\chapter{Exponentials and logarithms}
\section{Properties of logarithms and logarithmic functions}
\subsection{Special values and simplification formulas}
Two calculations of logarithms are especially simple and important: any number raised to \(0\)
is \(1\), and any number raised to \(1\) is the number itself. In the language of logarithms
this means:
\[\log_b 1 =0 \hspace{5ex} \mbox{ because } b^{0}=1\]
\[\log_b b =1 \hspace{5ex} \mbox{ because } b^{1}=b.\]
If we use the special bases \(10\) and \(e\), we find:
\[\log 1 =0 \hspace{5ex} \mbox{ because } 10^{0}=1\]
\[\log 10 =1 \hspace{5ex} \mbox{ because } 10^{1}=10,\]
and
\[\ln 1 =0 \hspace{5ex} \mbox{ because } e^{0}=1\]
\[\ln e =1 \hspace{5ex} \mbox{ because } e^{1}=e.\]
We summarize below these important special values:
| Special values for logarithms |
|---|
| \(\begin{array}{ccc} \begin{array}{c}
\log_b b =1 \\[1ex]
\log_b1 =0
\end{array}
&\begin{array}{c}
\log 10 =1 \\[1ex]
\log 1 =0
\end{array}&\begin{array}{c}
\ln e =1 \\[1ex]
\ln 1 =0
\end{array}\end{array}
\) |
We noticed in the previous section that an exponential function can never be zero. This means that
there is no exponent that used with any base will ever give zero. In the language of logarithms,
this means that the logarithm of zero, in any base, is undefined:
| \(\begin{array}{ccc}
\log_b 0 \mbox{ DNE } & \log 0 \mbox{ DNE } & \ln 0 \mbox{ DNE }
\end{array}
\)
|
In fact, if you try to calculate \(\ln 0\) with your calculator, you will get an error message.
The answer to the question \(b^? =b^x\) is, of course, \( ?=x\). In the language
of logarithms, this means that \(\log_b b^x =x\).
Also, since \(\log_b x\) is the exponent we need to give to \(b\) to get \(x\), if we do raise
\(b\) to that exponent we will get \(x\), or in other words \(b^{\log_b x}=x\).
So we have found the following important simplification formulas:
| Simplification formulas for logarithms |
|---|
| \(\begin{array}{ccc} \begin{array}{c}
\log_b b^x = x \\[2ex]
b^{\log_b x} = x
\end{array}
&\begin{array}{c}
\log 10^x=x \\[2ex]
10^{\log x} = x
\end{array}&\begin{array}{c}
\ln e^x =x \\[2ex]
e^{\ln x}=x
\end{array}
\end{array}
\) |
\subsection{Logarithmic functions}
We can now introduce a new basic parent function, the logarithmic function:
| Logarithmic function |
|---|
| A logarithmic function is one of the form |
| \(f(x) =\log_b x\) |
| where \(b\) is a positive number different from \(1\). |
In Unit 1.9 we discussed inverse functions. Given two functions \(f(x)\) and \(g(x)\), we
gave the following test to determine if they are inverses of each other:
\[(f\circ g)(x)=x \hspace{3ex} \mbox{ and } \hspace{3ex} (g\circ f)(x)=x.\]
If we let \(f(x)=b^x\) and \(g(x)=\log_b x\), then \((f\circ g)(x)=b^{\log_b x}\)
and \((g\circ f)(x)=\log_b b^x\). But the
the simplification formulas say \(b^{\log_b x}=x\) and \(\log_b b^x=x\).
So we conclude that:
| \(y=b^x\) and \(y=\log_b x\) are inverses of each other |
This means that the graph of \(y=\log_b x\) is the graph of \(y=b^x\) reflected across the
line \(y=x\), as in the figure below (for the case \(b>1\)):
\[\img{U3_4F0.png}{}{12em}{}\]
We can find domain, range, intercepts and asymptotes by exchanging the roles
of input and output:
| Logarithmic function \(y=\log_b x\) with \(b> 1\) |
|---|
| \begin{itemize}
\item Examples: \(y=\log_2 x\), \(y=\log x\), \(y=\ln x\)
\item Domain \(=(0,\infty)\), Range \(=(-\infty,\infty)\).
