Chapter 3 - Exponential Functions
| Friday, 27 August 2021 | |
| 4-minute read | |
| 795 words | |
3.1 Exponential Functions and Their Graphs
Linear vs Exponential Growth
Linear functions grow at a constant rate.
An exponential function grows by a multiple of its current value.
Example
Using the equation \(y = a(2)^{b(x-c)} + d\), change the values of \(a, b, c, d\) and determine the effects of them on the graph.
- \(a\) vertically dilates the graph
- \(b\) horizontally compresses the graph
- \(c\) horizontally translates the graph
- \(d\) vertically translates the graph
Euler's Number
Create the graph of \(y = (\frac{1}{x})^x\). What is the right end behavior of this graph?
\[ \lim_{x \to \infty} y \to 2.718... \]
This special value 2.718… equates to Euler's number \(e\).
\[ \lim_{x \to \infty} y \to e \]
Compound Interest Formula
\[ A = P(1 + \frac{r}{n})^{nt} \]
\(A\) = Total amount
\(P\) = Principle amount (starting amount)
\(r\) = Interest rate
\(n\) = Number of times applied per time period
\(t\) = Number of time periods elapsed
Continuous Compounding
\[ A = Pe^{rt} \]
All terms \(A\), \(P\), \(r\), and \(t\) have the same meaning as the compound interest formula.
This is because continuous compounding is derived from the compound interest formula.
Given we have \(A = P(1 + \frac{r}{n})^{nt}\), we can plug in \(n\) to an infinitely large number.
Remember Euler's number that \((1 + \frac{1}{x})^{x}\), where as \(x\) approaches infinity, the value approaches \(2.718...\)
TODO: Finish proofs
Examples
Nelson invests $12000 at a 3% interest rate for 4 years. Compute the value of his investment considering the following ypes of compounding:
- Anually — \(12000(1 + \frac{0.03}{1})^{1 \cdot 4} = \$13506.10\)
- Quarterly — \(12000(1 + \frac{0.03}{4})^{4 \cdot 4} = \$13523.90\)
- Continuously — \(12000 \cdot e^{0.03 \cdot 4} = \$13529.96\)
Let \(y\) represent a mass, in grams, of radioactive H-3 whose half-life is approximately 12 years. The quantity of Hydrogen-3 present after \(t\) years if \(y = 10(\frac{1}{2})^{\frac{t}{12}}\).
- What is the initial mass of Hydrogen-3? 10
- How much of the initial mass is present after 60 years? \(y = 10(\frac{1}{2})^{\frac{60}{12}} = 0.312\)
The approximate number of fruit flies in an experimental population after \(t\) hours is given by the equation
\[Q(t) = 30e^{0.04t}\]
- Find the initial number of fruit flies in the population. \(30\)
- How large is the population after 72 hours? \(30e^{0.04 \cdot 72} = 534\)
- When would you estimate the population of fruit flies to become greater than 500? \[500 = 30e^{0.04t}\] \[\frac{500}{30} = e^{0.04t}\] \[ln(\frac{500}{30}) = ln(e^{0.04t})\] \[0.04t = ln(\frac{500}{30})\] \[t = ln(\frac{500}{30}) \cdot \frac{1}{0.04} = 70.335\]
3.2 Logarithms and Their Graphs
A logarithm is the inverse of an exponential.
\[ y = a^x \]
\[ \log_a(y) = \log_a(a^x) = x \]
Properties
\[ \log_a 1 = 0 \]
\[ \log_a a = 1 \]
\[ \log_a(a^x) = x \]
\[ \log_a(\frac{1}{a^x}) = -x \]
If \(\log_a x = \log_a y\), then \(x = y\)
If \(y = \log_a x\), then \(a^y = x\)
Example
What is the domain of an untransformed logarithmic function?
\[ (0, \infty) \]
Types of Logarithms
\(\log x\) is the common logarithm. It is equal to \(\log_10 x\)
\(\ln x\) is the natural log. It is equal to \(\log_e x\)
Properties of \(\ln\)
\[ \ln 1 = 0 \]
\[ \ln e = 1 \]
\[ \ln e^x = x \]
If \(\ln x = \ln y\), then \(x = y\)
Examples
\[ \ln \frac{1}{e} = -1 \]
\[ e^{ln 5} = 5 \]
\[ 4 \ln 1 = 0 \]
\[ 2 \ln e = 2 \]
3.3 Properties of Logarithms
Proof \(\log_b(uv) = \log_b u + \log_b v\)
Let \(a^{log_a u + log_a v} = uv\)
\(a^u \cdot a^v = a^{u+v}\)
\(a^{log_a u} \cdot a^{log_a v} = uv\)
\(u \cdot v = uv\)
Therefore \(\log_b(uv) = \log_b u + \log_b v\), quod erat demonstrandum
Proof \(\log_b(u^n) = n \log_b u\)
Let \((a^{\log_a u})^n = u^n\)
Change of Base
\[ \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \]
Examples
- Expand \(3log(100x^4y^2)\) \(6 + 12\log x + 6\log y\)
- Condense \(\log_3(64) + 2\log_3(x) - 4\log_3(y)\) \(\log_3(64x^2y^{-4})\) OR \(\log_3(\frac{64x^2}{y^4})\)
Review
- The population of a town in 2007 is 113505 and is increasing at a rate of 1.2% per year. What will the population be in 2012? Set up the formula.
!||\(113505 \cdot (1.012)^5\)||
- A set of bacteria begins with 20 and doubles every 2 hours. How many bacteria would be present 15 hours after the experiment began? Set up the formula.
!||\(20(2)^{\frac{15}{2}}\)||
- The cost of manufactured goods is rising at the rate of inflation (2.3%) anually. Suppose an item costs $12 today. How much will it cost five years from now due to inflation?
!||\(12 \cdot (1.023)^5 = \$13.44\)||
- The cost of a new ATV is $7200. It depreciates at 18% per year. Find the value of the ATV when it is ten years old.
!||\(\$7200 \cdot (1 - 0.18)^{10} = \$989\)||