{3}{2} = 1.5).
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Problem: A radioactive substance has a half-life of 10 days. If you start with 80 grams, write an exponential decay model for the remaining mass (M(t)) after (t) days and calculate how much remains after 30 days.
- Solution: The half-life model is (M(t) = M_0 \cdot \left(\frac{1}{2}\right)^{t/h}), where (M_0) is the initial mass and (h) is the half-life. Here, (M_0 = 80) and (h = 10), so (M(t) = 80 \cdot \left(\frac{1}{2}\right)^{t/10}). After 30 days: (M(30) = 80 \cdot \left(\frac{1}{2}\right)^{30/10} = 80 \cdot \left(\frac{1}{2}\right)^3 = 80 \cdot \frac{1}{8} = 10) grams.
Conclusion
Mastering exponential functions and equations relies on a consistent set of foundational strategies: rewriting terms to share a common base, accurately distinguishing between growth and decay factors, and recognizing key graphical features like horizontal asymptotes and restricted ranges. As problems increase in complexity, you will inevitably encounter equations where bases cannot be easily matched, paving the way for logarithmic functions—the essential inverse tools for isolating variables in exponents. Regular practice with these core concepts not only strengthens algebraic fluency but also builds the analytical framework needed for advanced topics in calculus, finance, and scientific modeling. Always verify your solutions by substituting them back into the original equation, and remember to interpret your results within the context of the problem when dealing with real-world applications. With a solid command of exponentials, you are well-prepared to tackle the next stages of mathematical reasoning with confidence and precision.
Extending the Toolbox: Logarithms and Real‑World Modeling
When the bases of exponential expressions cannot be aligned through simple integer multiples, the next logical step is to introduce logarithms. The definition
[ \log_{b}a = c \quad\Longleftrightarrow\quad b^{c}=a]
provides a direct method for extracting the exponent that appears in the unknown position. For an equation of the form
[ a^{x}=k, ]
taking the logarithm with base (a) yields
[ x=\log_{a}k. ]
If the base is not convenient (e.g., (e) or 10), the change‑of‑base formula
[ \log_{a}k=\frac{\ln k}{\ln a} ]
allows the use of natural or common logarithms available on most calculators. Applying this technique to the equation
[ 5^{2x-1}= 125, ]
we first recognize that (125=5^{3}). Substituting gives
[ 5^{2x-1}=5^{3};\Longrightarrow;2x-1=3;\Longrightarrow;x=2. ]
If the right‑hand side were not an exact power of the base, we would instead write [ 2x-1=\log_{5}k\quad\Longrightarrow\quad x=\frac{\log_{5}k+1}{2} ]
and evaluate the logarithm numerically That alone is useful..
Graphical Insights
Visualizing exponential functions reinforces algebraic manipulation. The graph of
[ y = a^{x} ]
always passes through ((0,1)) and approaches the (x)-axis as (x\to -\infty) when (0<a<1). Adding a vertical shift (k) moves the entire graph up or down, altering the horizontal asymptote to (y=k). And for growth models ((a>1)), the curve rises without bound as (x) increases, while for decay models ((0<a<1)) it descends toward the axis. Here's the thing — scaling by a factor (c) stretches or compresses the graph vertically, affecting the rate at which the function approaches its asymptote. Recognizing these transformations aids in sketching accurate representations and interpreting key features such as intercepts and end behavior Took long enough..
Applications Beyond Pure Mathematics
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Finance – Compound interest problems often involve repeated multiplication by a growth factor. The formula
[ A = P\left(1+\frac{r}{n}\right)^{nt} ]
can be solved for time (t) by taking logarithms, yielding
[ t=\frac{\log!\bigl(\frac{A}{P}\bigr)}{n\log!\bigl(1+\frac{r}{n}\bigr)}. ]
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Biology – Population growth that follows a logistic pattern can be approximated locally by an exponential function. Determining the doubling time of a bacterial culture involves solving
[ N(t)=N_{0}e^{kt}=2N_{0}\quad\Longrightarrow\quad t=\frac{\ln 2}{k}. ]
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Physics – Radioactive decay, capacitor discharge, and cooling processes are all modeled by (N(t)=N_{0}e^{-kt}). The half‑life (T_{1/2}) satisfies
[ \frac{1}{2}=e^{-kT_{1/2}}\quad\Longrightarrow\quad T_{1/2}=\frac{\ln 2}{k}, ]
linking the decay constant directly to a measurable time interval.
Problem‑Solving Strategies - Identify the structure: Determine whether the equation can be rewritten with a common base or whether logarithms are required.
- Isolate the exponential term: Move constant factors to the opposite side before applying logarithmic operations.
- Apply the appropriate logarithm: Use natural logs for calculus‑oriented work, common logs for base‑10 calculations, or a base‑specific log when the base is known.
- Check for extraneous solutions: In equations involving absolute values or piecewise definitions, verify that each candidate satisfies the original expression.
- Interpret the answer: Translate the numerical result back into the context of the problem, ensuring units and realistic ranges are respected.
Looking Ahead
The techniques outlined above lay the groundwork for more sophisticated topics such as logarithmic differentiation, exponential regression, and the analysis of differential equations with variable coefficients. As you progress, you will encounter systems where multiple exponential terms interact, demanding simultaneous use of substitution and logarithmic manipulation. Mastery of these foundational ideas equips you to handle those
Building on this understanding, it becomes clear how essential these transformations are when tackling complex problems across disciplines. So whether you're analyzing economic models, predicting population trends, or solving physics-based scenarios, recognizing when a function needs scaling or when logarithms simplify the path forward is crucial. These tools not only refine your sketches but also deepen your grasp of how mathematical ideas translate into real-world solutions And it works..
By mastering these strategies, you equip yourself to approach unfamiliar challenges with confidence and precision. The ability to interpret graphs, adjust for scaling, and apply logarithmic functions effectively transforms abstract equations into actionable insights.
The short version: these concepts form a vital bridge between theoretical concepts and practical application, empowering you to work through diverse mathematical landscapes with clarity. Concluding this exploration, embracing these principles strengthens your analytical toolkit, ensuring you remain adaptable and informed in both academic and applied settings.
