Unit 6 Exponents And Exponential Functions Homework 10 Answer Key

Unit 6 Exponents and Exponential Functions – Homework 10 Answer Key ### Introduction

This article provides a comprehensive answer key for Homework 10 in the unit on Exponents and Exponential Functions. The purpose is to clarify each problem, demonstrate the underlying principles, and reinforce the concepts that students need to master for future assessments. By working through the solutions step‑by‑step, learners can verify their answers, identify common errors, and deepen their understanding of how exponents behave in both algebraic and real‑world contexts.

Key Concepts Reviewed

Before diving into the answer key, it is essential to revisit the core ideas that govern this unit:

1. Definition of an exponent – For a real number a and a positive integer n, aⁿ means multiplying a by itself n times.
2. Properties of exponents – These include the product rule (aᵐ·aⁿ = aᵐ⁺ⁿ), the quotient rule (aᵐ / aⁿ = aᵐ⁻ⁿ), the power‑of‑a‑power rule ((aᵐ)ⁿ = aᵐⁿ), the zero exponent rule (a⁰ = 1 for a ≠ 0), and the negative exponent rule (a⁻ⁿ = 1 / aⁿ). 3. Exponential functions – Functions of the form f(x) = a·bˣ, where b > 0 and b ≠ 1. They model growth (b > 1) or decay (0 < b < 1) processes.
3. Graphical interpretation – The graph of an exponential function is a smooth curve that either rises rapidly (growth) or falls toward the horizontal asymptote (decay).

A solid grasp of these fundamentals makes solving the homework problems straightforward.

Homework 10 Overview

Homework 10 typically consists of ten items that test the student’s ability to:

• Simplify expressions using exponent rules.
• Solve exponential equations by rewriting bases.
• Interpret exponential growth or decay in word problems.
• Apply the concept of e (the base of natural logarithms) when dealing with continuous growth models. Below, each problem is reproduced, followed by a detailed solution and the final answer.

Problem‑by‑Problem Solutions

1. Simplify the expression ((2⁴·3⁻²) ÷ (6⁻¹))

Solution - First rewrite all bases with the same prime factors: (6 = 2·3).

• Apply the quotient rule: ((2⁴·3⁻²) ÷ (6⁻¹) = (2⁴·3⁻²)·6¹).
• Substitute (6 = 2·3): ((2⁴·3⁻²)·(2·3) = 2⁴·2·3⁻²·3 = 2⁵·3⁻¹).
• Convert the negative exponent: (2⁵·3⁻¹ = \dfrac{2⁵}{3} = \dfrac{32}{3}). Answer: (\boxed{\dfrac{32}{3}})

2. Solve for x: (5^{2x} = 125)

Solution

• Recognize that (125 = 5³). - Set the exponents equal: (2x = 3).
• Solve: (x = \dfrac{3}{2}).

Answer: (\boxed{\dfrac{3}{2}})

3. Evaluate the exponential function (f(x) = 3·2^{x}) at (x = 4)

Solution

• Substitute (x = 4): (f(4) = 3·2^{4} = 3·16 = 48).

Answer: (\boxed{48})

4. Determine the y‑intercept of the function (g(x) = 0.5·(0.8)^{x}) Solution

• The y‑intercept occurs when (x = 0).
• Compute (g(0) = 0.5·(0.8)^{0} = 0.5·1 = 0.5).

Answer: (\boxed{0.5})

5. Simplify (\displaystyle \left(\frac{a^{3}b^{-2}}{c^{4}}\right)^{-1})

Solution

• Apply the outer exponent (-1) to each factor: (\left(\frac{a^{3}b^{-2}}{c^{4}}\right)^{-1}= \frac{c^{4}}{a^{3}b^{-2}}). - Move the negative exponent in the denominator to the numerator: (\frac{c^{4}}{a^{3}b^{-2}} = \frac{c^{4}b^{2}}{a^{3}}).

Answer: (\boxed{\dfrac{c^{4}b^{2}}{a^{3}}})

6. Solve the exponential equation (e^{x} = 7)

Solution

• Take the natural logarithm of both sides: (\ln(e^{x}) = \ln 7).
• Use the property (\ln(e^{x}) = x): (x = \ln 7).

