Download 13. Logarithmic functions

13. Logarithmic functions
We saw in the previous chapter that e x was a one–to–one function. This means that it has an inverse
function—this we will call lnÝ x Þ . It is also commonly referred to as logÝ x Þ or loge Ý x Þ (log to the base e of x).
Thus the solution of the equation
ex = 2
can be written as
x=
This is not necessarily very helpful, since we have to work out the value of ln 2 . However, most scientific
calculators have a button for this purpose.
More generally
y = ex
é
x = lnÝ y Þ .
(*)
i.e.,
if fÝ x Þ = e x then f ?1 Ý x Þ = lnÝ x Þ .
Since e x is always positive, we cannot solve the equation y = e x when y is either negative or zero—the range
of e x is 0,K . From the discussion on page 3 of Chapter 10 we have the following table.
Function
x
y=e ,
y = lnÝ x Þ ,
Domain Range
Ý ?K,K Þ
0,K
0,K
Ý ?K,K Þ
We deduce from equation (*) that
y = e lnÝy Þ
ln e
x
for any positive number y
= x for all x 5 R.
These are the results of f f ?1 Ý y Þ
= y and f ?1 fÝ x Þ = x. Note that because e x > 0 for all x, it follows that
lnÝ y Þ is not defined for negative values of y. So ln ?2 , for example, cannot be evaluated without the use
of complex numbers.
Since e 0 = 1 we deduce that
=
ln 1
1
and, since e = e,
lnÝ e Þ =
6
4
2
-6
-4
-2
0
2
x
4
6
-2
-4
-6
Graph of e x (dashed) and lnÝ x Þ (solid)
More generally, if y = b x (where b is a postive base), then x = logb Ý x Þ —“log to base b of x”. We shall deal
almost exclusively with natural logs (loge Ý x Þ , which is written lnÝ x Þ ). The natural log “ ln” could be changed to log
of any base in most of the examples.
The Laws of Logarithms
Important rules for working with logarithms (these are valid for all logs, regardless of the base, so we write
log rather than log10 ).
Suppose that p and q are positive real numbers and that r is any real number. Then we have the following
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rules for manipulating logs.
Theorem 13.1
a)
logÝpqÞ = logÝpÞ + logÝqÞ (the log of a product is the sum of the logs)
b)
logÝ pq Þ = logÝpÞ ? logÝqÞ (the log of a quotient is the difference of the logs)
c)
log p r = r logÝpÞ (the log of a power is the exponent times the log).
So, for example
ln 5
+ lnÝ x Þ = ln 5x
ln 5
? lnÝ x Þ = ln 5x
3ln 5
= ln 5 3
= ln 125 .
Proof Let y = pq then, since e lnÝy Þ = y,
e lnÝpq Þ =
However, p = e lnÝp Þ and q = e lnÝq Þ and so
(1)
pq =
where we have used e x 6 e t = e x+t . But this value of pq must coincide with that in equation (1) and so
e lnÝpq Þ =
The first rule then follows because e x is one–to–one: this means that when e x = e t then we must have x = t.
The second rule can be deduced from the first: let x = pq , so we need to find lnÝ x Þ . However, p = xq and so,
by the first rule
lnÝ p Þ =
which is rearranged to give rule b).
For rule c) we again argue from two directions, as for rule a). First, with y = p r ,
r
e lnÝp Þ=
but, since p = e lnÝp Þ ,
(2)
r
pr =
x
=
r
where we have used e
= e rx . But this value of p r must agree with that in equation (2) so, since e x is a
one–to–one function, rule c) follows.n
Note: there is no rule for expanding lnÝ p + q Þ .
Exercise: Prove that lnÝ pqr Þ = lnÝ p Þ + lnÝ q Þ + lnÝ r Þ for any three positive real numbers p, q, r.
Example 1. In Example 1 of Chapter 12 we saw that the number of doves at time t was 1000Ý1.1Þ t .
