1 Definition of logarithm (Section 4.2 , page 445)
2 Basic properties of logarithms (Section 4.2 , page 447)
3 Rules of exponents (Section 4.1 , page 425)
Today the most famous pharaoh of Ancient Egypt is King Nebkheperure Tutankhamun, popularly called “King Tut.” He was only 9 years old when he became a pharaoh. In 2005, a team of Egyptian scientists headed by Dr. Zahi Hawass determined that the pharaoh was 19 years old when he died (around 1346 b.c.).
Tutankhamun was a short-lived boy king who, unlike the great Egyptian kings Khufu (builder of the Great Pyramid), Amenhotep III (builder of temples throughout Egypt), and Ramesses II (prolific builder and usurper), accomplished nothing significant during his reign. In fact, little was known about him prior to Howard Carter’s discovery of his tomb (and the amazing treasures it held) in the Valley of the Kings on November 4, 1922. Carter, with his benefactor Lord Carnarvon at his side, entered the tomb’s burial chamber on November 26, 1922. Lord Carnarvon died seven weeks after entering the burial chamber, giving rise to the theory of the “curse” of King Tut.
Work on the tomb continued until 1933. The tomb contained a pristine mummy of an Egyptian king, lying intact in his original burial furniture. He was accompanied by a small slice of the royal world of the pharaohs: golden chariots, statues of gold and ebony, a fleet of miniature ships to accommodate his trip to the hereafter, his throne of gold, bottles of perfume, precious jewelry, and more. The “Treasures of Tutankhamun” exhibition, first shown at the British Museum in London in 1972, traveled to many countries, including the United States.
In Example 8, we discuss the age of a work of art found in King Tut’s tomb.
1 Learn the rules of logarithms.
Recall the basic properties of logarithms from Section 4.2 for a>0, a≠1:
In this section, we will discuss some important rules of logarithms that are helpful in calculations, simplifications, and applications.
Rules 1, 2, and 3 follow from the corresponding rules of the exponents:
For example, to prove the product rule for logarithms, we let
The corresponding exponential forms of these equations are
Then
This proves the product rule of logarithms. In Exercise 126, you are asked to prove the quotient rule and the power rule similarly by using the corresponding rules for exponents. The following table compares the rules of exponents and logarithms.
Given that log5 z=3 and log5 y=2, evaluate each expression.
log5(yz)
log5(125y7)
log5 √zy
log5(z1/30y5)
log5(yz)=log5y+log5zProduct rule=2+3=5Use the given values and simplify.
log5(125y7)=log5125+log5y7Product rule=log553+log5y7125=53=3+7 log5 yPower rule and loga a=1=3+7(2)=17Use the given values and simplify.
log5 √zy=log5 (zy)1/2Rewrite radical as exponent.=12 log5 zyPower rule=12(log5 z−log5 y)Quotient rule=12(3−2)=12Use the given values and simplify.
log5(z1/30y5)=log5z1/30+log5y5Product rule=130log5z+5 log5yPower rule=130(3)+5(2)Use the given values.=0.1+10=10.1Simplify.
Evaluate each expression for the given values of logarithms in Example 1 .
log5 (yz)
log5 (y2z3)
In many applications in more advanced mathematics courses, the rules of logarithms are used in both directions; that is, the rules are read from left to right and from right to left. For example, by the product rule of logarithms, we have
The expression log23+log2 x is the expanded form of log2 3x, while log2 3x is the condensed form, or the single logarithmic form of log2 3+log2 x.
Write each expression in expanded form.
log2 x2(x−1)3(2x+1)4
logc√x3y2z5
log2 x2(x−1)3(2x+1)4=log2x2(x−1)3−log2(2x+1)4Quotient rule=log2x2+log2(x−1)3−log2(2x+1)4Product rule=2log2x+3 log2(x−1)−4 log2(2x+1)Power rule
logc √x3y2z5=logc (x3y2z5)1/2√a=a1/2=12logc (x3y2z5)Power rule=12[logcx3+logcy2+logcz5]Product rule=12[3 logc x+2 logc y+5 logc z]Power rule=32logc x+logc y+52logc zDistributive property
Write each expression in expanded form.
