yes no
Note that
\[ b^{-x} = \left (b^{-1}\right )^x = \left (\frac {1}{b}\right )^x. \]
Exponential and logarithmic functions illuminated.
Exponential and logarithmic functions may seem somewhat esoteric at first, but they model many phenomena in the real-world.
1 What are exponential and logarithmic functions?
An exponential function is a function of the form
\[ f(x) = b^x \]
where \(b\neq 1\) is a positive real number.
The domain of an exponential function is \((-\infty ,\infty )\). Is \(b^{-x}\) an exponential function?
A logarithmic function is a function defined as follows
\[ \log _b(x) = y \qquad \text {means that}\qquad b^y = x \]
where \(b\ne 1\) is a positive real
number. The domain of a logarithmic function is \((0,\infty )\). In either definition above \(b\) is called the base.
Remember that with exponential and logarithmic functions, there is one very special base:
\[ e = 2.7182818284590\ldots \]
This is an irrational number that you will see frequently. The exponential with base \(e\), \(f(x) = e^x\) is often called the ‘natural exponential’ function. For the logarithm with base \(e\), we have a special notation, \(\ln (x)\) is ‘natural logarithm’ function.
1.1 Connections between exponential functions and logarithms
Let \(b\) be a positive real number with \(b\ne 1\).
- \(b^{\log _b(x)} = x\) for all positive \(x\)
- \(\log _b(b^x) = x\) for all real \(x\)
What exponent makes the following expression true?
\[ 3^x = e^{\left ( x \cdot \answer {\ln 3} \right )}. \]
2 What can the graphs look like?
2.1 Graphs of exponential functions
Here we see the the graphs of four exponential functions.
![[Picture]](digInExponentialAndLogarithmeticFunctions.online-aaf755887392fd061217cde0ece9bd78.svg)
Match the curves \(A\), \(B\), \(C\), and \(D\) with the functions
\[ e^x, \qquad \left (\frac {1}{2}\right )^{x}, \qquad \left (\frac {1}{3}\right )^{x}, \qquad 2^{x}. \]
One way to solve
these problems is to compare these functions along the vertical line \(x=1\),
![[Picture]](digInExponentialAndLogarithmeticFunctions.online-43a0645af5aaa9ef6451fd43b8e5d769.svg)
Note
\[ \left (\frac {1}{3}\right )^1 < \left (\frac {1}{2}\right )^1 < 2^1 < e^1. \]
Hence
we see:
- \(\left (\frac {1}{3}\right )^{x}\) corresponds to \(\answer [given]{B}\).
- \(\left (\frac {1}{2}\right )^{x}\) corresponds to \(\answer [given]{A}\).
- \(2^x\) corresponds to \(\answer [given]{D}\).
- \(e^x\) corresponds to \(\answer [given]{C}\).
2.2 Graphs of logarithmic functions
Here we see the the graphs of four logarithmic functions.
![[Picture]](digInExponentialAndLogarithmeticFunctions.online-9215c41b1eff75857ab69b5ef62e8e10.svg)
Match the
curves \(A\), \(B\), \(C\), and \(D\) with the functions
\[ \ln (x),\qquad \log _{1/2}(x), \qquad \log _{1/3}(x),\qquad \log _2(x). \]
First remember what \(\log _b(x)=y\) means:
![[Picture]](digInExponentialAndLogarithmeticFunctions.online-9f78b019a273fea9f83739d8bd21862a.svg)
Note again (this
is from the definition of a logarithm)
\[ \log _b(x) = y \qquad \text {means that}\qquad b^y = x. \]
Moreover, \(\ln (x) = \log _e(x)\)
where \(e= 2.71828\dots \). So now examine each of these functions along the horizontal line \(y=1\) \[ \left (\frac {1}{3}\right )^1 < \left (\frac {1}{2}\right )^1 < 2^1 < e^1. \]
Hence we see:
- \(\log _{1/3}(x)\) corresponds to \(\answer [given]{B}\).
- \(\log _{1/2}(x)\) corresponds to \(\answer [given]{A}\).
- \(\log _2(x)\) corresponds to \(\answer [given]{D}\).
- \(\ln (x)\) corresponds to \(\answer [given]{C}\).
3 Properties of exponential functions and logarithms
Working with exponential and logarithmic functions is often simplified by applying properties of these functions. These properties will make appearances throughout our work.
3.1 Properties of exponents
Let \(b\) be a positive real number with \(b\neq 1\).
- \(b^m\cdot b^n = b^{m+n}\)
- \(b^{-1} = \frac {1}{b}\)
- \(\left (b^m\right )^n = b^{mn}\)
What exponent makes the following true?
\[ 2^4 \cdot 2^3 = 2^{\answer {7}} \]
\[ (2^4) \cdot (2^3) = (2 \cdot 2\cdot 2 \cdot 2) \cdot (2 \cdot 2\cdot 2) \]
3.2 Properties of logarithms
Let \(b\) be a positive real number with \(b\neq 1\).
- \(\log _b(m\cdot n) = \log _b(m) + \log _b(n)\)
- \(\log _b(m^n) = n\cdot \log _b(m)\)
- \(\log _b\left (\frac {1}{m}\right ) = \log _b(m^{-1}) = -\log _b(m)\)
- \(\log _a(m) = \frac {\log _b(m)}{\log _b(a)}\)
What value makes the following expression true?
\[ \log _2\left (\frac {8}{16}\right ) = 3-\answer {4} \]
What makes the following expression true?
\[ \log _3(x) = \frac {\ln (x)}{\answer {\ln (3)}} \]
Solve the equation: \(\displaystyle 5^{2x-3} = 7\).
Since we can’t easily rewrite both sides as exponentials with the
same base, we’ll use logarithms instead. Above we said that \(\log _b(x) = y\) means that \(b^y = x\). That
statement means that each exponential equation has an equivalent logarithmic form
and vice-versa. We’ll convert to a logarithmic equation and solve from there.
\begin{align*} 5^{2x-3} &= 7\\ \log _{\answer {5}}\left ( \answer {7} \right ) &= 2x-3 \end{align*}
From here, we can solve for \(x\) directly.
\begin{align*} 2x &= \log _{5}\left (7\right ) + 3\\ x &= \frac {\log _{5}\left (7\right ) + 3}{2} \end{align*}
Solve the equation: \(\displaystyle e^{2x} = e^x + 6\).
Immediately taking logarithms of both sides will not help here,
as the right side has multiple terms. We know that logarithms behave well with
products and quotients, but not with sums. Notice that \(e^{2x} = \left (e^x\right )^2\). (This is a common trick
that you will likely see many times.)
\begin{align*} e^{2x} &= e^x + 6\\ \left (e^x\right )^2 &= e^x + 6\\ \left (e^x\right )^2 - e^x - 6 &= 0 \end{align*}
Our equation is really a quadratic equation in \(e^x\). The left-hand side factors as \(\left ( e^x - \answer {3}\right ) \left (e^x + \answer {2}\right )\), so we are dealing with
\[ e^x - \answer {3} = 0 \qquad \textrm {and} \qquad e^x+\answer {2} = 0.\]
For the first factor: \begin{align*} e^x &= \answer {3}\\ x &= \ln \left ( \answer {3}\right ). \end{align*}
From the second factor: \(\displaystyle e^x = \answer {-2}\). Recall from above that the range of the exponential function is \((0, \infty )\). There is no input to make the output a negative number, so \(e^x = -2\) has no solutions.
The solution to \(\displaystyle e^{2x} = e^x + 6\) is \(x = \answer {\ln (3)}\).