- 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*}
Section 4.4 Properties of Logarithms
Subsection Introduction
Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.
Each of these laws corresponds to one of three properties of logarithms.
Properties of Logarithms.
If \(x\text{,}\) \(y\text{,}\) \(b \gt 0\text{,}\) and \(b\ne 1\text{,}\) then
- \(\displaystyle \log_{b}{(xy)} = \log_{b}{x} + \log_{b}{y}\)
- \(\displaystyle \log_{b}\dfrac{x}{y} = \log_b x - \log_b y\)
- \(\displaystyle \log_b {x^k} = k \log_b x \)
We will consider proofs of the three properties of logarithms in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is an exponent.
- Property (1):\begin{equation*} \begin{aligned}[t] \log_{2}{32} \amp=\log_{2}{(4 \cdot 8)} \amp\amp=\log_2 4 +\log_2 8 \amp \text{ because } 2^{\alert{5}} \amp= 2^{\alert{2}} \cdot 2^{\alert{3}} \\ \alert{5} \amp\amp\amp = \alert{2} + \alert{3} \amp 32 \amp = 4 \cdot 8\\ \end{aligned} \end{equation*}
- Property (2):\begin{equation*} \begin{aligned}[t] \log_{2}{8} \amp=\log_{2}{\frac{16}{2}} \amp\amp=\log_2{16}-\log_2 {2} \amp \text{ because } 2^{\alert{3}} \amp= \frac{2^{\alert{4}}}{2^{\alert{1}}} \\ \alert{3} \amp\amp\amp = \alert{4} - \alert{1} \amp 8 \amp = \frac{16}{2}\\ \end{aligned} \end{equation*}
- Property (3):\begin{equation*} \begin{aligned}[t] \log_{2}{64} \amp=\log_{2}{(4)^3} \amp\amp=3\log_2 4 \amp \text{ because }\left(2^{\alert{2}}\right)^{\alert{3}} \amp= 2^{\alert{6}} \\ \alert{6} \amp\amp\amp = \alert{3}\cdot \alert{2} \amp (4)^{\alert{3}} \amp = 64\\ \end{aligned} \end{equation*}
Checkpoint 4.72. QuickCheck 1.
Which statement is true?
- \(\displaystyle \log (8+y)=\log 8+\log y\)
- \(\displaystyle \log (8y)=\log 8+\log y\)
- \(\displaystyle \log (8y)=\log 8 \cdot \log y\)
- \(\displaystyle \log \left(\dfrac{8}{y}\right)=\dfrac{\log 8}{\log y}\)
Answer.
\(\text{Choice 2}\)
Solution.
\(\log (8y)=\log 8+\log y\)
Subsection 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 following Example, we rewrite one log in terms of simpler logs.
Example 4.73.
Simplify \(~~\log_{b}\sqrt{{xy}}\text{.}\)
Solution.
First, we write \(\sqrt{xy}\) using a fractional exponent:
\begin{equation*}
\log_{b}{xy} = \log_{b}{(xy)^{1/2}}
\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, \(~~\log_{b}\sqrt{xy} = \frac{1}{2}(\log_{b}{x} + \log_{b}{y})\text{.}\)
Checkpoint 4.74. Practice 1.
Simplify \(\log_{b}\dfrac{x}{y^2}\text{.}\)
Answer: \(\log_b\)\(+\)\(\cdot\log_b\)
Answer 1.
\(x\)
Answer 2.
\(-2\)
Answer 3.
\(y\)
Solution.
\(\log_b x - 2\log_b y\)
Caution 4.75.
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 4.76.
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(\frac{x}{y}\right)^3
\end{equation*}
Checkpoint 4.77. Practice 2.
Express \(2\log_b x + 4\log_{b}{(x + 3)}\) as a single logarithm with a coefficient of \(1\text{.}\)
Answer: \(\log_b\)
Answer.
\(x^{2}\!\left(x+3\right)^{4}\)
Solution.
\(\log_b x^2(x+3)^4\)
Checkpoint 4.78. QuickCheck 2.
