Chapter 6 Notes, Precalculus 3e Stitz/Zeager
6 Exponential and Logarithmic Functions
6.1 Introduction to Exponential and Logarithmic Functions
Exponential Functions
Graphs of exponential functions
Consider \(f(x)=2^x\) and \(g(x) = 4^x\).
Let’s look at a table of \(x\) and \(y\) values for these functions.
| \(x\) | \(f(x)\) | \(g(x)\) |
|---|---|---|
| -2 | 1/4 | 1/16 |
| -1 | 1/2 | 1/4 |
| 0 | 1 | 1 |
| 1 | 2 | 4 |
| 2 | 4 | 16 |
| 3 | 8 | 64 |
We can graph this data on the \(xy\)-axes.

What about the graph of \(f(x) = 2^{-x}\)?

In general we have two basic shapes. \(y = a^x\) and \(y = a^{-x}\). If we plot them on the same set of axes we can see they are very similar:

The Natural Base \(e\)
\(e\) is just a number
Since \(e\) is a number we can use it in the exponential function \(f(x)=e^x\). Why? Because it works in many situations.
Properties of exponents
Let \(a\) and \(b\) be positive numbers with \(a \neq 1\), \(b \neq 1\) and let \(x\) and \(y\) be real numbers. Then:
- Exponent Laws:
- \(a^x a^y = a^{x+y}\)
- \((a^x)^y = a^{xy}\)
- \((ab)^x = a^x b^x\)
- \(\displaystyle \left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}\)
- \(\displaystyle \frac{a^x}{a^y} = a^{x-y}\)
- \(a^x = a^y\) if and only if \(x=y\).
Logarithms
Let \(a > 0\), \(a \neq 1\) and \(x > 0\) then \[y=\log_a (x) \Longleftrightarrow x=a^y\]
The function \(f(x) = \log_a (x)\) is called the logarithmic function with base \(a\)
OR
“log base \(a\)”
In words: \(\log_a (x)=y\) The logarithm of a number (\(x\)) is the power (\(y\)) to which the base (\(a\)) must be raised to get that number (\(x\))
Graphing logarithms
Suppose we have the function \(y = f(x) = b^x\).
Then the inverse function is \(x = b^y\) OR \(y = f^{-1}(x) = \log_b(x)\). So the logarithm is the inverse of the exponential function.