\item \(y\)-intercept: None
\item \(x\)-intercept: \((1,0)\)
\item Horizontal asymptote: None
\item Vertical asymptote: \(x=0\) (the \(y\)-axis)
\item Always increasing
\item One-to-one: Yes
\item Shape of the graph:
\(\img{U3_4F1.png}{-5em}{10em}{}\)
\end{itemize} |
Sometimes it is necessary to re-write a number using exponents, especially when
dealing with radicals (square roots, cube roots, etc.) and reciprocals. After all, logarithms are all
about exponents, so you can expect that writing \(3^{1/2}\) instead of \(\sqrt{3}\) and
\(x^{-2}\) instead of \(1/x^2\) will be useful.
\begin{example} \
\begin{enumerate}
\item
Suppose we want to find \(\log_5 \sqrt{5}\). All we need to do is re-write the square root as a
fractional exponent, and use the first simplification formula:
\[\log_5\sqrt{5} =\log_55^{1/2}=\frac{1}{2}.\]
\item In a similar way to find \(\displaystyle \ln\frac{1}{\sqrt{e}}\) we first rewrite it using
a negative, fractional exponent, and then use the simplification formula:
\[\ln\frac{1}{\sqrt{e}}=\ln e^{-1/2}=-\frac{1}{2}.\]
\item Cube roots or higher roots are done in the same way:
\[\log \sqrt[3]{10} = \log 10^{1/3} = \frac{1}{3}\]
\[\log_2 \frac{1}{\sqrt[5]{2}}=\log_2 2^{-1/5} = -\frac{1}{5}.\]
\end{enumerate}
\end{example}
\subsection{Properties of the logarithms}
Recall the properties of the exponents:
\[x^mx^n=x^{m+n}, \hspace{2ex} \frac{x^m}{x^n} = x^{m-n}, \hspace{2ex} (x^m)^n =x^{mn}.\]
These properties are expressed in exponential form. The equivalent logarithmic form gives us
three important and useful properties for logarithms, that we will label (A), (B) and (C):
| Properties of logarithms |
|---|
| \(\begin{array}{cc}
\mbox{(A)}& \hspace{2ex} \log_b(xy) = \log_b x + \log_b y \\[1ex]
\mbox{(B)} & \hspace{2ex} \displaystyle \log_b\left(\frac{x}{y}\right) = \log_b x -\log_by \\[2ex]
\mbox{(C)} & \hspace{2ex}\log_b x^y = y \log_b x
\end{array}
\)
|
Using these properties, we can sometimes find exact values for expressions containing
logarithms.
\begin{example}
Suppose we want to find the exact value of \(\log_6 12 + \log_6 3\). Clearly we cannot
solve the equations \(6^?=12\) and \(6^?=3\). We could use the change of base formula,
use a calculator to find an approximate value for both, and add. But there is a better way:
using property (A), we can re-write:
\[\log_6 12 +\log_6 3 =\log_6(12\cdot 3) = \log_6(36),\]
and \(\log_6(36)\) is easily found to be \(2\), because \(6^2=36\). So we conclude that
\(\log_6 12 + \log_6 3=2\).
\end{example}
\begin{example}
Now we look for the exact value of \[\log 40-2\log 2.\]
Using property (C) on the second
term, we can re-write this as
\[\log 40 - \log 2^2\]
or
\[\log 40 -\log 4.\]
Now use property (B) to get:
\begin{eqnarray*}
\log 40 -\log 4&= & \log\left(\frac{40}{4}\right)\\
& = & \log 10\\
& = & 1
\end{eqnarray*}
\end{example}
\begin{example}
Sometimes it is useful to factor a number, and then use property (A). So to find
\(\log_4 8\), we can factor the \(8\):
\[\begin{array}{rcll}
\log_4 8 & = & \log_4 (4\cdot 2) \\[2ex]
&=& \log_4 4 + \log_4 2 \\[2ex]
&=& \displaystyle 1 + \frac{1}{2} &\mbox{because \(4^1=4\) and \(4^{1/2}=\sqrt{4}=2\)}\\[2ex]
&=& \displaystyle\frac{3}{2}
\end{array}
\]
\end{example}
In the next example we will use a basic and important fact of whole numbers: each whole number is
the product of prime numbers. This means that if we know the logarithm of the prime numbers,
then we can also find the logarithm of any other whole number.
\begin{example}
Suppose we know that \(\log_b 2=0.60551\), \(\log_b 3=0.95971\).
Then we can find the logarithm (in base \(b\)) of any number that is made with the prime numbers \(2,3\). And we don't need to know what \(b\) is. For example,
\[24=(8)(3)=(2^3)(3)\] and so
\begin{array}{rcll}
\log_b24 &= &\log_b(2^33) & \\
&=&\log_b(2^3)+\log_b 3 & \mbox{by property (A)}\\
&=&3\log_b2+\log_b3 & \mbox{by property (C)}\\
&=&3(0.60551)+0.95971 & \\
&=& 2.7762.&&
\end{array}
\end{example}
Properties (A), (B), (C) can be used to expand the logarithm of a complicated expression
into simpler parts that contain no exponents and no variables in denominators,
as in the following example.
\begin{example}
We are given the expression
\[\ln\left(\frac{x^2\sqrt{z}}{ey^3}\right)\]
and we want to expand it into simple terms.