Which expression is equivalent to \(3\log_2 H-\log_2 T\text{?}\)
- \(\displaystyle 3\log_2 (H-T)\)
- \(\displaystyle \log_2 (H^3-T)\)
- \(\displaystyle \log_2 \dfrac{H^3}{T}\)
- \(\displaystyle \log_2\left (\dfrac{H}{T}\right)^3\)
Answer.
\(\text{Choice 3}\)
Solution.
\(\log_2 \dfrac{H^3}{T}\)
Subsection Solving Exponential Equations
By using Property (3), we can now solve exponential equations in which the base is not \(10\text{.}\) For example, 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, we cannot evaluate \(\log_{5}{7}\text{;}\) there is no log base \(5\) button on the calculator. If we want a decimal approximation for the solution, we begin by taking the base \(10\) logarithm of both sides, 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 \(7\) ) ÷ LOG \(5\) ) ENTER
to find that \(x \approx 1.2091\text{.}\)
Caution 4.79.
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*}
Checkpoint 4.80. QuickCheck 3.
How can we solve \(4^x=60\text{?}\)
- A) Divide both sides by 4.
- B) Take log base 10 of both sides.
- C) Take the fourth root of 60.
- D) This equation has no solution.
Answer.
\(\text{B) Take ... both sides.}\)
Solution.
Take log base 10 of both sides.
Example 4.81.
Solve \(~~1640 = 80 \cdot 6^{0.03x}\)
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 4.82.
In Example 4.81, do not try to simplify
\begin{equation*}
80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \blert{\text{ Incorrect!}}
\end{equation*}
Remember that the order of operations tells us to compute the power \(6^{0.03x}\) before multiplying by \(80\text{.}\)
We summarize our method for solving exponential equations as follows.
Steps for Solving Exponential Equations.
- Isolate the power on one side of the equation.
- Take the log base \(10\) of both sides.
- Simplify by applying Log Property (3).
- Solve for the variable.
Checkpoint 4.83. Practice 3.
Solve \(5(1.2)^{2.5x} = 77\)
\(x=\)
Hint.
\(\blert{\text{Divide both sides by 5.}}\)
\(\blert{\text{ the log of both sides.}}\)
\(\blert{\text{Apply Property (3) to simplify the left side.}}\)
\(\blert{\text{Solve for}~x.}\)
Answer.
\(\frac{\log\!\left(\frac{77}{5}\right)}{2.5\log\!\left(1.2\right)}\)
Solution.
\(x=\dfrac{\log 15.4}{2.5\log 1.2}\approx 5.999\)
Checkpoint 4.84. Pause and Reflect.
Explain why the distributive law does not apply to the expression \(\log (a+b)\text{.}\)
Subsection Applications
By using the properties of logarithms, we can now solve equations that arise in exponential growth and decay models, no matter what base the exponential function uses.
Example 4.85.
The population of Silicon City was \(6500\) in \(1990\) and has been tripling every \(12\) years. When will the population reach \(75,000\text{?}\)
Solution.
The population of Silicon City grows according to the formula
\begin{equation*}
P(t) = 6500 \cdot 3^{t/12}
\end{equation*}
where \(t\) is the number of years after \(1990\text{.}\) We want to find the value of \(t\) for which \(P(t) = 75,000\text{;}\) that is, we want to solve the equation
\begin{equation*}
\begin{aligned}[t]
6500 \cdot 3^{t/12} \amp = 75,000\amp\amp \blert{\text{ Divide both sides by 6500.}}\\
3^{t/12} \amp = \frac{150}{13}
\end{aligned}
\end{equation*}
Now we take the base \(10\) logarithm of both sides and solve for \(t\text{.}\)
\begin{equation*}
\begin{aligned}[t]
\log_{10}{(3^{t/12})} \amp = \log_{10}\left(\frac{150}{13}\right)
\amp\amp \blert{\text{Apply Property (3).}}\\
\frac{t}{12}\log_{10}{3} \amp= \log_{10}\left(\frac{150}{13}\right)
\amp\amp \blert{\text{Divide by }\log_{10}{3}\text{; multiply by 12.}}\\
t \amp = \frac{12 \left(\log_{10}{\frac{150}{13}}\right)}{\log_{10}{3}}\\
\amp\approx 26.71
\end{aligned}
\end{equation*}
The population of Silicon City will reach \(75,000\) about \(27\) years after \(1990\text{,}\) or in \(2017\text{.}\)
Checkpoint 4.86. Practice 4.
