Supplemental Videos.
The main topics of this section are also presented in the following videos:
Section 4.2 Properties of Logarithms
Recall that an exponential function is a function of the form
\begin{equation*}
P=P(t)=a(\alert{b})^t.
\end{equation*}
In this section we will discuss logarithmic functions, which are inverses of exponential functions. In particular, the function
\begin{equation*}
t=\log_{\alert{b}}\left(\frac{P}{a}\right)
\end{equation*}
is the inverse of the exponential function above. For \(a=1\text{,}\) the following applet illustrates this inverse relationship:
While this inverse relationship is how we will think of logarithms in practice, let’s give a formal definition:
The Logarithm.
Let \(b\neq 1 \) be a positive number, then the function
\begin{equation*}
f(t)=\log_b(t)
\end{equation*}
is called a logarithm with base \(b \text{.}\)
Upon inputting a value \(t\) the function \(\log_b(t)\) will tell you the power of \(b \) which will yield \(t \text{.}\)
Due to the relationship between logarithms and exponentials, we often say that the equations
\begin{equation*}
x=\log_b(y)\ \ \ \ \text{and}\ \ \ \ b^x=y
\end{equation*}
are equivalent.
Caution 4.20.
One of the more common bases of a logarithm is base \(10 \text{.}\) Since it is used so often, we have developed a short hand notation for a logarithm of base \(10\text{.}\) This short hand is shown below:
\begin{equation*}
\log_{10}(y)=\log(y).
\end{equation*}
In other words, we simply drop the subscript when referring to base \(10 \text{.}\)
Subsection 4.2.1 Properties of Logarithms
Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Recall the following laws of exponents:
-
To multiply two powers with the same base, add the exponents and leave the base unchanged.\begin{equation*} a^m \cdot a^n = a^{m+n} \end{equation*}
-
To divide two powers with the same base, subtract the exponents and leave the base unchanged.\begin{equation*} \frac{a^m}{a^n} = a^{m-n} \end{equation*}
-
To raise a power to a power, keep the same base and multiply the exponents.\begin{equation*} \left(a^m\right)^n = a^{mn} \end{equation*}
Each of these laws corresponds to one of three properties of logarithms.
Properties of Logarithms.
If \(x,y,b>0\text{,}\) and \(b\neq 1\text{,}\) then
-
\(\displaystyle \log_b(xy)=\log_b(x)+\log_b(y), \)
-
\(\displaystyle \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y), \)
-
\(\displaystyle \log_b(x^k)=k\cdot \log_b(x), \)
-
\(\displaystyle \log_b(b^y)=y, \)
-
\(\displaystyle b^{\log_b(x)}=x. \)
We will examine the properties of logarithms closer in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is the inverse function of an exponential function.
-
Property (1):\begin{equation*} \log_{2}({32})=\log_{2}{(4 \cdot 8)}=\log_2(4) +\log_2(8)= \alert{2} + \alert{3}=\alert{5} \end{equation*}
-
Property (2):\begin{equation*} \log_{2}\left({\frac{16}{2}}\right)=\log_2(16)-\log_2 (2)=\alert{4}-\alert{1}=\alert{3} \end{equation*}
-
Property (3):\begin{equation*} \log_{2}({64})=\log_{2}{(4^3)}=3\log_2 (4)= \alert{3} \cdot \alert{2}= \alert{6} \end{equation*}
Subsection 4.2.2 Using the Properties of Logarithms
Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In the next example, we rewrite one logarithm in terms of simpler logarithms.
Example 4.21.
Simplify \(\log_{b}\sqrt{{xy}}\text{.}\)
Solution.
First, we write \(\sqrt{xy}\) using a fractional exponent:
\begin{equation*}
\log_{b}{\sqrt{xy}} = \log_{b}{\big((xy)^{1/2}\big)}.
\end{equation*}
Then we apply Property (3) to rewrite the exponent as a coefficient:
\begin{equation*}
\log_{b}{((xy)^{1/2})} = \frac{1}{2}\log_{b}{(xy)}.
\end{equation*}
Finally, by Property (1) we write the log of a product as a sum of logs:
\begin{equation*}
\frac{1}{2}(\log_{b}({xy}) = \frac{1}{2}(\log_{b}({x}) + \log_{b}({y}).
\end{equation*}
Thus, \(\log_{b}\sqrt{xy} = \frac{1}{2}(\log_{b}({x}) + \log_{b}({y})\text{.}\)
Example 4.22.