\[
\begin{array}{rcll}
\displaystyle \ln\left(\frac{x^2\sqrt{z}}{ey^3}\right) & = &
\ln\left(x^2\sqrt{z}\right) - \ln \left(ey^3\right)
& \mbox{Use property (B)}\\[2ex]
& = & \ln x^2 +\ln \sqrt{z} -\left( \ln e + \ln y^3 \right) & \mbox{Use property (A)}\\[2ex]
&=& \ln x^2+\ln z^{1/2} -\ln e -\ln y^3 & \mbox{Re-write \(\sqrt{z} \) using exponent and drop parentheses}\\[2ex]
& = & \displaystyle 2 \ln x +\frac{1}{2}\ln z-1-3\ln y & \mbox{Use property (C) and the special value \(\ln e =1\)}
\end{array}
\]
So the solution to the problem is:
\[\ln\left(\frac{x^2\sqrt{z}}{ey^3}\right)=2 \ln x +\frac{1}{2}\ln z-1-3\ln y.\]
Pay attention to the use of parentheses in the second step,
to make sure to get the negative sign right.
\end{example}
Sometimes it is necessary to reverse the process: we need to condense an expression containing
several logarithms into a single logarithm.
\begin{example}
We are given the expression
\[3\log (x-1)+2\log(y+1)-\frac{1}{2}\log(x+1) \]
and we want to condense it into a single logarithm.
\[
\begin{array}{rcll}
& & \displaystyle 3\log (x-1)+2\log(y+1)-\frac{1}{2}\log(x+1) & \\[2ex]
&=& \log(x-1)^3 +\log(y+1)^2-\log(x+1)^{1/2} & \mbox{Use property (C)}\\[2ex]
& = & \log\left( (x-1)^3 (y+1)^2\right)- \log\sqrt{x+1} & \mbox{Use property (A)
and re-write \((x+1)^{1/2}\) as \(\sqrt{x+1}\)}\\[2ex]
&=& \displaystyle \log\left( \frac{(x-1)^3(y+1)^2}{\sqrt{x+1}} \right) & \mbox{Use property (B)}
\end{array}
\]
So we found the answer for our problem:
\[3\log (x-1)+2\log(y+1)-\frac{1}{2}\log(x+1) =\log\left( \frac{(x-1)^3(y+1)^2}{\sqrt{x+1}} \right) .\]
\end{example}
Problems
\problem
Find the exact value of the following logarithms. Do not use a calculator.
\[
\begin{array}{cccc}
\mbox{a. } \log_4 4^t &\mbox{b. } \ln e^{x^2} &\mbox{c. } 10^{\log (x+3)}
& \mbox{d. } \displaystyle \log\frac{1}{\sqrt{10}}\\
\mbox{e. } \ln \sqrt[3]{e} &\mbox{f. } \log_8 1 &\mbox{g. } \log_3 3 &
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
Use the simplification formula \(\log_b b^x=x\):
\[\log_4 4^t =t.\]
\item
Use the simplification formula \(\ln e^x=x\):
\[\ln e^{x^2} = x^2.\]
\item
Use the simplification formula \(10^{\log x}=x\):
\[10^{\log (x+3)} = x+3.\]
\item
\[\begin{array}{rcll}
\displaystyle \log\frac{1}{\sqrt{10}} & = & \log 10^{-1/2} & \mbox{re-write in exponential notation}\\[2ex]
& = & \displaystyle -\frac{1}{2}& \mbox{use the simplification formula \(\log 10^x=x\).}
\end{array}
\]
\item
\[\begin{array}{rcll}
\ln \sqrt[3]{e} & = & \ln e^{1/3} & \mbox{re-write in exponential notation}\\[2ex]
& = & \displaystyle \frac{1}{3}& \mbox{use the simplification formula \(\log 10^x=x\).}
\end{array}
\]
\item
Use the special value formula \(\log_b 1 =0\):
\[\log_8 1=0.\]
\item
Use the special value formula \(\log_b b =1\):
\[\log_3 3 = 1.\]
\end{enumerate}
\end{sol}
\mproblem
Find the exact value of the following logarithms. Do not use a calculator.
\[
\begin{array}{cccc}
\mbox{(a) } \log 10^{1-x} &\mbox{(b) } \log_3 3^{2t} &\mbox{(c) } e^{\ln (x+y)} & \mbox{(d) }\displaystyle \ln\frac{1}{\sqrt[3]{e}}\\
\mbox{(e) } \log_4 \sqrt[5]{4} &\mbox{(f) } \log_2 2 &\mbox{(g) } \log_5 1 &
\end{array}
\]
\problem
Find the exact value of the expressions.