Traffic on U.S. highways is growing by 2.7% per year. (Source: Time, Jan. 25, 1999)
-
Write a formula for the volume, \(V\text{,}\) of traffic as a function of time, using \(V_0\) for the current volume. [Note: Enter “V0” to get \(V_0\text{.}\)]\(V=\)
-
How long will it take the volume of traffic to double? Hint: Find the value of \(t\) that gives \(V = 2V_0\text{.}\)Answer: about years
Answer 1.
\(V_0\cdot 1.027^{t}\)
Answer 2.
\(\frac{\log\!\left(2\right)}{\log\!\left(1.027\right)}\)
Solution.
- \(\displaystyle V(t) = V_0(1.027)^t\)
- about \(26\) years
Checkpoint 4.87. Pause and Reflect.
How does Property (3) of logarithms help us solve exponential equations?
Subsection Compound Interest
The amount of money in an account that earns interest compounded annually grows exponentially according to the formula
\begin{equation*}
A(t) = P(1 + r )^t
\end{equation*}
(See Section 4.1 to review compound interest.) Many accounts compound interest more frequently than once a year. If the interest is compounded \(n\) times per year, then in \(t\) years there will be \(nt\) compounding periods, and in each period the account earns interest at a rate of \(\dfrac{r}{n}\text{.}\) The amount accumulated is given by a generalization of our earlier formula.
Compound Interest.
The amount \(A(t)\) accumulated (principal plus interest) in an account bearing interest compounded \(n\) times annually is
\begin{equation*}
A(t)=P\left(1+ \frac{r}{n}\right)^{nt}
\end{equation*}
where
| \(P\) | is the principal invested, | ||
| \(r\) | is the interest rate, | ||
| \(t\) | is the time period, in years. |
Example 4.88.
Rashad deposited $\(1000\) in an account that pays \(4\)% interest. Calculate the amount in his account after \(5\) years if the interest is compounded
- semiannually
- quarterly
- monthly
Solution.
- Semiannually means "twice a year,"" so we use the formula for compound interest with \(P = 1000\text{,}\) \(r = 0.04\text{,}\) \(n = 2\text{,}\) and \(t = 5\text{.}\)\begin{equation*} \begin{aligned}[t] A(5) \amp = 1000\left(1 + \frac{0.04}{2}\right)^{2(5)} \\ \amp = 1000(1.02)^{10} = 1218.99 \end{aligned} \end{equation*}If interest is compounded semiannually, the balance in the account after \(5\) years is $\(1218.99\text{.}\)
- Quarterly means "4 times a year,"" so we use the formula for compound interest with \(P = 1000\text{,}\) \(r = 0.04\text{,}\) \(n = 4\text{,}\) and \(t = 5\text{.}\)\begin{equation*} \begin{aligned}[t] A(5) \amp = 1000\left(1 + \frac{0.04}{4}\right)^{4(5)} \\ \amp = 1000(1.01)^{20} = 1220.19 \end{aligned} \end{equation*}If interest is compounded quarterly, the balance in the account after \(5\) years is $\(1220.19\text{.}\)
- There are \(12\) months in a year, so we use the formula for compound interest with \(P = 1000\text{,}\) \(r = 0.04\text{,}\) \(n = 12\text{,}\) and \(t = 5\text{.}\)\begin{equation*} \begin{aligned}[t] A(5) \amp = 1000\left(1 + \frac{0.04}{12}\right)^{12(5)}\\ \amp = 1000(1.003)^{60} = 1221.00 \end{aligned} \end{equation*}If interest is compounded monthly, the balance in the account after \(5\) years is $\(1221\text{.}\)
Checkpoint 4.89. QuickCheck 4.
Which formula gives the anount when interest is compounded quarterly?