Suppose we need to know the time at which the population reaches 3700. For this we need to solve the
equation
3700 = 1000Ý1.1Þ t
ì
Ý1.1Þ t = 3.7
to find the value of t. The answer is to take logs of both sides:
ln 1.1 t
= ln 3.7
and to use rule c):
from which we find
t=
Example 2. If the money in our account grows at 2% per year and we began with £5000 the amount we have
at year n will be gÝ n Þ = 5000 1.02 n .(See Example 2 of Chapter 12). The money will double when
5000 1.02
n
=
i.e.,
1.02
n
=
whose solution is obtained by taking logs of both sides: ln 1.02 n
= ln 2 , so
n=
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NB With a growth rate of 2%, it always takes the same number of years to double, regardless of starting amount.
Example 3. From Example 3 of Chapter 12 the number of dogs in the population at year t was
p t = 0.955 t 1500. The population will have halved when
i.e.,
0.955
t
t
=
.955
1500 =
1.
2
Taking logs of both sides then gives
t=
NB With a decay rate of 4.5%, it always takes the same number of years to halve, regardless of starting amount. This time is
known as the “ half–life” of the population.
Example 4. On page 3 of Chapter 12 we saw that 10 u 2 3.3219... , where did this exponent come from?
Solving 10 = 2 x by first taking logs of both sides gives
ln 10
=
x=
and, with a calculator,
ln 10
ln 2
= 3. 321928095... .
Example 5. Solve 2 x 6 3 2?x = 5.
We take logs of both sides to obtain, with the aid of rule a),
We now apply rule c) twice on the left to give
Collecting terms in x on the left and moving the remaining ones to the right gives
x
Example 6. Solve 2 x = 5 4 x .
3
We take logs of both sides and use rules a) and b) on the right, then use rule c) on both sides
xln 2
=
Collecting terms in x on the left and moving the remaining ones to the right gives
Example 7. Solve 2 e 6 e x = e 2 6 2 x .
We take logs of both sides and use rule a) then use rule c)
where we have used lnÝ e Þ = 1.Collecting terms in x on the left and moving the remaining ones to the right
gives
Example 8. Solve 2a x = b 2x .
Example 9. Solve the equation e x ? e ?x = 4.
x
?x
¾ This time we let y = e then, since e
= 1x , the equation becomes
e
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from which we find
y=
Then, since e x = y, we have, on taking natural logs, x = lny. However, 2 ? 5 < 0 so there is no real value for
x corresponding to y = 2 ? 5. Hence, the only solution is
x = ln 2 + 5 .
y
= 1 ? 2x for y.
1?y
Using the fact that e p = e q when p = q, we “take the exponential” of each side to give (we use the notation
exp Ý x Þ instead of e x because the exponent is cumbersome)
y
exp ln
= e 1?2x
1?y
and, since exp lnÝ p Þ = p, this simplifies to
Example 10. Solve ln
Next multiply both sides by the denominator then collect all y terms on the left:
Then
y=
3
Example 11. Simplify lnÝ pq Þ ? 2lnÝ p Þ + ln qp
, where p and q are positive quantities.
By rule c)
2lnÝ p Þ =
then, with rule b)
lnÝ pq Þ ?
then, with rule a)
3
lnÝ pq Þ ? 2lnÝ p Þ + ln qp
Example 12. Simplify 2lnÝ xy Þ ?
=
1
2
ln x 2 y
+ 4ln
x
where x and y are positive real numbers.
2lnÝ xy Þ =
1
2
ln x 2 y
4ln
x
=
=
and so
2lnÝ xy Þ ?
1
2
ln x 2 y
+ 4ln
x
=
Example 13. Solve lnx 2 + lnx = ln27, x 5
0,K
2
lnx + lnx =
and so the equation becomes
and we have
x=
since 27 = 3 3 .
Example 14. Solve the equation lnÝx ? 2Þ + lnÝ2x + 4Þ = ?2.
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First note that the maximal domain of lnÝ x Þ is x 5
interval 2,K .
0,K
so any solution to this equation must lie in the
Now combine the two terms on the left using the Rule: the sum of logs is the log of the product:
ln
Take the exponential of both sides
exp ln
so, since exp lnÝ y Þ
= y, we obtain the quadratic equation
which simplifies to
x2 =
There appear to be two solutions of this equation
x=±
However, as mentioned at the start of this solution, the natural logarithm is only defined for positive values of
its argument. Both x ? 2 and 2x + 4 must be positive in the original equation,
lnÝx ? 2Þ + lnÝ2x + 4Þ = ?2, so we have to discard the negative solution because Ý?2.01685 ? 2Þ < 0.