ln 2x−1x+4
log √4xyz
Write each expression in condensed form.
log 3x−log 4y
2 ln x +12 ln (x2+1)
2 log25+log29−log275
13 [ln x +ln (x+1)−ln (x2+1)]
log 3x−log 4y=log (3x4y)Quotient rule
2 ln x +12 ln (x2+1)=ln x2+ln (x2+1)1/2Power rule=ln (x2√x2+1)Product rule; √a=a1/2
2 log2 5+log2 9−log2 75=log2 52+log2 9−log2 75Power rule=log2 (25⋅9)−log2 7552=25; Product rule=log225⋅975Quotient rule=log2 325⋅975=93=3
13[ln x +ln (x+1)−ln (x2+1)]=13[ln x(x+1)−ln (x2+1)]Productrule=13ln[x(x+1)x2+1]Quotient rule=ln3√x(x+1)x2+1Power rule; a1/3=3√a
Write the expression in condensed form:
Be careful when using the rules of logarithms to simplify expressions. For example, there is no property that allows you to rewrite loga (x+y).
In general,
To illustrate this statement, let x=100, y=10, and a=10. Then the value of the left side of (11) is
The value of the right side of (11) is
Therefore, log (100+10)≠log 100+log 10.
To avoid common errors, be aware that in general,
2 Estimate a large number.
In estimating the magnitude of a positive number K, we look for the number of digits needed to express K. If K has a fractional part, then “number of digits” means the number of digits to the left of the decimal point. For example, 357.29 and 231.4796 are viewed as 3-digit numbers in this context.
A number K written in the form K=s×10n, where 1≤s<10 and n is an integer, is said to be in scientific notation. For example, 432.1=4.321×102 and 0.56=5.6×10−1 are both in scientific notation.
The exponent n represents the magnitude of K and is closely related to the common logarithm. In fact, we have
This says that the exponent n in K=s×10n is the largest integer less than or equal to log K. We note that a number m is a 3-digit number if
Similarly, a number K has (n+1) digits if n≤log K<n+1.
For example, the number 8765.43, which is a 4-digit number, has its common logarithm between 3 and 4. Conversely, if log N is approximately 57.3 then N is a 58-digit number.
Write an estimate of the number K=e700 in scientific notation.
Because log K lies between the integers 304 and 305, the number K requires 305 digits to the left of the decimal point. Also, by definition of the common logarithm, we have
Repeat Example 4 for K=(234)567.
3 Change the base of a logarithm.
Calculators usually come with two types of log keys: a key for natural logarithms (base e, LN) and a key for common logarithms (base 10, LOG). These two types of logarithms are frequently used in applications. Sometimes, however, we need to evaluate logarithms for bases other than e and 10. The change-of-base formula helps us evaluate these logarithms.
Suppose we are given logb x and we want to find an equivalent expression in terms of logarithms with base a. We let
In exponential form, we have
Then
Equation (14) is the change-of-base formula. By choosing a=10 and a=e, we can state the change-of-base formula as follows:
Compute log513 by changing to
common logarithms and
natural logarithms.
log513=log 13log 5Change to base 10.≈1.59369Use a calculator.
log513=ln 13ln 5Change to base e.≈1.59369Use a calculator.
Find log315.
Find the following.
An equation of the form y=c+b log x whose graph contains the points (2, 3) and (4, −5).
An equation of the form y=c+b logax whose graph contains the points (2, 3) and (4, −5) where a, b, and c are integers.
Substitute (2, 3) and (4, −5) in the equation y=c+b log x to obtain:
Find b: Subtract equation (16) from equation (15) to obtain
Find c:
Use the change-of-base formula to find a, b, and c with integer values.
Then a=2, b=−8, and c=11. (Answers may vary.)
Repeat Example 6 for the graph that contains the points (3, 3) and (9, 1).
The change-of-base formula can be used to produce a variety of integer values for a and b in part b of Example 6. Because 8 log2x=4 log4x we could also chose a=4, b=−4.
4 Apply logarithms to growth and decay.
We express the exponential growth (or decay) formula (page 437) in an equivalent logarithmic form.
Growth occurs when k>0 and decay occurs when k<0.