- \(\displaystyle P(1+4r)^t\)
- \(\displaystyle P(1+4r)^{4t}\)
- \(\displaystyle P\left(1+\dfrac{r}{4}\right)^{4t}\)
- \(\displaystyle P\left(1+\dfrac{r}{4}\right)^{\frac{t}{4}}\)
Answer.
\(\text{Choice 3}\)
Solution.
\(P\left(1+\dfrac{r}{4}\right)^{4t}\)
Note 4.90.
In Example 4.88, you can see that the larger the value of \(n\text{,}\) the greater the value of \(A\text{,}\) keeping the other parameters fixed. More frequent compounding periods result in a higher account balance.
Checkpoint 4.91. Practice 5.
Calculate the amount in Rashad’s account after 5 years if the interest is compounded daily. (See Example 4.88. There are 365 days in a year.)
Answer: $
Answer.
\(1000\cdot \left(1+\frac{0.04}{365}\right)^{365\cdot 5}\)
Solution.
$1221.39
Subsection Solving Formulas
The techniques for solving exponential equations can also be used to solve formulas involving exponential expressions for one variable in terms of the others.
Example 4.92.
Solve \(~~2C = Cb^{kt}~~\) for \(t\text{.}\) (Assume that \(C\) and \(k \ne 0\text{.}\))
Solution.
First, we divide both sides by \(C\) to isolate the power.
\begin{equation*}
b^{kt} = 2
\end{equation*}
Next, we take the log base \(10\) of both sides.
\begin{equation*}
\begin{aligned}[t]
\log b^{kt} \amp = \log 2 \\
kt \log b \amp = \log 2 \amp\amp \blert{\text{Apply Log Property (3).}}
\end{aligned}
\end{equation*}
Finally, we divide both sides by \(k \log b\) to solve for \(t\text{.}\)
\begin{equation*}
t = \frac{\log 2}{k \log b}
\end{equation*}
Checkpoint 4.93. Practice 6.
Solve \(A = P(1 + r )^t\) for \(t\text{.}\)
\(t=\)
Answer.
\(\frac{\log\!\left(\frac{A}{P}\right)}{\log\!\left(1+r\right)}\)
Solution.
\(t=\dfrac{\log(A/P)}{\log(1+r)}\)
Subsection Section Summary
Subsubsection Vocabulary
Look up the definitions of new terms in the Glossary.
- Compounding period
Subsubsection CONCEPTS
Properties of Logarithms.
If \(x\text{,}\) \(y\text{,}\) \(b \gt 0\text{,}\) and \(b\ne 1\text{,}\) then- \(\displaystyle \log_{b}{(xy)} = \log_{b}{x} + \log_{b}{y}\)
- \(\displaystyle \log_{b}\dfrac{x}{y} = \log_b x - \log_b y\)
- \(\displaystyle \log_b {x^k} = k \log_b x \)
-
We can use the properties of logarithms to solve exponential equations with any base.
Steps for Solving Exponential Equations.
- Isolate the power on one side of the equation.
- Take the log base \(10\) of both sides.
- Simplify by applying Log Property (3).
- Solve for the variable.
-
The amount in an account earning interest compounded \(n\) times per year is an exponential function of time.
Compound Interest.
The amount \(A(t)\) accumulated (principal plus interest) in an account bearing interest compounded \(n\) times annually is\begin{equation*} A(t)=P\left(1+ \frac{r}{n}\right)^{nt} \end{equation*}where\(P\) is the principal invested, \(r\) is the interest rate, \(t\) is the time period, in years.
Subsubsection STUDY QUESTIONS
- The properties of logs are really another form of which familiar laws?
- Which log property allows us to solve an exponential equation whose base is not \(10\text{?}\)
- Explain why \(12\cdot 10^{3x}\) is not the same as \(120^{3x}\text{.}\)
- Which of the following expressions are equivalent?\begin{equation*} \log\left(\dfrac{x}{4} \right) ~~~~~ \dfrac{\log x}{\log 4} ~~~~~~~ \log(x-4) ~~~ \log x - \log 4 \end{equation*}
- Which of the following expressions are equivalent?\begin{equation*} \log(x+2) ~~~~~ \log x+\log 2 ~~~~~ \log(2x) ~~~~~ (\log 2) (\log x) \end{equation*}
- Which of the following expressions are equivalent?\begin{equation*} \log x^3 ~~~~~(\log 3)(\log x) ~~~~~ 3\log x ~~~~~ \log 3^x \end{equation*}
Subsubsection SKILLS
Practice each skill in the Homework problems listed.
