Unit 7 Exponential And Logarithmic Functions Homework 4 Answers

Unit 7 exponential and logarithmic functions homework 4 answers provides step‑by‑step solutions that help students master the core ideas of growth, decay, and the inverse relationship between exponentials and logarithms. By working through these answers, learners can see how to apply properties such as the product, quotient, and power rules, convert between exponential and logarithmic forms, and solve real‑world problems involving compound interest, half‑life, and pH calculations. The following guide breaks down each type of problem typically found in Homework 4, explains the reasoning behind every step, and highlights common pitfalls to avoid.

1. Core Concepts Reviewed in Homework 4

Before diving into the answers, it is useful to recall the main topics covered in Unit 7, Lesson 4:

• Exponential functions of the form (f(x)=a\cdot b^{x}) where (a\neq0) and (b>0, b\neq1).
• Logarithmic functions as the inverse of exponentials: (y=\log_{b}x) means (b^{y}=x).
• Change‑of‑base formula: (\displaystyle \log_{b}x=\frac{\log_{k}x}{\log_{k}b}) for any convenient base (k) (usually 10 or (e)).
• Properties of logarithms: product (\log_{b}(MN)=\log_{b}M+\log_{b}N), quotient (\log_{b}\left(\frac{M}{N}\right)=\log_{b}M-\log_{b}N), and power (\log_{b}(M^{p})=p\log_{b}M).
• Solving exponential and logarithmic equations by isolating the term, rewriting with a common base, or applying logarithms to both sides.
• Applications: compound interest (A=P\left(1+\frac{r}{n}\right)^{nt}), continuous growth (A=Pe^{rt}), half‑life (N(t)=N_{0}\left(\frac{1}{2}\right)^{t/T}), and pH (=-\log_{[H^{+}]}).

Understanding these ideas makes it easier to follow the answer key and to adapt the methods to similar problems.

2. Walkthrough of Typical Homework 4 Problems

Below are representative questions from Homework 4, each accompanied by a detailed solution that mirrors the answer key. The explanations stress why each step is taken, not just what to do.

2.1. Solving Simple Exponential Equations

Problem: Solve (3\cdot 2^{x}=48).

Solution:

1. Isolate the exponential term – divide both sides by 3:
   [ 2^{x}= \frac{48}{3}=16. ]
2. Express the constant as a power of the base – note that (16=2^{4}).
3. Set the exponents equal (since the bases are identical and the function is one‑to‑one):
   [ x=4. ]

Answer: (x=4) That's the part that actually makes a difference..

Why it works: The exponential function (2^{x}) is strictly increasing, so equal outputs imply equal inputs.

2.2. Using Logarithms to Solve Exponential Equations

Problem: Solve (5^{2x-1}=125) But it adds up..

Solution:

1. Recognize that (125=5^{3}).
2. Rewrite the equation with the same base:
   [ 5^{2x-1}=5^{3}. ]
3. Equate the exponents:
   [ 2x-1=3 ;\Longrightarrow; 2x=4 ;\Longrightarrow; x=2. ]

Answer: (x=2).

Alternative method (using logs): Take (\log) of both sides, apply the power rule, and solve for (x). Both routes give the same result; the same‑base method is quicker when the numbers are powers of the base It's one of those things that adds up..

2.3. Solving Logarithmic Equations

Problem: Solve (\log_{4}(x+3)=2).

Solution:

1. Convert the logarithmic statement to its exponential form using the definition (\log_{b}y=z \iff b^{z}=y):
   [ 4^{2}=x+3. ]
2. Compute (4^{2}=16).
3. Isolate (x):
   [ x=16-3=13. ]
4. Check the domain: The argument of a log must be positive; (x+3=16>0), so the solution is valid.

Answer: (x=13).

Common mistake: Forgetting to verify that the resulting (x) keeps the argument inside the log positive. Always perform a domain check.

2.4. Applying the Change‑of‑Base Formula

Problem: Evaluate (\log_{7}50) using a calculator (base 10 or natural log) Nothing fancy..

Solution:

1. Apply the change‑of‑base formula with base 10:
   [ \log_{7}50=\frac{\log_{10}50}{\log_{10}7}. ]
2. Use a calculator: (\log_{10}50\approx1.69897) and (\log_{10}7\approx0.84510).
3. Divide:
   [ \frac{1.69897}{0.84510}\approx2.010. ]

Answer: (\log_{7}50\approx2.01).

