\chapter{Exponentials and logarithms}
\section{Logarithms }
We can think of ``logarithm'' as another word for exponent.
For example the equation
\[3^2=9 \hspace{5ex} \mbox{(exponential form)}\]
can be described as
\begin{center}
When the base is \(3\), the exponent needed to get \(9\) is \(2\)
\end{center}
or, in the language of logarithms,
\begin{center}
The logarithm in base \(3\) of \(9\) is \(2\)
\end{center}
or in math notation:
\[\log_3 9=2 \hspace{5ex} \mbox{(logarithmic form)}.\]
So the definition of logarithm is
| Definition of logarithm |
|---|
| \(\log_b x\)
is the exponent to which we need to raise \(b\) to get \(x\). |
\begin{example}
\[\log_2 8 =3 \hspace{5ex} \mbox{ because } 2^3=8\]
\[\log_3 9 =2 \hspace{5ex} \mbox{ because } 3^2=9\]
\[\log_4 64 =3 \hspace{5ex} \mbox{ because } 4^3=64\]
\end{example}
As we know the exponent can be any number (positive, negative, zero, fractional). So the
logarithm can be any number. The next example shows negative, fractional and zero logarithms.
\begin{example}
\[\log_3 \frac{1}{9} =-2 \hspace{5ex} \mbox{ because } 3^{-2}=\frac{1}{9}\]
\[\log_4 2 =\frac{1}{2} \hspace{5ex} \mbox{ because } 4^{1/2}=\sqrt{4}=2\]
\[\log_5 1 =0 \hspace{5ex} \mbox{ because } 5^0=1\]
\end{example}
\subsection{Special bases for logarithms}
All the logarithms we evaluated so far were exact values, such as \(\log_5 25=2\), because \(5^2=25\).
But there are many numbers that cannot be obtained by raising a given base to a whole number
or a fraction.
\begin{example}
What is \(\log_{10} 7\)? We need an exponent that given to \(10\)
will produce \(7\). In other words, we need to find the question mark in the equation:
\[10^? = 7.\]
But \(10^0=1\) and \(10^1=10\), so the answer must be some decimal number between \(0\) and \(1\).
Using a calculator, we can experiment with exponents between \(0\) and \(1\):
\(10^{0.5}=3.162\ldots \)
So \(0.5\) is too small. If we try \(0.9\), we get
\(10^{0.9}=7.943\ldots \)
so \(0.9\) is too large, and \(\log 7\) must be between \(0.5\) and \(0.9\).
We can continue this way, trying for example \(0.7\), until we find an exponent \(y\) that gives
us \(10^y=7\) with enough precision. But a scientific calculator will do all the work for us.
All scientific
calculators have a \(\log\) key, with no base shown. This means that the base is understood
to be \(10\). So \(10\) is a special base for logarithms:
\[\log \hspace{1ex} \mbox{ means } \hspace{1ex} \log_{10},\]
and
| \(\log x\) is the exponent that we need to give to \(10\) in order to get \(x\). |
So using the \(\log\) key on a calculator we can find the right fractional exponent
\(y\) (usually with up to \(10\)
decimal places of accuracy) that gives \(10^y=7\):
\[\fbox{\(\log\)} \hspace{1ex} \fbox{7} \hspace{1ex} \fbox{=} \hspace{1ex} 0.84509804\]
As expected, \(\log 7\) is between \(0.5\) and \(0.9\).
The answer is an irrational number, with infinitely many decimals, and the calculator only shows the first few decimal places.
We can check that this answer is right by raising \(10\) to it. Using the calculator again, we find:
\[10^{0.84509804} = 7.\]
In reality, if the calculator could show us more decimal places, we would find that
\[10^{0.84509804}=6.99999999977\ldots\]
and
\[10^{0.84509805}=7.00000016\ldots\]
So the exact answer for \(\log 7\) is an irrational number between \(0.84509804\)
and \(0.84509805\).