- Use the properties of logarithms to simplify expressions: #1–24, #45–52
- Solve exponential equations using logs base 10: #25–36
- Solve problems about exponential models: #37–44
- Solve problems about compound interest: #53–58
- Solve formulas involving exponential expressions: #59–64
Exercises Homework 4.4
1.
- Simplify \(10^2 \cdot 10^6\text{.}\)
- Compute \(\log 10^2\text{,}\) \(\log 10^6\text{,}\) and \(\log(10^2\cdot 10^6)\text{.}\) How are they related?
2.
- Simplify \(\dfrac{10^9}{10^6} \text{.}\)
- Compute \(\log 10^9\text{,}\) \(\log 10^6\text{,}\) and \(\log\left(\dfrac{10^9}{10^6} \right) \text{.}\) How are they related?
3.
- Simplify \(\dfrac{b^8}{b^5} \text{.}\)
- Compute \(\log_b b^8\text{,}\) \(\log_b b^5\text{,}\) and \(\log_b\left(\dfrac{b^8}{b^5} \right) \text{.}\) How are they related?
4.
- Simplify \(b^4 \cdot b^3\text{.}\)
- Compute \(\log_b b^4\text{,}\) \(\log_b b^3\text{,}\) and \(\log_b(b^4\cdot b^3)\text{.}\) How are they related?
5.
- Simplify \(\left(10^3 \right)^5 \text{.}\)
- Compute \(\log \left(10^3 \right)^5\) and \(\log\left(10^3 \right) \text{.}\) How are they related?
6.
- Simplify \(\left(b^2 \right)^6 \text{.}\)
- Compute \(\log_b \left(b^2 \right)^6\) and \(\log_b\left(b^2 \right) \text{.}\) How are they related?
Exercise Group.
For Problems 7-14, use the properties of logarithms to expand each expression in terms of simpler logarithms. Assume that all variable expressions denote positive numbers.
7.
- \(\displaystyle \log_b 2x \)
- \(\displaystyle \log_b\dfrac{x}{2} \)
8.
- \(\displaystyle \log_b\dfrac{2x}{x-2} \)
- \(\displaystyle \log_b x(2x+3) \)
9.
- \(\displaystyle \log_3 (3x^4) \)
- \(\displaystyle \log_5 1.1^{1/t} \)
10.
- \(\displaystyle \log_b (4b)^t \)
- \(\displaystyle \log_2 5(2^x) \)
11.
- \(\displaystyle \log_b \sqrt{bx} \)
- \(\displaystyle \log_3 \sqrt[3]{x^2+1} \)
12.
- \(\displaystyle \log_{10} \sqrt{\dfrac{2L}{R^2}} \)
- \(\displaystyle \log_{10} 2\pi\sqrt{\dfrac{l}{g}} \)
13.
- \(\displaystyle \log P_0 (1-m)^t \)
- \(\displaystyle \log_4 \left(1+\dfrac{r}{4}\right)^{4t} \)
14.
- \(\displaystyle \log_3 \dfrac{a^2-2}{a^5} \)
- \(\displaystyle \log \dfrac{a^3b^2}{(a+b)^{3/2}} \)
Exercise Group.
For Problems 15-20, combine into one logarithm and simplify. Assume all expressions are defined.
15.
- \(\displaystyle \log_b 8 - \log_b 2\)
- \(\displaystyle 2 \log_4 x + 3 \log_4 y\)
16.
- \(\displaystyle \log_b 5 + \log_b 2\)
- \(\displaystyle \dfrac{1}{4} \log_5 x - \dfrac{3}{4} \log_5 y\)
17.
- \(\displaystyle \log 2x+ 2\log x -\log\sqrt{x} \)
- \(\displaystyle \log(t^2-16)-\log(t+4) \)
18.
- \(\displaystyle \log x^2 + \log x^3-5\log x\)
- \(\displaystyle \log(x^2-x)-\log\sqrt{x^3} \)
19.
