Introduction
Unit 7 of most high‑school mathematics curricula focuses on exponential and logarithmic functions, two powerful tools that model growth, decay, and many real‑world phenomena. This article provides clear, step‑by‑step homework answers and explains the underlying concepts so you can not only copy the solutions but also understand why each step works. Students often encounter homework that asks for graphing, solving equations, and applying these functions to word problems. By mastering these techniques, you’ll be prepared for quizzes, exams, and future courses such as calculus or statistics.
1. Core Concepts Review
1.1 Exponential Functions
An exponential function has the form
[ f(x)=a\cdot b^{x} ]
where
- (a) is the initial value (the y‑intercept).
- (b) is the base, (b>0) and (b\neq1).
If (b>1) the function shows growth; if (0<b<1) it shows decay.
Key properties
- Domain: ((-\infty,\infty))
- Range: ((0,\infty))
- Horizontal asymptote: (y=0)
1.2 Logarithmic Functions
A logarithmic function is the inverse of an exponential function:
[ y=\log_{b}(x) \quad \Longleftrightarrow \quad b^{y}=x ]
- Base (b) must satisfy the same conditions as above.
- Domain: ((0,\infty))
- Range: ((-\infty,\infty))
- Vertical asymptote: (x=0)
1.3 Important Formulas
| Concept | Formula |
|---|---|
| Change of base | (\displaystyle \log_{b}(x)=\frac{\log_{k}(x)}{\log_{k}(b)}) (choose (k=10) or (e)) |
| Exponential growth/decay | (P(t)=P_{0},e^{kt}) where (k>0) for growth, (k<0) for decay |
| Solving (b^{x}=c) | (x=\log_{b}(c)) |
| Solving (\log_{b}(x)=c) | (x=b^{c}) |
2. Typical Homework Problems and Detailed Answers
2.1 Problem 1 – Graphing an Exponential Function
Question: Sketch the graph of (f(x)=3\cdot 2^{x-1}+4). Identify the vertical shift, horizontal shift, and asymptote It's one of those things that adds up..
Answer Steps:
- Rewrite the function in standard form (f(x)=A\cdot b^{(x-h)}+k).
[ f(x)=3\cdot 2^{x-1}+4 \quad \Rightarrow \quad A=3,; b=2,; h=1,; k=4 ] - Horizontal shift: (h=1) → shift right 1 unit.
- Vertical shift: (k=4) → shift up 4 units.
- Asymptote: The parent exponential (y=2^{x}) has asymptote (y=0). After the vertical shift, the new asymptote is (y=4).
- Plot key points:
- When (x=1): (f(1)=3\cdot 2^{0}+4=3+4=7) (this is the new y‑intercept).
- When (x=2): (f(2)=3\cdot 2^{1}+4=6+4=10).
- When (x=0): (f(0)=3\cdot 2^{-1}+4=1.5+4=5.5).
- Draw the curve: Start just above the asymptote (y=4), pass through the plotted points, and continue upward as (x) increases.
Result: The graph is a right‑shifted, upward‑shifted exponential curve with asymptote (y=4) Most people skip this — try not to. Surprisingly effective..
2.2 Problem 2 – Solving an Exponential Equation
Question: Solve (5^{2x-3}=125) It's one of those things that adds up..
Answer Steps:
-
Recognize that (125=5^{3}) Still holds up..
-
Rewrite the equation:
[ 5^{2x-3}=5^{3} ]
-
Because the bases are equal, set the exponents equal:
[ 2x-3=3 ]
-
Solve for (x):
[ 2x=6 \quad \Rightarrow \quad x=3 ]
Answer: (x=3).
2.3 Problem 3 – Solving a Logarithmic Equation
Question: Find (x) such that (\log_{2}(x+5)-\log_{2}(x-1)=3).
Answer Steps:
-
Use the logarithm property (\log_{b}A-\log_{b}B=\log_{b}\left(\frac{A}{B}\right)):
[ \log_{2}!\left(\frac{x+5}{x-1}\right)=3 ]
-
Convert to exponential form:
[ \frac{x+5}{x-1}=2^{3}=8 ]
-
Multiply both sides by ((x-1)):
[ x+5=8(x-1) ]
-
Expand and solve:
[ x+5=8x-8 \quad\Rightarrow\quad 5+8=8x-x \quad\Rightarrow\quad 13=7x \quad\Rightarrow\quad x=\frac{13}{7} ]
-
Check domain: Both arguments of the logs must be positive.
- (x+5 = \frac{13}{7}+5 = \frac{13+35}{7}= \frac{48}{7}>0)
- (x-1 = \frac{13}{7}-1 = \frac{13-7}{7}= \frac{6}{7}>0)
The solution is valid.
Answer: (x=\dfrac{13}{7}) Simple as that..
2.4 Problem 4 – Real‑World Exponential Growth
Question: A bacteria culture doubles every 4 hours. If the initial count is 250 cells, how many cells will be present after 24 hours?
Answer Steps:
-
Determine the growth factor per hour. Since it doubles every 4 h, the base per 4 h is (b=2).
-
Write the model using the number of 4‑hour intervals:
[ P(t)=250\cdot 2^{,t/4} ]
where (t) is measured in hours.
-
Plug (t=24):
[ P(24)=250\cdot 2^{24/4}=250\cdot 2^{6}=250\cdot 64=16,000 ]
Answer: After 24 hours the culture contains 16,000 cells Simple, but easy to overlook..
2.5 Problem 5 – Logarithmic Decay in Radioactive Dating
Question: The half‑life of a certain isotope is 5,200 years. If a sample currently contains 12 % of the original amount, how old is the sample?