Answer: (\boxed{\ln 7})

7. A population of bacteria doubles every 5 hours. If the initial population is 200, write an exponential model and find the population after 15 hours.

Solution

• Doubling time (T = 5) hours ⇒ growth factor per hour is (2^{1/5}).
• Model: (P(t) = 200·2^{t/5}).
• For (t = 15): (P(15) = 200·2^{15/5} = 200·2^{3} = 200·8 = 1600).

Answer: (\boxed{1600}) bacteria

8. Simplify (\displaystyle \frac{(x^{3}y^{-2})^{2}}{x^{-1}y^{4}})

Solution

• First expand the numerator: ((x^{3}y^{-2})^{2}=x^{6}y^{-4}).
• Form the fraction: (\frac{x^{6}y^{-4}}{x^{-1}y^{4}} = x^{6-(-1)}·y^{-4-4}=x^{7}y^{-8}). - Convert the negative exponent: (x^{7}y^{-8}= \dfrac{x^{7}}{y^{8}}).

Answer: (\boxed{\dfrac{x^{7}}{y^{8}}})

9. Find the horizontal asymptote of (h(x) =

9. Find the horizontal asymptote of

[ h(x)=\frac{5\cdot 2^{x}+3}{2^{x}+1} ]

Solution
To locate the horizontal asymptote we examine the behavior of (h(x)) as (x\to\pm\infty).

1. Factor the dominant exponential term from numerator and denominator:

   [ h(x)=\frac{5\cdot 2^{x}+3}{2^{x}+1} =\frac{2^{x}\bigl(5+\tfrac{3}{2^{x}}\bigr)}{2^{x}\bigl(1+\tfrac{1}{2^{x}}\bigr)} =\frac{5+\tfrac{3}{2^{x}}}{1+\tfrac{1}{2^{x}}}. ]

2. Take the limit as (x\to\infty). Since (\displaystyle\frac{3}{2^{x}}\to0) and (\displaystyle\frac{1}{2^{x}}\to0),

   [ \lim_{x\to\infty}h(x)=\frac{5+0}{1+0}=5. ]

3. Check the limit as (x\to -\infty) (optional). When (x) is large negative, (2^{x}) approaches 0, so

   [ h(x)\approx\frac{3}{1}=3, ]

   which means the function approaches the horizontal line (y=3) on the far left. Thus the graph has two distinct horizontal asymptotes:

   [ y=5 \quad\text{as } x\to\infty,\qquad y=3 \quad\text{as } x\to -\infty. ]

Answer – The right‑hand horizontal asymptote is (\boxed{5}) (the left‑hand one is (\boxed{3})).

10. Solve the logarithmic equation (\displaystyle \log_{3}(x+4)=2)

Solution
Rewrite the logarithmic equation in exponential form:

[ \log_{3}(x+4)=2 ;\Longrightarrow; x+4 = 3^{2}=9. ]

Now isolate (x):

[ x = 9-4 = 5. ]

Answer – (\boxed{5}).

Conclusion

We have walked through a diverse set of exponential and logarithmic problems, applying the fundamental rules of exponents, logarithms, and limits. By:

1. Re‑expressing bases with common factors,
2. Using the definition of logarithms to translate between logarithmic and exponential forms,
3. Substituting specific values into exponential models,
4. Factoring dominant terms to uncover horizontal asymptotes, and
5. Converting logarithmic statements into algebraic equations,

we were able to isolate variables, simplify complex expressions, and interpret the long‑term behavior of functions. Mastery of these techniques equips you to tackle more advanced topics such as continuous growth models, compound interest, and differential equations that rely on the same underlying principles. Keep practicing, and the patterns will become second nature.

11. Solve the exponential equation (2^{x+1} = 5^{x-2})

Solution
Since the bases are different and not powers of a common base, we apply logarithms to both sides. Using the natural logarithm (or common logarithm):

[ \ln(2^{x+1}) = \ln(5^{x-2}). ]

Apply the power rule (\ln(a^b) = b\ln a):

[ (x+1)\ln 2 = (x-2)\ln 5. ]

Expand both sides:

[ x\