The only valid solution is x = 4 + 12 e ?2 u 2.02. This is confirmed by drawing the graph of the functions
y = lnÝx ? 2Þ + lnÝ2x + 4Þ and y = ?2 (shown below), since they cross just to the right of x = 2.
4
2
-2
2
x
4
6
0
-2
-4
-6
-8
-10
Example 15. Does the equation
lnÝx ? 2Þ ? lnÝx + 1Þ = 5
have any real solutions?
Again, note that any solution must lie in the interval 2,K . Next , combining the log terms using the rule: the
difference of logs is the log of the quotient:
=
ln
¾
and taking exponentials of both sides (we use the notation exp Ý y Þ instead of e y because the
exponent is cumbersome)
exp
= e5 .
ln
Using inverse relation exp lnÝ x Þ
= x gives
= e5
and so
x=
This would make x ? 2 = and
x+1 =
However, the natural log is only defined for positive values of its argument, so the conclusion is that there
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are no real solutions of the equation. The graphs of the functions y = lnÝx ? 2Þ ? lnÝx + 1Þ and y = 5 are
shown below and do not cross.
4
2
-2
-1
1
2
x
3
4
5
0
-2
-4
-6
-8
-10
One should be wary when using Scientific Notebook to plot logarithmic functions. When asked to plot the
function lnÝx ? 2Þ ? lnÝx + 1Þ it first simplifies using the rule of logs and actually plots the function ln x?2
x+1
which is shown below. It is evident that the equation ln x?2
does have a real solution. What is it?
x+1
20
10
-4
0
-2
2
4
x
-10
-20
The domain of fÝ x Þ = lnÝx ? 2Þ ? lnÝx + 1Þ is x 5
?K,?1 W 2,K .
2,K
while the domain of ln
x?2
x+1
is
The difficulties involved with logs of negative numbers are often avoided because the argument is enclosed in
| 6 |.
Example 16. Prove that ln|x | is an even function.
With fÝ x Þ = ln|x | and bearing in mind the definition of the modulus (or absolute value) function
|x | =
x if x ³ 0
?x if x < 0
which means that |x | is even: |?x | = |x |, we find
fÝ ?x Þ =
and so ln|x | is an even function.
4
2
-4
-2
0
2
x
4
-2
-4
Graph of ln|x |,
Also, ln|x | ¸ ?K as x ¸ 0 so ln 0
x 5 R\ 0 .
is undefined and the maximal domain of ln|x | is x 5 R\ 0 .
Example 17. Find all real solutions of the equation
ln 3x ? 2 ? ln x + 2 = 3.
The presence of the absolute values means that the expression ln 3x ? 2 ? ln x + 2 is defined for all
x 5 R\ 2/3,?2 : we have to exclude x = ?2 and x = 2/3 since both values lead to ln 0 which is
undefined. Since
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ln 3x ? 2 ? ln x + 2 = ln
we have so solve
= e3 .
There are two possibilities,
= e3
which leads to
u
x=
or
= ?e 3
which leads to
u
x=
That there are two solutions is confirmed by the graphs of y = ln 3x ? 2 ? ln x + 2 and y = 3.
10
8
6
4
2
-4
-3
-2
x
-1
0
1
2
-2
-4
-6
-8
-10
The course text : Calculus by Anton , Section 4.2 has further exercises & examples.
Common Logarithms
Logarithms were invented by John Napier () of Edinburgh but it was Henry Briggs (1561–1630) who was
responsible for publicising them and ensuring that they became widely used in scientific calculations. It was
on his suggestion that Napier adjusted his definitions so that he used a base b = 10. That is,
the inverse function of
fÝ x Þ = 10 x is f ?1 Ý x Þ = log10 Ý x Þ
y = 10 x é x = log10 Ý y Þ .
These are now called common logs and were an extremely powerful tool for multiplication and division
before electronic calculators became available. This is illustrated in the next example.