Note: The same result is obtained with natural logs: (\displaystyle \frac{\ln 50}{\ln 7}) And that's really what it comes down to..

2.5. Compound Interest Application

Problem: If $2,000 is invested at an annual rate of 4.5% compounded quarterly, what is the balance after 6 years?

Solution:

1. Identify the variables: (P=2000), (r=0.045), (n=4) (quarterly), (t=6) And it works..

2. Plug into the compound‑interest formula (A=P\left(1+\frac{r}{n}\right)^{nt}):
   [ A=2000\left

3. Plug into the compound‑interest formula (A=P\left(1+\frac{r}{n}\right)^{nt}):
   [ A=2000\left(1+\frac{0.045}{4}\right)^{4\cdot 6} =2000\left(1+0.01125\right)^{24}. ]

4. Evaluate the base and exponent:
   [ 1.01125^{24}\approx 1.30765. ]

5. Multiply by the principal:
   [ A\approx 2000 \times 1.30765 \approx 2615.30. ]

Answer: After 6 years, the balance is approximately $2,615.30 Simple, but easy to overlook..

Tip: When working with compound interest, always confirm that the compounding frequency (n) matches the time period of the rate (r). Mismatched units are a common source of error No workaround needed..

2.6. Exponential Growth Model

Problem: A population of bacteria doubles every 3 hours. If the initial population is 500, find the population after 18 hours Nothing fancy..

Solution:

1. Write the growth model: (P(t)=P_0 \cdot 2^{t/d}), where (P_0=500), doubling time (d=3) hours, and (t=18) hours.
2. Substitute the values:
   [ P(18)=500 \cdot 2^{18/3}=500 \cdot 2^{6}. ]
3. Simplify:
   [ 2^{6}=64 \quad\Rightarrow\quad P(18)=500 \times 64 = 32{,}000. ]

Answer: The population after 18 hours is 32,000 bacteria Simple, but easy to overlook..

Insight: Exponential growth problems often reduce to counting how many doubling (or halving) periods fit into the total time Worth keeping that in mind. Simple as that..

2.7. Newton’s Law of Cooling

Problem: A cup of coffee at (95^{\circ}\text{C}) is left in a room at (20^{\circ}\text{C}). After 5 minutes, the temperature drops to (70^{\circ}\text{C}). What is the temperature after 15 minutes?

Solution:

1. Apply Newton’s Law of Cooling:
   [ T(t)=T_s+(T_0-T_s)e^{-kt}, ]
   where (T_s=20), (T_0=95), and (T(5)=70).
2. Find the decay constant (k):
   [ 70=20+(95-20)e^{-5k} ;\Rightarrow; 50=75e^{-5k} ;\Rightarrow; e^{-5k}=\frac{2}{3}. ]
   Taking natural logs:
   [ -5k=\ln!\left(\tfrac{2}{3}\right) ;\Rightarrow; k=-\tfrac{1}{5}\ln!\left(\tfrac{2}{3}\right)\approx0.0811. ]
3. Predict the temperature at (t=15) minutes:
   [ T(15)=20+75e^{-0.0811(15)} =20+75e^{-1.2165} \approx20+75(0.296) \approx42.2. ]

Answer: After 15 minutes, the coffee is approximately (42.2^{\circ}\text{C}).

Application: This model is widely used in forensics, food safety, and engineering to estimate cooling or warming rates.

2.8. pH Calculation in Chemistry

Problem: Calculate the pH of a solution with a hydrogen‑ion concentration of (3.2\times10^{-5}) M.

Solution:

1. Recall the pH definition: (\text{pH}=-\log_{10}[H^+]).
2. Substitute the concentration:
   [ \text{pH}=-\log_{10}(3.2\times10^{-5}). ]
3. Use logarithm properties:
   [ \text{pH}=-(\log_{10}3.2+\log_{10}10^{-5}) =-(\log_{10}3.2-5) =5-\log_{10}3.2. ]
4. Evaluate: (\log_{10}3.2\approx0.505), so
   [ \text{pH}\approx5-0.505=4.495. ]

Answer: The pH is approximately 4.50, indicating an acidic solution.

Real‑world note: pH values are crucial in fields ranging from environmental science to pharmaceuticals