The half-life of any quantity whose value decreases with time is the time required for the quantity to decay to half its initial value.
Radioactive substances undergo exponential decay. Krypton-85, for example, has a half-life of about ten years; this means that any initial quantity of krypton-85 will dissipate or decay to half that amount in about ten years.
If h is the half-life of a substance undergoing exponential decay at the rate k, then any mass A0 decays to 12A0 in time h. See Figure 4.18. This fact leads to a half-life formula.
In an experiment, 18 grams of the radioactive element sodium-24 decayed to 6 grams in 24 hours. Find its half-life to the nearest hour.
To find the half-life of sodium-24, we use the formula
In an experiment, 100 grams of the radioactive element strontium-90 decayed to 66 grams in 15 years. Find its half-life to the nearest year.
Ordinary carbon, called carbon-12 (C12), is stable and does not decay. However, carbon-14 (C14) is a form of carbon that decays radioactively with a half-life of 5700 years. Sunlight constantly produces carbon-14 in Earth’s atmosphere. When a living organism breathes or eats, it absorbs carbon-12 and carbon-14. After the organism dies, no more carbon-14 is absorbed; so the age of its remains can be calculated by determining how much carbon-14 has decayed. The method of radiocarbon dating was developed by the American scientist W. F. Libby.
In 1960, a group of specialists from the British Museum in London investigated whether a piece of art containing organic material found in Tutankhamun’s tomb had been made during his reign or (as some historians claimed) whether it belonged to an earlier period. We know that King Tut died in 1346 b.c. and ruled Egypt for ten years. What percent of the amount of carbon-14 originally contained in the object should be present in 1960 if the object was made during Tutankhamun’s reign?
Because the half-life h of carbon-14 is approximately 5700 years, we have
Substituting this value of k in the exponential form, we obtain
The time t that elapsed between King Tut’s death and 1960 is
Therefore, the percent of the original amount of carbon-14 remaining in the object (after 3306 years) is 66.897%.
Because King Tut began ruling Egypt ten years before he died, the time t1 that elapsed from the beginning of his reign to 1960 is 3316 years.
Therefore, a piece of art made during King Tut’s reign would have between 66.816% and 66.897% of carbon-14 remaining in 1960.
Repeat Example 8 , assuming that the art object was made 194 years before King Tut’s death.
loga MN=_+_.
_=loga M−loga N.
loga Mr=_.
The change-of-base formula using base e is loga M=_.
True or False. loga (u+v)=loga u+loga v.
True or False. log u− log v= log u log v.
True or False. ln(2√x)=12ln(2x).
True or False. log x10=log x−1.
In Exercises 9–20, given that log x=2, log y=3, log 2≈0.3, and log 3≈0.48, evaluate each expression without using a calculator.
log 6
log 4
log 5 [Hint: 5=102]
log (3x)
log (2x)
log x2
log (2x2y)
log xy3
log3√x2y4
log (logx2)
log 3√48
log2 3
In Exercises 21–44, write each expression in expanded form. Assume that all expressions containing variables represent positive numbers.
ln [x(x−1)]
ln x−2x+3
ln x+1(x−2)2
ln x(x+1)(x−1)2
loga√xy3
loga3x2√y
loga3√xy
loga3√x2y5
log24√xy28
log 3√x2y100
log √x2+1x+3
log x2√x+1
log3(x−1)(x+1)x2−4
log4 (x2−9x2−6x+8)2/3
logbx2y3z
logb √xyz
ln [x√x−1x2+2]
ln[√x−2 3√x+1x2+3]
ln [(x+1)2(x−3)√x+4]
ln [2x+3(x+4)2(x−3)4]
ln [(x+1)√x2+2x2+5]
ln [3√2x+1(x+1)(x−1)2(3x+2)]
ln [x3(3x+1)4√x2+1(x+2)−5(x−3)2]
ln [(x+1)1/2(x2−2)2/5(2x−1)3/2(x2+2)−4/5]
In Exercises 45–60, write each expression in condensed form.
log2 x+log2 7
log2 x−log2 3
log (3x+2)− log x
2 log x + log y
12ln x +2 ln y
3 log x +4 log y
12log x−log y +log z
12(log x +log y)
15(log2 z+2 log2 y)
13(log x−2 log y +3 log z)
ln x +2 ln y +3 ln z
2 ln x−3 ln y +4 ln z
3 log x− log y +12log (x2+4)
log x + log (x2+1)− log (x−1)− log (2x2+3)
2 ln x−12ln (x2+1)
2 ln x +12ln (x2−1)−12ln (x2+1)
In Exercises 61–68, write an estimate of each number in scientific notation.
e500
π650
324756
723416
Which is larger, 234567 or 567234?