Answer Steps:
-
Use the decay model (A(t)=A_{0},e^{-kt}) where (k=\frac{\ln 2}{\text{half‑life}}) Not complicated — just consistent..
[ k=\frac{\ln 2}{5200}\approx\frac{0.6931}{5200}\approx1.333\times10^{-4} ]
-
Set up the proportion (A(t)/A_{0}=0.12):
[ 0.12=e^{-kt} ]
-
Take natural logs of both sides:
[ \ln 0.12=-kt \quad\Rightarrow\quad t=-\frac{\ln 0.12}{k} ]
-
Compute:
[ \ln 0.But 12\approx -2. Now, 1203 \quad\Rightarrow\quad t=\frac{2. 1203}{1.
Answer: The sample is approximately 15,900 years old.
2.6 Problem 6 – Converting Between Log Bases
Question: Express (\log_{5}(23)) using common logarithms (base 10) Small thing, real impact..
Answer Steps:
-
Apply the change‑of‑base formula:
[ \log_{5}(23)=\frac{\log_{10}(23)}{\log_{10}(5)} ]
-
If a calculator is allowed, compute the two common logs:
- (\log_{10}(23)\approx 1.3617)
- (\log_{10}(5)\approx 0.6990)
-
Divide:
[ \frac{1.3617}{0.6990}\approx 1.948 ]
Answer: (\log_{5}(23)\approx 1.95) (using base 10 logs) Most people skip this — try not to..
2.7 Problem 7 – Inverse Function of an Exponential
Question: Find the inverse of (f(x)=4e^{2x}-7).
Answer Steps:
-
Replace (f(x)) with (y):
[ y=4e^{2x}-7 ]
-
Solve for (x):
- Add 7: (y+7=4e^{2x})
- Divide by 4: (\dfrac{y+7}{4}=e^{2x})
- Apply natural log: (\ln!\left(\dfrac{y+7}{4}\right)=2x)
- Divide by 2: (x=\dfrac{1}{2}\ln!\left(\dfrac{y+7}{4}\right))
-
Swap (x) and (y) to write the inverse function:
[ f^{-1}(x)=\frac{1}{2}\ln!\left(\frac{x+7}{4}\right) ]
Answer: (f^{-1}(x)=\dfrac{1}{2}\ln!\bigl(\frac{x+7}{4}\bigr)).
3. Common Mistakes and How to Avoid Them
| Mistake | Why It Happens | Correct Approach |
|---|---|---|
| Treating (\log_{b}(a)=b^{a}) | Confusing the definition of logarithm with exponentiation. On top of that, | Remember: (\log_{b}(a)=c) iff (b^{c}=a). |
| Ignoring domain restrictions | Plugging negative numbers into logs. | Always check that arguments of logs are positive before solving. |
| Mismatching bases when simplifying | Using (\log_{2}(8)=3) but then applying base 10 rules. | Keep the same base throughout a single equation, or use the change‑of‑base formula. |
| Forgetting the horizontal/vertical shift | Graphing (y=2^{x}+3) as if asymptote were still (y=0). | The asymptote moves with the vertical shift: (y=k) for (y=a\cdot b^{x}+k). |
| Dividing by zero when using change of base | Accidentally choosing a base where (\log_{k}(b)=0). | Choose (k) such that (\log_{k}(b)\neq0); common choices are 10 or (e). |
4. Frequently Asked Questions (FAQ)
Q1: Can I use a calculator for every step?
A: While calculators speed up numeric work, you should still understand the algebraic manipulations (e.g., isolating the variable, applying log properties). Exams often limit calculator use, so practice the manual steps Most people skip this — try not to. Still holds up..
Q2: What is the difference between natural logs and common logs?
A: Natural logs have base (e) ((\ln)), useful in continuous growth/decay models. Common logs have base 10 ((\log)), often appear in engineering contexts. Both obey the same rules; only the base changes.
Q3: How do I know whether a problem involves growth or decay?
A: Look at the base (b). If (b>1) (or the exponent’s coefficient (k>0) in (e^{kt})), the function grows. If (0<b<1) (or (k<0)), it decays Small thing, real impact..
Q4: Why do exponential and logarithmic functions have opposite shapes?
A: They are inverses. The reflection of an exponential graph across the line (y=x) yields its logarithmic counterpart, swapping horizontal and vertical asymptotes Simple, but easy to overlook..
Q5: Is (\log_{b}(1)=0) always true?
A: Yes. Since any non‑zero base raised to the 0 power equals 1, the logarithm of 1 is always 0, regardless of the base (as long as the base meets the standard conditions).
5. Tips for Acing Unit 7 Homework
- Create a quick reference sheet of key formulas, especially the change‑of‑base and exponential‑decay equations.
- Practice domain checks after each solution; a missed domain restriction can turn a correct algebraic answer into an invalid one.
- Sketch rough graphs before solving equations; visualizing the curve often reveals extraneous roots or helps decide which branch of a function to use.
- Convert all exponentials to the same base when possible. To give you an idea, rewrite (9^{x}) as ((3^{2})^{x}=3^{2x}) to match a base of 3 in another term.
- Use natural logs for continuous models (population, cooling, radioactive decay) because the derivative of (e^{kt}) is simply (ke^{kt}), simplifying calculus later on.
6. Conclusion
Unit 7’s exponential and logarithmic functions may initially appear intimidating, but the homework answers presented here demonstrate that each problem follows a logical sequence of properties and algebraic steps. By internalizing the core concepts, mastering the key formulas, and vigilantly checking domains, you can solve graphing tasks, exponential and logarithmic equations, and real‑world application problems with confidence. Keep this guide handy, practice regularly, and the patterns will become second nature—setting a solid foundation for advanced mathematics and scientific coursework.