- \(\displaystyle 3-3\log 30 \)
- \(\displaystyle \dfrac{1}{3}\log_6(8w^6) \)
20.
- \(\displaystyle 2-\log_4 (16z^2) \)
- \(\displaystyle 1-2\log_3 x \)
Exercise Group.
For Problems 21-24, use the three logs below to find the value of each expression.
\begin{equation*}
\log_b 2 = 0.6931, ~~\log_b 3 = 1.0986, ~~\log_b 5 = 1.6094
\end{equation*}
(Hint: For example, \(\log_b 15 = \log_b 3 + \log_b 5\text{.}\))
21.
- \(\displaystyle \log_b 6\)
- \(\displaystyle \log_b\dfrac{2}{5} \)
22.
- \(\displaystyle \log_b 10\)
- \(\displaystyle \log_b\dfrac{3}{2} \)
23.
- \(\displaystyle \log_b 9\)
- \(\displaystyle \log_b\sqrt{50} \)
24.
- \(\displaystyle \log_b 25\)
- \(\displaystyle \log_b 75 \)
Exercise Group.
For Problems 25-36, solve the equation by using logarithms base \(10\text{.}\) Round answers to four decimal places.
25.
\(2^x=7 \)
26.
\(3^x=4 \)
27.
\(3^{x+1}=8 \)
28.
\(2^{x-1}=9 \)
29.
\(4^{x^2}=15 \)
30.
\(3^{x^2}=21 \)
31.
\(4.26^{-x}=10.3 \)
32.
\(2.13^{-x}=8.1 \)
33.
\(25\cdot 3^{2.1x}=47 \)
34.
\(12\cdot 5^{1.5x}=85 \)
35.
\(3600=20\cdot 8^{-0.2x} \)
36.
\(0.06=50\cdot 4^{-0.6x} \)
37.
If raw meat is allowed to thaw at \(50\degree\)F, Salmonella grows at a rate of \(9\%\) per hour.
- Write a formula for the amount of Salmonella present after \(t\) hours, if the initial amount is \(S_0\text{.}\)
- Health officials advise that the amount of Salmonella initially present in meat should not be allowed to increase by more than \(50\%\text{.}\) How long can meat be left to thaw at \(50\degree\)F?
38.
Starting in 1998, the demand for electricity in Ireland grew at a rate of \(5.8\%\) per year. In 1998, \(20,500\) gigawatts were used. (Source: Electricity Supply Board of Ireland)
- Write a formula for electricity demand in Ireland as a function of time.
- If demand continues to grow at the same rate, when would it reach \(30,000\) gigawatts?
39.
The concentration of a certain drug injected into the bloodstream decreases by \(20\%\) each hour as the drug is eliminated from the body. The initial dose creates a concentration of \(0.7\) milligrams per milliliter.
- Write a function for the concentration of the drug as a function of time.
- The minimum effective concentration of the drug is \(0.4\) milligrams per milliliter. When should the second dose be administered?
- Verify your answer with a graph.
40.
A small pond is tested for pollution, and the concentration of toxic chemicals is found to be \(80\) parts per million. Clean water enters the pond from a stream, mixes with the polluted water, then leaves the pond so that the pollution level is reduced by \(10\%\) each month.
- Write a function for the concentration of toxic chemicals as a function of time.
- How long will it be before the concentration of toxic chemicals reaches a safe level of \(25\) parts per million?
- Verify your answer with a graph.
41.
According to the National Council of Churches, the fastest growing denomination in the United States in \(2004\) was the Jehovah’s Witnesses, with an annual growth rate of \(1.82\%\text{.}\)
- The Jehovah’s Witnesses had \(1,041,000\) members in \(2004\text{.}\) Write a formula for the membership in the Jehovah’s Witnesses as a function of time, assuming that the church continues to grow at the same rate.
- When will the Jehovah’s Witnesses have \(2,000,000\) members?
42.
In \(2004\text{,}\) the Presbyterian Church had \(3,241,000\) members, but membership was declining by \(4.87\%\) annually.