\end{example}
The \(\fbox{log}\) calculator key allows us to find any logarithm with base 10. But what about other bases?
\begin{example}
Suppose we want to find \(\log_2 10\). In other words, we want to solve for ? in the equation
\[2^?=10.\]
Since \(2^3=8\) and \(2^4=16\), the answer must be a number between \(3\) and \(4\).
To solve this problem, we first need
to study another special base for logarithms.
\end{example}
On your calculator, you will find a second key marked by \(\fbox{ln}\). The
\textit{n} stands for \textit{ natural}, and it means that the base is understood to be the
number \(e\). So
\[\ln \hspace{1ex} \mbox{ means } \hspace{1ex} \log_{e},\]
and \(\ln x\) is called the \textit{ natural logarithm } of \(x\):
| Natural logarithm |
|---|
| \(\ln x\) is the exponent we need to give to \(e\) in order to get \(x\). |
\begin{example}
What is \(\ln 10\)? We need to answer the question: \(e^? = 10\). This is rather hard to do without
a calculator. We know that \(e^1=e=2.718\ldots\) so \(\ln 10\) must be larger than \(1\).
If we try \(2\) and \(3\) as exponents,
we find \(e^2=7.38\ldots...\) and \(e^3=20.08\ldots.\). So \(\ln 10\) must be between \(2\) and
\(3\). Using the
\(\ln \) key, we find \(\ln 10=2.302585093\ldots.\), and in fact using this as exponent we find
\[e^{2.302585093}=10.\]
But again this is only because the calculator cannot show us more decimals, because in reality
\[e^{2.302585093}=10.00000000005\ldots\]
and
\[e^{2.302585092}=9.999999990\ldots.\]
So the exact value of \(\ln 10\) is an irrational number between
\(2.302585092\) and \(2.302585093\).
\end{example}
The natural logarithm can be used to find the logarithm of any number in any base, via the
\textit{change of base formula}:
| Change of base formula |
|---|
| \(\log_b x=\dfrac{\ln x}{\ln b}\). |
\begin{example}
In example 3.2.4 we found that \(\log_2 10\) is a number between \(3\) and \(4\). We can now be more precise, using the change of base formula and a calculator:
\[\log_2 10 =\frac{\ln 10}{\ln 2}=\frac{2.302585093}{0.693147181}=3.321928095.\]
We can check the answer:
\[2^{3.321928095}=10.\]
\end{example}
\subsection{Logarithmic and exponential forms for equations}
We began this unit saying that logarithm is just another word for exponent. This concept
is summarized in the \textit{translation formulas} that allow us
to go from the exponential form of an equation to the logarithmic form:
| Translation formulas |
|---|
|
\(
\begin{array}{lcl}
\log_bx =y \hspace{1ex} & \mbox{is equivalent to} & \hspace{1ex} b^y=x\\[1ex]
\log x =y \hspace{1ex} & \mbox{is equivalent to} & \hspace{1ex} 10^y=x\\[1ex]
\ln x =y \hspace{1ex} & \mbox{is equivalent to} &\hspace{1ex} e^y=x
\end{array}
\)
|
These formulas are often all that is needed to solve some equations that at first sight seem
difficult to solve. We will soon see examples later in this chapter.
As an aid to memorize the translation formulas, remember BAB, that means Base, Across, Back. So for example
\(5^t=a\) becomes \(\log_5 a = t\) because \(\mbox{Base}=5\), \(\mbox{Across}=a\) and \(\mbox{Back}=t\).