Simplify \(\log_{b}({xy^2})\text{.}\)
Solution.
\begin{equation*}
\begin{aligned} \log_b(xy^2) \amp = \log_b(x)+\log_b(y^2) \amp\amp\text{Property 1} \\ \amp = \log_b(x)+2\log_b(y) \amp\amp\text{Property 3} \end{aligned}
\end{equation*}
Caution 4.23.
Be careful when using the properties of logarithms! Compare the statements below:
-
\(\log_{b}{(2x)} = \log_{b}({2}) + \log_{b}({x}) \ \ \ \text{ by Property 1,}\)but\(\log_{b}{(2 + x)} \ne \log_{b}({2}) + \log_{b}({x}).\)
-
\(\log_{b}{\left(\dfrac{x}{5}\right)}= \log_b (x) - \log_b (5) \ \ \ \text{ by Property 2,}\)but\(\log_{b}{\left(\dfrac{x}{5}\right)} \ne \dfrac{\log_b (x)}{\log_b (5)}.\)
We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.
Example 4.24.
Express \(3(\log_b (x) - \log_b (y))\) as a single logarithm with a coefficient of \(1\text{.}\)
Solution.
We begin by applying Property (2) to combine the logs.
\begin{equation*}
3(\log_b (x) - \log_b (y)) = 3 \log_{b}\left(\frac{x}{y}\right)
\end{equation*}
Then, using Property (3), we replace the coefficient \(3\) by an exponent \(3\text{.}\)
\begin{equation*}
3 \log_{b}\left(\frac{x}{y}\right)=\log_{b}\left(\left(\frac{x}{y}\right)^3\right)
\end{equation*}
Example 4.25.
Express \(2\log_b (x) + 4\log_{b}{(x + 3)}\) as a single logarithm with a coefficient of \(1\text{.}\)
Answer.
\begin{equation*}
\begin{aligned} 2\log_b(x)+4\log_b(x+3) \amp= \log_b(x^2)+\log_b\big((x+3)^4\big) \amp\amp\text{Property 3} \\ \amp= \log_b(x^2(x+3)^4) \amp\amp\text{Property 1} \end{aligned}
\end{equation*}
Subsection 4.2.3 The Natural Exponential Function
There is another base for logarithms and exponential functions that is often used in applications. This base is an irrational number called \(e\text{,}\) where
\begin{equation*}
e \approx 2.71828182845.
\end{equation*}
The number \(e\) is essential for many advanced topics, and it is often called the natural base .
The base \(e\) logarithm of a number \(x\text{,}\) or \(\log_ e(x)\text{,}\) is called the natural logarithm of \(x\) and is denoted by \(\ln(x)\text{.}\)
The Natural Logarithm.
The natural logarithm is the logarithm base \(e\text{.}\)
\begin{equation*}
\ln(x) = \log_{e}({x}), ~~~~ x\gt 0
\end{equation*}
We use natural logarithms in the same way that we use logs to other bases. The properties of logarithms that we studied above also apply to logarithms base \(e\text{.}\)
Subsection 4.2.4 Solving Exponential Equations
Suppose we want to solve the equation
\begin{equation*}
5^x = 7.
\end{equation*}
We could rewrite the equation in logarithmic form to obtain the exact solution
\begin{equation*}
x = \log_{5}({7}).
\end{equation*}
However, sometimes we are stuck in a situation where we cannot evaluate \(\log_{5}({7})\text{.}\) For example, some calculators do not have a log base \(5\) button. So, if we want a decimal approximation for the solution, we begin by taking the base \(10\) logarithm of both sides of the original equation, even though the base of the power is not \(10\text{.}\) This gives us
\begin{equation*}
\log_{10}{(5^x)} = \log_{10}({7}).
\end{equation*}
Then we use Property (3) to rewrite the left side as
\begin{equation*}
x \log_{10}({5}) = \log_{10}({7}).
\end{equation*}
Note how using Property (3) allows us to solve the equation: The variable, \(x\text{,}\) is no longer in the exponent, and it is multiplied by a constant, \(\log_{10}({5})\text{.}\) To finish the solution, we divide both sides by \(\log_{10}({5})\) to get
\begin{equation*}
x = \frac{\log_{10}({7})}{\log_{10}({5})}.
\end{equation*}
On your calculator, enter the sequence
LOG( \(7\) ) \(\div\) LOG ( \(5\) ) ENTER
to find that \(x \approx 1.2091\text{.}\)
Caution 4.26.