- Write a formula for the membership in the Presbyterian Church as a function of time, assuming that the membership continues to decline at the same rate.
- When will the Presbyterian Church have \(2,000,000\) members?
43.
Sodium-24 is a radioactive isotope that is used in diagnosing circulatory disease. It decays into stable isotopes of sodium at a rate of \(4.73\%\) per hour.
- Technicians inject a quantity of sodium-24 into a patient’s bloodstream. Write a formula for the amount of sodium-24 present in the bloodstream as a function of time.
- How long will it take for \(75\%\) of the isotope to decay?
44.
The population of Afghanistan is growing at \(2.6\%\) per year.
- Write a formula for the population of Afghanistan as a function of time.
- In 2005, the population of Afghanistan was \(29.9\) million. At the given rate of growth, how long would it take the population to reach \(40\) million?
Exercise Group.
For Problems 46-52, evaluate each expression. Which (if any) are equal?
45.
- \(\displaystyle \log_2(4\cdot 8) \)
- \(\displaystyle (\log_2 4)(\log_2 8) \)
- \(\displaystyle \log_2 4 + \log_2 8 \)
46.
- \(\displaystyle \log_2(16+16) \)
- \(\displaystyle \log_2 16+\log_2 16 \)
- \(\displaystyle \log_2 2 + \log_2 16 \)
47.
- \(\displaystyle \log_3(27^2) \)
- \(\displaystyle (\log_3 27)^2 \)
- \(\displaystyle \log_3 27 + \log_3 27 \)
48.
- \(\displaystyle \log_3(3 \cdot 27) \)
- \(\displaystyle \log_3 3 +\log_3 27 \)
- \(\displaystyle \log_3 3 \cdot \log_3 27 \)
49.
- \(\displaystyle \log_{10}\left(\dfrac{240}{10} \right) \)
- \(\displaystyle \dfrac{\log_{10} 240}{\log_{10} 10} \)
- \(\displaystyle \log_{10} (240-10) \)
50.
- \(\displaystyle \log_{10}\left(\dfrac{1}{2}\cdot 80 \right) \)
- \(\displaystyle \dfrac{1}{2} \log_{10} 80 \)
- \(\displaystyle \log_{10} \sqrt{80} \)
51.
- \(\displaystyle \log_{10} (75-15) \)
- \(\displaystyle \log_{10} 75-\log_{10}{15} \)
- \(\displaystyle \dfrac{\log_{10}75}{\log_{10}15} \)
52.
- \(\displaystyle \log_{10}\left(8\cdot 25 \right) \)
- \(\displaystyle \log_{10} (25^8) \)
- \(\displaystyle \log_{10} 8+25 \)
Exercise Group.
For Problems 53–58, use the formula for compound interest,
\begin{equation*}
A=P\left(1+\dfrac{r}{n} \right)^{nt}
\end{equation*}
53.
What rate of interest is required so that \(\$1000\) will yield \(\$1900\) after \(5\) years if the interest rate is compounded monthly?
54.
What rate of interest is required so that \(\$400\) will yield \(\$600\) after \(3\) years if the interest rate is compounded quarterly?
55.
How long will it take a sum of money to triple if it is invested at \(10\%\) compounded daily?
56.
How long will it take a sum of money to increase by a factor of \(5\) if it is invested at \(10\%\) compounded quarterly?
57.
- Suppose you invest \(\$1000\) at \(12\%\) annual interest for \(5\) years. In this problem, we will investigate how the number of compounding periods,\(n\text{,}\) affects the amount, \(A\text{.}\) Write \(A\) as a function of \(n\text{,}\) with \(P = 1000\text{,}\) \(r = 0.12\text{,}\) and \(t = 5\text{.}\)
- Use your calculator to make a table of values for \(A\) as a function of \(n\text{.}\) What happens to \(A\) as \(n\) increases?
- What value of \(n\) is necessary to produce an amount \(A\gt 1818\text{?}\) To produce \(A\gt 1820\text{?}\) To produce \(A\gt 1822\text{?}\)
- Graph the function \(A(n)\) in the window\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 52\\ {\text{Ymin}} \amp = 1750 \amp\amp {\text{Ymax}} = 1850 \end{align*}Describe the graph: Is it increasing or decreasing? Concave up or down? Does it appear to have an asymptote? Give your best estimate for the asymptote.