Since the two functions are inverses of each other then \[b^{\log_b(x)}=x \qquad \text{AND} \qquad \log_b(b^x) = x\]
Properties of Logarithms
Let \(b\) be a positive real number with \(b \neq 1\), and let \(x\) be any real number. Then:
- \(\log_b(1) = 0\) \(\qquad\) i.e. \(b^0 = 1\)
- \(\log_b(b) = 1\) \(\qquad\) i.e. \(b^1 = b\)
- \(\log_b(b^x) = x\) \(\qquad\) i.e. \(b^x = b^x\)
- \(b^{\log_b(x)} = x\) if \(x>0\)
The natural logarithm
This is the same as before but now we use base \(e\) where \(e\) is the number we found in the previous section. Since the log base \(e\) shows up so often we call it the natural log. \[\log_e(x) = \ln(x)\]
We also use log base 10 very often so we abbreviate that as \[\log_{10}(x) = \log(x)\]
Your calculator follows the same convention.
- \(\log(1000)\)
- \(\log_9(243)\)
- \(\log_b(b^{-3})\)
- \(\ln(e^{-2})\)
- \(\log(345)\)
- \(\log(4/5)\)
- \(3 \ln(1+\sqrt{3})\)
- Rewrite \(\displaystyle \log_{64}8=\frac{1}{2}\) in exponential form.
- Rewrite \(\displaystyle 4^{-2}=\frac{1}{16}\) in logarithmic form.
- Evaluate \(\displaystyle \log_{8}2\).
- Use the properties to evaluate \(\displaystyle \ln\left(\frac{1}{e^{21}}\right)\)
- Find the domain of \(g(x)=\ln(13-x)\)
- Use the properties to evaluate \(-73 \ln(e)\)
- Use the properties to evaluate \(e^{\ln(55)}\)
6.2 Properties of Logarithms
Change of Base Formula
Let \(a, b, x\) be positive real numbers with \(a \neq 1, b\neq 1\). Then \[\log_a(x)=\frac{\log_b(x)}{\log_b(a)} \qquad \text{(For any $b$)}\]
For the calculator you can use either base 10 or base \(e\). \[\log_a(x)=\frac{\log(x)}{\log(a)} \qquad \text{OR} \qquad \log_a(x)=\frac{\ln(x)}{\ln(a)}\]
- \(\log_{10}(10z) =\)
- \(\log_6\left(\frac{y}{2}\right) =\)
- \(\log_5(\sqrt[3]{5}) = \log_5\left(5^{1/3}\right) =\)
- \(\displaystyle \ln\left(\frac{x^2-1}{x^3}\right)\)
- \(\displaystyle \log\left(\frac{\sqrt{x} y^4}{z^4}\right)\)
- \(\ln(x-2) - \ln(x+2) =\)
- \(\log_b w + \log_b x - \log_b y =\)
- \(4[\ln z + \ln(z + 5)] - 2\ln(z-5) =\)
- Simplify by using common logarithms: \(\log_2(29)\)
- Simplify by using natural logarithms: \(\log_4 7\)
- Write the expression in terms of \(\ln 5\) and \(\ln 3\): \(\displaystyle \ln\frac{5}{81}\)
- Expand the logarithmic expression: \(\displaystyle \ln\frac{\sqrt[7]{x-1}}{2}\)
- Use the properties of logarithms to expand the expression: \(\log_6 (12xy^3)\)
- Write as a single logarithm: \(\displaystyle \frac{1}{3}[\log_6 x + \log_6 (x+7)]\)
- Write as a single logarithm: \(\displaystyle \frac{1}{8}[\log_8 (x-4) + 5 \log_8 (x^2+9)]\)
6.3 Exponential Equations
Recall: \[a^x=a^y ~ \Longleftrightarrow ~ x = y\] and \[\log_ax = \log_a y~ \Longleftrightarrow ~ x = y\]
- \(64^{x+3}=16\)
- \(3^x = 243\)
- \(\displaystyle \left(\frac{1}{4}\right)^x=64\)
- \(2^{x-3} = 32\)
- \(e^{9x} = 15\)
- \(2+e^{x+2} = 32\)
- \(e^{2x} = 50\)
- \(6^{5x}=3000\)
- \(\displaystyle \frac{119}{e^{6x} - 14} = 7\)
6.4 Logarithmic Equations
\(\log\left( \dfrac{1}{\sqrt[7]{u^3}} \right)\)
\(-\dfrac{3}{7} \log\left( u \right)\)
\(\log\left( \dfrac{1}{u^{3/7}} \right)\)
\(-\dfrac{1}{7}\log\left( u^3 \right)\)
\(-\log\left( u^{3/7} \right)\)
- \(\ln x - \ln 5 = 0\)
- \(\log_x 625 = 4\)
- \(\ln (2x-1) = 0\)
- \(5+5\ln x = 30\)
- \(\log 2x - \log 8x^2 = 3\)
- \(\log_6 6^{2x-1}\)
- \(\ln e^{x^4}\)
- \(\ln 4x = 1\)
- \(\log_6 6x + \log_6(x+5) = 2\)
- \(\log_6 x^2 - \log_6 (x+1) = 2\)
6.5 Exponential and Logarithmic Models
Finance: Compound interest.
\(P\) is some principal amount of money.
\(r\) is the interest rate per year.
Simple interest:
After 1 year we have \(P_1 = P +rP = P(1+r)\)
After 2 year we have \(P_2 = P_1 +rP_1 = P_1(1+r)=P(1+r)(1+r)=P(1+r)^2\)
In fact at year \(t\) you have \[P_t = P(1+r)^t\]
If you compound each month then you have to add 1/12 of the interest every month and you get \[P_t = P\left(1+\frac{r}{12}\right)^{12t}, \qquad \text{where } t = \text{number of years}\]
In general if you compound \(n\) times per year you have \[P_t = P\left(1+\frac{r}{n}\right)^{nt}\]
If we compound continuously that mean \(n \rightarrow \infty\) so \[P_t = Pe^{rt}\]
- \(t=1\) year
- \(t = 10\) years
- \(t = 50\) years
The formula for interest compounded continuously is \[P_t = P_0 e^{rt}\]
To double the investment we want to find the value of \(t\) when \(P_t = 2P_0\).
step 1: Find \(k\) by solving the formula \(280 = 250 e^{k(10)}\).
step 2: Find \(t\) for \(N = 2(250) = 500\). (double the original population)
Decay Model: \[A = A_0 e^{-kt}\] where \(A_0\) is the original amount of material, \(k\) is the decay constant and \(A\) is the amount of material left after \(t\) years.
Step 1: Find \(k\) using the half life. \(\frac{1}{2} A_0 = A_0 e^{-k 5730}\)
Step 2: Find \(t\) for \(0.15 A_0 = A_0 e^{-kt}\).