Do not confuse the expression \(\dfrac{\log_{10}({7})}{\log_{10}({5})}\) with \(\log_{10}{\left(\dfrac{7}{5}\right)}\text{;}\) they are not the same! Property (2) allows us to simplify \(\log{\left(\dfrac{x}{y}\right)}\text{,}\) but not \(\dfrac{\log (x)}{\log (y)}\text{.}\) We cannot rewrite \(\dfrac{\log_{10}({7})}{\log_{10}({5})}\text{,}\) so we must evaluate it as \((\log (7)/(\log (5))\text{.}\) You can check on your calculator that
\begin{equation*}
\dfrac{\log_{10}({7})}{\log_{10}({5})}\ne \log_{10}{\left(\dfrac{7}{5}\right)}= \log_{10}({1.4})\text{.}
\end{equation*}
Example 4.27.
Solve \(1640 = 80 \cdot 6^{0.03x}\text{.}\)
Solution.
First we divide both sides by \(80\) to obtain
\begin{equation*}
20.5 = 6^{0.03x}.
\end{equation*}
Next, we take the base \(10\) logarithm of both sides of the equation and use Property (3) of logarithms to get
\begin{equation*}
\log_{10}{20.5} = \log_{10}{(6^{0.03x})}= 0.03x \log_{10}{6}.
\end{equation*}
On the right side of the equation, \(x\) is multiplied by two constants, \(0.03\) and \(\log_{10}{6}\text{.}\) So, to solve for \(x\) we must divide both sides of the equation by \(0.03 \log_{10}{6}\text{.}\) We use a calculator to evaluate the answer:
\begin{equation*}
x = \frac{\log_{10}{20.5}}{0.03 \log_{10}{6}}\approx 56.19.
\end{equation*}
(On your calculator, remember to enclose the denominator, \(0.03 \log_{10}{6}\text{,}\) in parentheses.)
Caution 4.28.
In Example 4.27, do not try to simplify
\begin{equation*}
80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \text{ Incorrect!}
\end{equation*}
Remember that the order of operations tells us to compute the power \(6^{0.03x}\) before multiplying by \(80\text{.}\)
Example 4.29.
Solve \(5(1.2)^{2.5x} = 77\text{.}\)
Solution.
First we divide both sides by \(5\text{.}\)
\begin{equation*}
1.2^{2.5x}=15.4
\end{equation*}
Next, we take the base \(10\) logarithm on both sides and use Property (3):
\begin{equation*}
\begin{aligned} \log_{10}(1.2^{2.5x}) \amp= \log_{10}(15.4) \\ (2.5x)\log_{10}(1.2) \amp= \log_{10}(15.4) \end{aligned}
\end{equation*}
Finally, solving for \(x\) we have
\begin{equation*}
\begin{aligned} (2.5x)\log_{10}(1.2) \amp= \log_{10}(15.4) \\ x \amp= \frac{\log_{10}(15.4)}{2.5\log_{10}(1.2)}\approx 5.999 \end{aligned}
\end{equation*}
Exercises 4.2.5 Exercises
1. Logarithms to Exponents.
Rewrite the following using exponents instead of logs.
a)
\(\ln(5) \approx 1.609\)
-
\(\displaystyle e^{5} \approx 1.609\)
-
\(\displaystyle 5^{1.609} \approx e\)
-
\(\displaystyle e^{1.609} \approx 5\)
-
\(\displaystyle 1.609^{e} \approx 5\)
b)
\(\log(1) = 0\)
-
\(\displaystyle e^1 = 0\)
-
\(\displaystyle 1^0 = 10\)
-
\(\displaystyle 10^0 = 1\)
-
\(\displaystyle 10^1 = 0\)
c)
\(\log(A) = B\)
-
\(\displaystyle B^A = 10\)
-
\(\displaystyle A^{10} = B\)
-
\(\displaystyle 10^B = A\)
-
\(\displaystyle 10^A = B\)
2. Definition of Logarithms.
Complete the following statements by filling in the blanks.
a) If \(y = \log(100)\text{,}\) then \(\underline{\hspace{0.5 in}}^y = 100\text{.}\)
b) \(\log(5.5)\) is the power of \(\underbrace{\underline{\hspace{0.5 in}}}_{A}\) that gives \(\underbrace{\underline{\hspace{0.5 in}}}_{B}\text{.}\)
A:
B:
c) \(\log(\underbrace{\underline{\hspace{0.5 in}}}_{A})\) is the power of \(\underbrace{\underline{\hspace{0.5 in}}}_{B}\) that gives 600.