Which is larger, 43218765 or 87654321?
Find the number of digits in 17200⋅5367.
Find the number of digits in 67200÷23150.
In Exercises 69–76, use the change-of-base formula and a calculator to evaluate each logarithm.
log25
log411
log1/23
log√3 12.5
log√5 √17
log15 123
log27+log43
log29−log√2 5
In Exercises 77–84, find the value of each expression without using a calculator.
log3√3
log1/44
log3(log28)
2log2 2
52 log5 3+log5 2
e3 ln 2−2 ln 3
log 4+2 log 5
log2160−log25
In Exercises 85–92, find an equation of the form y=c+b loga x (a, b, and c integers) whose graph contains the two given points.
(10, 1) and (1, 2)
(4, 10) and (2, 12)
(e, 1) and (1, 2)
(3, 1) and (9, 2)
(5, 4) and (25, 7)
(4, 3) and (8, 5)
(2, 4) and (4, 9)
(1, 1) and (5, 7)
In Exercises 93–96, determine the half-life of each substance that decays from A0 to A in time t.
A0=50,A=23,t=12 years
A0=200,A=65,t=10 years
A0=10.3,A=3.8,t=15 hours
A0=20.8,A=12.3,t=40 minutes
In Exercises 97–100, assume the exponential growth model A(t)=A0ekt or ln(A(t)A0)=kt and a world population of 7 billion in 2011.
Rate of population growth. If the world adds about 90 million people in 2012, what is the rate of growth?
Population. Use the rate of growth from Exercise 97 to estimate the year in which the world population will be
12 billion.
20 billion.
Population growth. If the population must stay below 20 billion during the next 100 years, what is the maximum acceptable annual rate of growth? (Round your answer to six decimal places.)
Population decline. If the world population must decline below 5 billion during the next 25 years, what must be the annual rate of decline? (Round your answer to six decimal places.)
Continuous compounding. If $1000 is deposited in a bank at 10% interest compounded continuously, how many years will it take for the money to grow to $3500?
Continuous compounding. At what rate of interest compounded continuously will an investment double in six years?
Half-life. Iodine-131, a radioactive substance that is effective in locating brain tumors, has a half-life of only eight days. A hospital purchased 20 grams of the substance but had to wait five days before it could be used. How much of the substance was left after five days?
Half-life. Plutonium-241 has a half-life of 13 years. A laboratory purchased 10 grams of the substance but did not use it for two years. How much of the substance was left after two years?
Half-life. Tritium is used in nuclear weapons to increase their power. It decays at the rate of 5.5% per year. Calculate the half-life of tritium.
Half-life. Sixty grams of magnesium-28 decayed into 7.5 grams after 63 hours. What is the half-life of magnesium-28?
Effective medicine dosage. Use the following model in Exercises 107–110: The concentration C(t) of a drug administered to a patient intravenously jumps to its highest level almost immediately. The concentration subsequently decays exponentially according to the law
where C(0)=C0. The physician administering the drug would like to have
where m is the concentration below which the drug is ineffective and M is the concentration above which the drug is dangerous.
Drug concentration. Suppose that immediately after a certain drug is administered, the concentration is 5 milligrams per milliliter. Ten hours later the concentration drops to 1.5 milligrams per milliliter. Determine the value of k for this drug.
Drug concentration. For the drug of Exercise 107, suppose the maximum safe level is M=12 milligrams per milliliter and the minimum effective level is m=3 milligrams per milliliter. If the initial concentration is M, find the maximum possible time between doses of this drug.