This works also in the other direction:
\(\log_5 a=t\) becomes \(5^t=a\) because \(\mbox{Base}=5\), \(\mbox{Across}=t\) and \(\mbox{Back}=a\).
\begin{example} \
\begin{itemize}
\item Going from exponential to logarithmic form:
\[5^t=a \hspace{2ex} \Longrightarrow \hspace{2ex} \log_5 a=t\]
\[10^{x-2}=y+1 \hspace{2ex} \Longrightarrow \hspace{2ex} \log(y+1) = x-2\]
Note the use of parentheses: \(\log(y+1) \) is not the same as \(\log y +1\), and only the first
is correct for the previous equation.
\[x^2+1 = e^{u/2} \hspace{2ex} \Longrightarrow \hspace{2ex} \ln(x^2+1) = \frac{u}{2}\]
\item Going from logarithmic to exponential form:
\[\log_3(x+4)=2y-1 \hspace{2ex} \Longrightarrow \hspace{2ex} 3^{2y-1}=x+4\]
\[\log(x^2+y^2)=t+1 \hspace{2ex} \Longrightarrow \hspace{2ex} 10^{t+1}=x^2+y^2\]
\[\ln(1-t)=5x \hspace{2ex} \Longrightarrow \hspace{2ex} e^{5x}=1-t\]
\end{itemize}
\end{example}
Problems
\problem
Find the following logarithms:
\[
\begin{array}{cccc}
\mbox{(a) } \log_4 16 & \mbox{(b) } \log_3 27 & \mbox{(c) } \displaystyle \log_2\frac{1}{8} & \mbox{(d) } \log_9 3
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
\(\log_4 16=2\), because \(4^2=16\).
\item
\(\log_3 27=3\), because \(3^3=27\).
\item
\(\displaystyle \log_2\frac{1}{8}=-3\), because \(\displaystyle 2^{-3}=\frac{1}{8}\).
\item
\(\displaystyle \log_9 3 =\frac{1}{2}\), because \(9^{1/2}=\sqrt{9}=3\).
\end{enumerate}
\end{sol}
\mproblem
Find the following logarithms:
\[
\begin{array}{cccc}
\mbox{(a) } \log_6 36 & \mbox{(b) } \log_2 16 & \mbox{(c) } \displaystyle \log_3\frac{1}{27}
& \mbox{(d) } \log_{25} 5
\end{array}
\]
\problem
Without using a calculator, find two whole numbers that must contain the given logarithm between them.
Then use a calculator to find the numerical value of the logarithms, and check that the
answer is right.
\[
\begin{array}{ccc}
\mbox{(a) } \log 25 & \mbox{(b) } \log 0.5 & \mbox{(c) } \log 5
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item \(\log 25\).
The base of the logarithm is \(10\). Since \(10^1=10\) and \(10^2=100\), \(\log 25\) must be between
\(1\) and \(2\), in other words:
\[1 < \log 25 < 2.\]
Using a calculator, we find \(\log 25=1.397940009\), and then we check that
\(10^{1.397940009}=25.00000002\). The fact that we do not find \(25\) exactly is just because
the real value of \(\log 25\) is somewhere between \(1.397940008\) and \(1.397940009\), as we can
check by computing \(10^{1.397940008}=24.99999996\).
\item
\(\log 0.5\).
Since \(10^0=1\), \(\log 0.5\) must be less than \(0\). But \(10^{-1}=1/10=0.1\), so
\(\log 0.5\) must be more than \(-1\). So \(\log 0.5\) must be between \(-1\) and \(0\), or
\[-1 < \log 0.5 < 0.\]
Using a calculator, we find \(\log 0.5 = -0.301029996\), and we check that \(10^{-0.301029996}=0.5\).
\item
\(\log 5\).
Since \(10^0=1\) and \(10^1=10\), \(\log 5\) must be between \(0\) and \(1\), or
\[0 < \log 5 < 1.\]
Using a calculator, we find \(\log 5=0.698970004\), and we check that
\(10^{0.698970004}=4.999999996\).
\end{enumerate}
\end{sol}
\mproblem
Without using a calculator, find two whole numbers that must contain the given logarithm between them.