A:
B:
d) If \(10^m = n\text{,}\) then \(\log(n) = \underline{\hspace{0.5 in}}\text{.}\)
e) \(\ln(610)\) is the power of \(\underbrace{\underline{\hspace{0.5 in}}}_{A}\) that gives \(\underbrace{\underline{\hspace{0.5 in}}}_{B}\)
A:
B:
f) \(\ln(\underbrace{\underline{\hspace{0.5 in}}}_{A})\) is the power of \(\underbrace{\underline{\hspace{0.5 in}}}_{B}\) that gives 0.64.
A:
B:
g) If \(e^y = 15\text{,}\) then \(\ln(\underline{\hspace{0.5 in}}) = y\text{.}\)
3. Exponents to Logarithms.
Rewrite the following using logs instead of exponents.
a) \(e^{13} \approx 442413.392\)
-
\(\displaystyle \ln(13) \approx 442413.392\)
-
\(\displaystyle \log_{13} (442413.392) \approx e\)
-
\(\displaystyle \ln (442413.392) \approx 13\)
-
\(\displaystyle e_{442413.392} \approx 13\)
b) \(10^{-4}=\frac{1}{0.0001}\)
-
\(\displaystyle \log_{0.0001}(-4) = 10\)
-
\(\displaystyle \log \left(\frac{1}{0.0001}\right) = -4\)
-
\(\displaystyle \ln \left(\frac{1}{0.0001}\right) = -4\)
-
\(\displaystyle \log(-4) = 0.0001\)
c) \(e^t=H\)
-
\(\displaystyle \ln(t) = H\)
-
\(\displaystyle \log_{t} (H) = e\)
-
\(\displaystyle \ln(Ht) = e\)
-
\(\displaystyle \ln(H) = t\)
4. Evaluating Logarithms.
Evaluate the following without using a calculator.
a) \(e^{\ln 15}\) =
b) \(\ln\left(e^{6}\right)\) =
c) \(\ln\left(\frac{1}{\sqrt[3]{e}}\right)\) =
5. Evaluating Logarithms.
Evaluate the following without using a calculator.
a) \(10^{\log 13}\) =
b) \(\log (10^{4})\) =
c) \(\log \left(\sqrt[4]{10}\right)\) =
d) \(\log \left(\frac{1}{10^{6}}\right)\) =
6. Evaluating Logarithms.
Evaluate the following without using a calculator.
a) \(\log(1000)\) =
b) \(\log(10000)-\log(10)\) =
c) \(\log(10000)+\log\left(\frac{1}{10}\right)\) =
d) What do you notice about your answers to (a)-(c)?
-
They are all the same answer.
-
Two of the three are the same answer.
-
They are all different answers.
7. Evaluating Logarithms.
A googol is the number 1 followed by 100 zeros, or \(10^{100}\text{.}\) A googolplex is the number 1 followed by a googol zeros, or \(10^{\text{googol}}\text{.}\) Evaluate:
a) \(\sqrt{\log \left(\text{googol}\right)}\) =
b) \(\log \sqrt{\text{googol}}\) =
c) \(\log \left(\log \left(\text{googolplex}\right)\right)\) =
8. Applications of Logarithms.
The population of Endor (in millions) is given by \(P(t) = 5 e^{0.0011 t}\text{,}\) where \(t\) is the time in years since 2000. When will Endor’s population reach 10 million?
In the year
Hint.
We are asked to find the value of \(t\) that gives us \(P(t) = 10\) million when plugged into the equation \(P(t) = 5 e^{0.0011 t}\text{.}\) That is, we want to solve
\begin{equation*}
10 = 5 e^{0.0011 t}
\end{equation*}
for t.
To reverse the operation of using an exponential formula, you will need to apply the natural logarithm \(\ln\) to both sides of the equation after some other simplification.
Make sure you round your answer down, because we want to know what year this happens in. For example, if we had an answer like \(t = 4.2\text{,}\) we would round this down to \(t = 4\) since this would mean the event happened in the fourth year after 2000. Adding this rounded value of \(t\) to 2000 gets us what we want.
9. Applications of Logarithms.
Suppose Jocelyn invests $2500 at a nominal interest rate of 3\(\%\text{,}\) compounded annually. How many years will she have to wait until the account balance is $5500? (Round to two decimal places.)
years
Hint.
We are asked to find the value of \(t\) that gives us \(A(t) = 5500\) dollars when plugged into the equation \(A(t) = 2500 (1.03)^{t}\text{.}\) That is, we want to solve
\begin{equation*}
5500 = 2500 (1.03)^{t}
\end{equation*}
for t.