Half-life. The half-life of Valium (diazepam) is 36 hours. Suppose a patient receives 16 milligrams of the drug at 8 a.m. How much Valium is in the patient’s blood at 4 p.m. the same day?
Half-life. The half-life of aspirin in the blood is 12 hours. Estimate the time required for the aspirin to decay to 90% of the original amount ingested.
In Exercises 111–114, use the following amortization formula:
where P=the payment, r=the annual interest rate, M=the mortgage amount, t=the number of years, and n=the number of payments per year.
Finding P. What is the monthly payment on a mortgage of $120,000 with a 6% interest rate for 20 years? How much interest will be paid over 20 years?
Finding t. The Garcia family wanted to take out a mortgage for $150,000 at 8% interest with monthly payments. The family can afford monthly payments of $1200. How long would they have to make payments to pay off their mortgage, and how much interest would they be paying?
Finding M. The First National Bank offers Andy an 8.5% interest rate on a 30-year mortgage to be paid back in monthly payments. The most Andy can afford to pay in monthly payments is $850. What mortgage amount can Andy afford?
Finding r. Lisa is 40 years old and needs to take out a mortgage of $120,000. She can afford monthly payments up to $850. She wants her mortgage paid off when she retires at age 65. What interest rate on the mortgage will satisfy all her requirements? [Hint: Start with r=7% and compute the monthly payment, increasing r by 0.001 each time.]
Show that logb (√x2+1−x)+logb (√x2+1+x)=0.
Show that logb (√x+1+√x)=−logb (√x+1−√x).
Show that (logb a)(loga b)=1.
Show that log ab+log ba=0.
Write log2x=ln xln 2 by the change-of-base formula. Then use a graphing calculator to sketch the graph of y=log2 x.
Sketch the graph of y=log5(x+3).
Simplify each expression.
log (ab)+log (ba)+log (ca)+log (ac)
log (a2bc)+log (b2ca)+log (c2ab)
log2 3⋅log3 4
loga b⋅logb c⋅logc a
Find the number of digits in N if log2(log2 N)=4.
Find the domain of f(x)=log4(log5(log3(18x−x2−77)))
Let f(x)=loga x. Show that
f(1x)=−f(x)=log1/ax.
f(x+h)−f(x)h=loga (1+hx)1/h, h≠0.
State whether each of the following is true or false for all permissible values of the variable.
log(x+2)=log x +log 2
log 2x=log x +log 2
log2 3x=3 log2 x
(log3 2x)4=4 log3 2x
log5x2=2 log5 |x|
log x3=log x−log 3
log x4=log xlog 4
ln(1x)=−ln x
log2x2=(log2 x)(log2 x)
log |10x|=1+log |x|
Prove the quotient rule for logarithms:
Prove the power rule for logarithms: loga Mr=r loga M.
If log (a+b3)=12(log a +log b), show that a2+b2−7ab=0.
Evaluate 11+logc ab+11+loga bc+11+logb ca.
[Hint: Convert to common logarithms and then simplify.]
What went wrong? Find the error in the following argument:
By the power rule for logarithms, log x2=2 log x. However, the graphs of f(x)=log x2 and g(x)=2 log x are not identical. Explain why.
Prime Number. As of 2008, the largest known prime number was 243112609−1. How many digits does this prime number have?
[Hint: First, argue that if p=2m−1 is a prime number, then p and 2m have the same number of digits.]
Find the domain of g(x)=1log(1−x)+1ln(x+2).
In Exercises 133–136, simplify each expression and write your result without exponents.
11⋅30
−4⋅5x⋅5−x
4x⋅2−2x+1
(7x)2⋅(72)−x
In Exercises 137–140, let t=5x and write each expression in the form at2+bt+c=0, where a, b, and c are real numbers.
52x−5x=−1
3⋅52x−2⋅5x=7
5x+3⋅5−x5x=14
5−x+5x5−x−5x=12
In Exercises 141–144, solve for x.
2x−(11+x)=8x+(7+2x)
4x−1+(6x−2)=3−(5x+1)
x2+3x−1=3
2x2−7x=3x+48
In Exercises 145–148, solve for x.
2x<7+x
5x−2≤19−(1−x)
12x>30−3x
4x−17≥6−(5+2x)