Then use a calculator to find the numerical value of the logarithms, and check that the
answer is right.
\[
\begin{array}{ccc}
\mbox{(a) } \log 350 & \mbox{(b) } \log 3 & \mbox{(c) } \log 0.05
\end{array}
\]
\problem
Use your calculator and the change of base formula (if needed) to find the following
logarithm, then check the answer:
\[
\begin{array}{ccc}
\mbox{(a) } \ln 15 & \mbox{(b) } \log_3 45 & \mbox{(c) } \log_2 0.1
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
\(\ln 15\)
The base of the logarithm is \(e\). Since \(e^1=2.718\ldots\), \(e^2=7.38\ldots\) and
\(e^3=20.08\ldots\), the value of \(\ln 15\) must be between \(2\) and \(3\), or
\[2< \ln 15 < 3.\]
Using a calculator, we find \(\ln 15 = 2.708050201\), and we check that \(e^{2.708050201}=15\).
\item
\(\log_3\)
Using the change of base formula, we find
\[\log_3 45 =\frac{\ln 45}{\ln 3}=\frac{3.80666249}{1.098612289}=3.464973521\]
and \(3^{3.464973521}=45.00000001\).
\item
\(\log_2 0.1\)
\[\log_2 0.1=\frac{\ln 0.1}{\ln 2}=\frac{-2.302585093}{0.693147181}=-3.321928095\]
and \(2^{-3.321928095}=0.1\).
\end{enumerate}
\end{sol}
\mproblem
Use your calculator and the change of base formula (if needed) to find the following
logarithm, then check the answer:
\[
\begin{array}{ccc}
\mbox{(a) } \ln 43 & \mbox{(b) } \log_5 50 & \mbox{(c) } \log_3 0.8
\end{array}
\]
\problem \
\begin{enumerate}
\item
Re-write the equations in logarithmic form:
\[
\begin{array}{ccc}
\mbox{i. } 5^{3x+2}=1-y & \hspace{2ex}\mbox{ii. } t^2+4=e^{x-1} &
\hspace{2ex} \mbox{iii. }10^{t-1}=x+y
\end{array}
\]
\item
Re-write the equations in exponential form:
\[
\begin{array}{ccc}
\mbox{i. } \ln(r+3) = 7+x & \hspace{2ex} \mbox{ii. } \log_3(x+1)=2y
\hspace{2ex} \mbox{iii. } 3x+1=\log(1-t)
\end{array}
\]
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
- \(\hspace{2ex}5^{3x+2}=1-y \hspace{2ex} \Longrightarrow \hspace{2ex} \log_5(1-y)=3x+2\)
make sure to include the parentheses for \(\log_5(1-y)\)
- \(\hspace{2ex}t^2+4=e^{x-1} \hspace{2ex} \Longrightarrow \hspace{2ex} \ln(t^2+4)=x-1\)
- \(\hspace{2ex}10^{t-1}=x+y \hspace{2ex} \Longrightarrow \hspace{2ex} \log (x+y)=t-1\)
\item
-
\(\hspace{2ex}\ln(r+3) = 7+x \hspace{2ex} \Longrightarrow \hspace{2ex} e^{7+x}=r+3 \)
-
\(\hspace{2ex}\log_3(x+1)=2y \hspace{2ex} \Longrightarrow \hspace{2ex} 3^{2y}=x+1\)
-
\(\hspace{2ex}3x+1=\log(1-t) \hspace{2ex} \Longrightarrow \hspace{2ex} 10^{3x+1} =1-t\)
\end{enumerate}
\end{sol}
\mproblem \
\begin{enumerate}
\item
Re-write the equations in logarithmic form:
\[
\begin{array}{ccc}
\mbox{(i) } e^{4u-2}=x+2 & \hspace{2ex}\mbox{(ii) } 10^{3t}=x-1 & \hspace{2ex} \mbox{(iii) } 2^{2t+1}=6x
\end{array}
\]
\item
Re-write the equations in exponential form:
\[
\begin{array}{ccc}
\mbox{(i) } \log(y+5) = 1+u & \hspace{2ex} \mbox{(ii) } \log_7(w-4)=y+5
& \hspace{2ex} \mbox{(iii) } 2u+3=\ln(1+x)
\end{array}
\]
\end{enumerate}