58.
- In this problem we will repeat Problem 49 for \(4\%\) interest. Write \(A\) as a function of \(n\text{,}\) with \(P = 1000\text{,}\) \(r = 0.04\text{,}\) and \(t = 5\text{.}\)
- Use your calculator to make a table of values for \(A\) as a function of \(n\text{.}\) What happens to \(A\) as \(n\) increases?
- What value of \(n\) is necessary to produce an amount \(A\gt 1218\text{?}\) To produce \(A\gt 1220\text{?}\) To produce \(A\gt 1221.40\text{?}\)
- Graph the function \(A(n)\) in the window\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 52\\ {\text{Ymin}} \amp = 1210 \amp\amp {\text{Ymax}} = 1225 \end{align*}Describe the graph: Is it increasing or decreasing? Concave up or down? Does it appear to have an asymptote? Give your best estimate for the asymptote.
Exercise Group.
For Problems 59-64, solve the formula for the specified variable.
59.
\(N = N_0a^{kt}\text{,}\) for \(k\)
60.
\(Q=Q_0 b^{t/2} \text{,}\) for \(t\)
61.
\(A = A_0(10^{kt}-1)\text{,}\) for \(t\)
62.
\(B=B_0 (1-10^{-kt}) \text{,}\) for \(t\)
63.
\(w = pv^q\text{,}\) for \(q\)
64.
\(l=p^a q^b \text{,}\) for \(b\)
Exercise Group.
In Problems 65–68 we use the laws of exponents to prove the properties of logarithms.
65.
We will use the first law of exponents, \(a^p\cdot a^q = a^{p+q}\text{,}\) to prove the first property of logarithms.
- Let \(m = \log_b x\) and \(n = \log_b y\text{.}\) Rewrite these equations in exponential form:\begin{equation*} x=\fillinmath{XXXXXXXXXXXXXXX}~~~ \text{ and } ~~~y=\fillinmath{XXXXXXXXXXXXXXX} \end{equation*}
- Now consider the expression \(\log_b (xy)\text{.}\) Replace \(x\) and \(y\) by your answers to part (a).
- Apply the first law of exponents to your expression in part (b).
- Use the definition of logarithm to simplify your answer to part (c).
- Refer to the definitions of \(m\) and \(n\) in part (a) to finish the proof.
66.
We will use the second law of exponents, \(\dfrac{a^p}{a^q}=a^{p-q} \text{,}\) to prove the second property of logarithms.
- Let \(m = \log_b x\) and \(n = \log_b y\text{.}\) Rewrite these equations in exponential form:\begin{equation*} x=\fillinmath{XXXXXXXXXXXXXXX}~~~ \text{ and } ~~~y=\fillinmath{XXXXXXXXXXXXXXX} \end{equation*}
- Now consider the expression \(\log_b (\frac{x}{y})\text{.}\) Replace \(x\) and \(y\) by your answers to part (a).
- Apply the second law of exponents to your expression in part (b).
- Use the definition of logarithm to simplify your answer to part (c).
- Refer to the definitions of \(m\) and \(n\) in part (a) to finish the proof.
67.
We will use the third law of exponents, \((a^p)^q = a^{pq}\text{,}\) to prove the third property of logarithms.
- Let \(m = \log_b x\text{.}\) Rewrite this equation in exponential form:\begin{equation*} x=\fillinmath{XXXXXXXXXXXXXXX} \end{equation*}
- Now consider the expression \(\log_b (x^k)\text{.}\) Replace \(x\) by your answers to part (a).
- Apply the third law of exponents to your expression in part (b).
- Use the definition of logarithm to simplify your answer to part (c).
- Refer to the definitions of \(m\) in part (a) to finish the proof.
68.
- Use the laws of exponents to explain why \(\log_b 1 = 0\text{.}\)
- Use the laws of exponents to explain why \(\log_b b^x = x\text{.}\)
- Use the laws of exponents to explain why \(b^{\log_b x} = x\text{.}\)