\[
\begin{array}{cc}
\mbox{a. }\log 20 +\log 5 & \mbox{b. } \log_9 108 - 2\log_9 2
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
Using property (A),
\[\begin{array}{rcl}
\log 20 +\log 5&=&\log 100\\[2ex]
&=& 2
\end{array}
\]
\item
\[
\begin{array}{rcll}
\log_9 108 - 2\log_9 2&=& \log_9108 -\log_9 2^2 & \mbox{by property (C)}\\[2ex]
&=& \log_9 \frac{108}{4} & \mbox{by property (B)}\\[2ex]
&=& \log_9 27 & \mbox{simplify}\\[2ex]
&=& \log_9 (9\cdot 3) & \mbox{factor the \(27\)}\\[2ex]
&=& \log_9 9 +\log_9 3 &\mbox{use property (A)}\\[2ex]
& = &\displaystyle 1+ \frac{1}{2} & \mbox{because \(9^{1/2}=\sqrt{9}=3\)}\\[2ex]
&=& \displaystyle \frac{3}{2}
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Find the exact value of the expressions.
\[
\begin{array}{cc}
\mbox{(a) } \log_8 2 +\log_8 32 & \mbox{(b) } \log_4 32 - 2\log_4 2
\end{array}
\]
\problem
Expand the expression and simplify:
\[\log\left( \frac{t^3 z^4}{10\sqrt{x}} \right) \]
\begin{sol}
\[
\begin{array}{rcll}
& & \displaystyle \log\left( \frac{t^3 z^4}{10\sqrt{x}} \right) & \\[2ex]
&=& \log\left( t^3z^4\right) -\log \left(10\sqrt{x}\right) &
\mbox{Use property (B)}\\[2ex]
& = & \log t^3 +\log z^4 -(\log 10 + \log x^{1/2}) & \mbox{Use property (A)
and re-write \(\sqrt{x}\) as \(x^{1/2}\)}\\[2ex]
&=& \displaystyle 3\log t +4\log z -\log 10 -\frac{1}{2} \log x& \mbox{Use property (C)}\\[2ex]
&=& \displaystyle 3\log t +4\log z -1 -\frac{1}{2} \log x & \mbox{Use special value \(\log 10=1\)}
\end{array}
\]
\end{sol}
\mproblem
Expand the expression and simplify:
\[\log_3\left( \frac{(x+2)^4(x+3)}{9\sqrt{x+1}} \right) \]
\problem
Condense the expression into a single logarithm:
\[4\ln x -\frac{1}{2}\ln(x-1) +3\ln (x+1)\]
\begin{sol}
\[
\begin{array}{rcll}
& & \displaystyle 4\ln x -\frac{1}{2}\ln(x-1) +3\ln (x+1) & \\[2ex]
&=& \displaystyle \ln x^4-\ln(x-1)^{1/2} +\ln(x+1)^3 & \mbox{Use property (C)}\\[2ex]
& = & \displaystyle \ln\left( \frac{x^4}{\sqrt{x-1}}\right) +\ln(x+1)^3 & \mbox{Use property (B)
and re-write \((x-1)^{1/2}\) as \(\sqrt{x-1}\)}\\[2ex]
&=& \displaystyle \ln \left( \frac{x^4}{\sqrt{x-1}} (x+1)^3\right)& \mbox{Use property (A)}\\[2ex]
& = & \displaystyle \ln \left( \frac{x^4 (x+1)^3}{\sqrt{x-1}} \right)&
\mbox{Re-write as single fraction}
\end{array}
\]
\end{sol}
\mproblem
Condense the expression into a single logarithm:
\[3\log (t-4) +\frac{1}{2}\log(t+2) -5\log (t+1)\]
\problem
Suppose that we know that \(\ln x = 12\) and \( \ln y =-7\). Find the exact value of
\(\ln \left(e y^2\sqrt{x}\right)\).
\begin{sol}
\[
\begin{array}{rcll}
\ln \left(e y^2\sqrt{x}\right) & = & \ln e+ \ln y^2 + \ln x^{1/2} & \mbox{Use property (A)
and re-write \(\sqrt{x}\) as \(x^{1/2}\)}\\[2ex]
&=& 1+ \displaystyle 2\ln y +\frac{1}{2}\ln x & \mbox{use the special value \(\ln e =1\) and property (C)}\\[2ex]
&=& 1+ \displaystyle 2(-7) +\frac{1}{2}(12) & \mbox{substitute the known values for \(\ln x\) and \(\ln y\)}\\[2ex]
&=& 1-14+6 & \\[2ex]
&=& -7
\end{array}
\]
\end{sol}
\mproblem
Suppose that we know that \(\log x = -2\) and \( \log y =8\). Find the exact value of
\(\log \left(10 x^3\sqrt{y}\right)\).