To eliminate the exponential part of this equation, you will need to use a logarithm. We recommend using common or natural log and applying to both sides to solve.
10. Properties of Logarithms.
Match each expression with its equivalent expression for \(A, B > 0\text{.}\)
a) \(\ln(AB)\)
-
\(\displaystyle 0\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle 1\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle \log(A^{3t})\)
-
\(\displaystyle \log(A)-\log(B)\)
b) \(\log\left(\frac{A}{B}\right)\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle \log(A)-\log(B)\)
-
\(\displaystyle \log(A^{3t})\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 0\)
-
\(\displaystyle 1\)
c) \(\log(A^{2}) - \log (B)\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 1\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle 0\)
-
\(\displaystyle \log(A^{3t})\)
-
\(\displaystyle \log(A)-\log(B)\)
d) \(t \log (A^{3})\)
-
\(\displaystyle \log(A)-\log(B)\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 0\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 1\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle \log(A^{3t})\)
e) \(\log(1)\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 0\)
-
\(\displaystyle \log(A)-\log(B)\)
-
\(\displaystyle 1\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle \log(A^{3t})\)
f) \(\ln(e)\)
-
\(\displaystyle \log(A)-\log(B)\)
-
\(\displaystyle \log(A^{3t})\)
-
\(\displaystyle 0\)
-
\(\displaystyle 1\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
g) \(\ln(\sqrt{A})\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle \log(A)-\log(B)\)
-
\(\displaystyle \frac{\ln(A)}{2}\)
-
\(\displaystyle 0\)
-
\(\displaystyle \ln (A) + \ln (B)\)
-
\(\displaystyle 1\)
-
\(\displaystyle 2\log\left(\frac{A}{\sqrt{B}}\right)\)
-
\(\displaystyle \log(A^{3t})\)
Hint.
Recall the following properties of logarithms to help you answer this question:
\begin{equation*}
\log_a(x\cdot y)=\log_a(x)+\log_a(y)
\end{equation*}
\begin{equation*}
\log_a(x/ y)=\log_a(x)-\log_a(y)
\end{equation*}
\begin{equation*}
\log_a(x^y)=y\log_a(x)
\end{equation*}
\begin{equation*}
\log_a(a^x)=x
\end{equation*}
\begin{equation*}
a^{\log_a(x)}=x.
\end{equation*}
Some exponent rules may also be helpful:
\begin{equation*}
(a\cdot b)^m=a^m\cdot b^m
\end{equation*}
\begin{equation*}
\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}
\end{equation*}
\begin{equation*}
(b^m)^n=b^{m\cdot n}
\end{equation*}
\begin{equation*}
\sqrt[n]{b}=b^{1/n}
\end{equation*}
Finally, you may find it useful to do some process of elimination to first take out any automatically incorrect options. Then you should try to rewrite each statement using the rules given above.
11. Solving Exponents with Logarithms.
Solve for \(x\) in the equations below. [Remember to use \(\text{log10}()\) or \(\text{logten}()\) if you need a base 10 logarithm, and use \(\log()\) or \(\ln()\) if you need a natural logarithm (base \(e\)).]
a)
\(e^x=25\)
\(x =\)
b)
\(400 = 200(10)^x\)
\(x =\)
c)
\(e^{-0.2 x}=\frac{1}{2}\)
\(x =\)
d)
\(\ln(50e^{0.99x})=20\)
\(x =\)
e)
\(\log(x^{4})+\log(x)=4\)
\(x =\)
f)
\(\log(x^{4})-\log(2 x)=2\)
\(x =\)
Hint.
Recall the following properties of logarithms to help you answer this question:
\begin{equation*}
\log_a(x\cdot y)=\log_a(x)+\log_a(y)
\end{equation*}
\begin{equation*}
\log_a(x/ y)=\log_a(x)-\log_a(y)
\end{equation*}
\begin{equation*}
\log_a(x^y)=y\log_a(x)
\end{equation*}
\begin{equation*}
\log_a(a^x)=x
\end{equation*}
\begin{equation*}
a^{\log_a(x)}=x.
\end{equation*}
You can always rewrite an exponential function into a logarithm, or vice versa, by using the following equivalence:
\begin{equation*}
\log_a(y)=x
\end{equation*}
and
\begin{equation*}
a^x=y.
\end{equation*}
Again, recall that \(\log_e(y)=\ln(y)\) and \(\log_{10}(y)=\log(y)\text{.}\)