Example Because of the laws of logarthims given earlier, common logarithms were used to speed up
arithmetic calculations. For example, to multiply 1234.56 × 34.567
log10 Ý1234.56 × 34.567Þ = log10 Ý1234.56Þ + log10 Ý34.567Þ
= 3.0915 + 1.5387 = 4.6302
4. 630 2
and so 1234.56 × 34.567 = 10
= 42677. 6, (which could be found by looking up a table of values of the
inverse function). Logarithms convert a problem of multiplication into the easier tasks of looking up tables
and adding:
look up the logs of the two numbers in a book of tables, add the results together, then look up the inverse
(called “anti-logs”).
Similarly, division is reduced to subtraction.
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Common logs are especially important when measuring quantities that vary greatly in magnitude, since the
common log of the numbers 10 ?6 , 10 ?3 ,1, 10 3 ,10 6 , for example, give the modestly sized numbers ?6, ?3, 0,
3 and 6. Practical examples of measuring units that use common logs are given on page 8.
Example Moore’s Law.
Below is a table of the speeds of the fastest supercomputer of the time—taken from SIAM News, Nov. 2001.
The speed is given in “flops”: floating point operations per second which, loosely speaking, is the time it
takes to perform a calculation such as “3x + 5” (i.e., one multiplication and one addition).
Year
Speed log10 speed
1950
7.0×10 2
2.84
1951
9.0×10 2
2.95
1960
6.0×10
4
4.78
1.0×10
6
6.00
6.0×10
6
6.78
6.0×10
6
6.78
1.0×10
8
8.00
1.0×10
9
9.00
2.0×10
9
9.30
1987 1.0×10
10
10.0
1991 1.3×10
11
11.1
1992 1.3×10
11
11.1
1998 1.3×10
12
12.1
2000 3.0×10
12
12.4
1965
1970
1971
1976
1982
1985
It can be seen that the speed varies enormously, from 700 flops to 3 × 10 11 flops. The common log of the
speed however, varies only from 2.8 to 12.4. This shows up more dramatically when we plot this data. The
top graph above plots speed against time and we can see that it has almost no information content. The
speeds at earlier years are too small to register on the scale.
What we then do is to plot log10 speed against time. This is shown in the bottom graph and, more clearly,
on page 11 where the names of the computers are also given. The data points lie pretty well on a straight line
and, with a little effort, one can find that the equation of the straight line through the data is approximately
log10 speed
where x = year ?1950. Since 10 log 10
speed
u 2. 78 + 0.2x
= speed, we find speedu 10 2. 78+0.2x u 602. × 10 0.2x . That is,
speed u 600 × 10
year ?1950 /5
Þ
To get a better grip on this equation, we observe that number of years it takes for the speed to double is t,
where
10 t/5 = 2
i.e.,
t = 5log10 2
u 1. 51.
This is Moore’s Law: Computer speed doubles roughly every 18months.
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Practical Applications
Logarithms still have important uses now as the following example show.
DECIBEL (dB)
(excerpt from Audio Dictionary by Glenn While)
Literally, one tenth of a bel. The bel is named after Alexander Graham Bell, and the number of bels is defined
as the common logarithm of the ratio of two powers (as in electrical power).
Signal Intensity
Sound level in dB = 10log10
Reference Intensity
A rule of thumb that is useful to remember is that 10 dB is a power ratio of 10. Any time a power is increased
tenfold, it is increased by 10 dB; thus, a 200W power amplifier will put out 10 dB more electrical power than
a 20W amplifier, and its sound power output will also increase by 10 dB.
pH value in Chemistry:
pH = ? log10 ßH +à = negative log of hydrogen ion concentration
pH = 7 is neutral, pH < 7 is acid (Smaller number, more acidic) and pH > 7 is alkaline (Larger number, more
alkaline.)
Earthquakes and the Richter Scale
Seismologists use a Magnitude scale to express the seismic energy released by each earthquake. Here are the
typical effects of earthquakes in various magnitude ranges:
Earthquake Severity
(Richter scale)
Effect
< 3.5
Generally not felt, but recorded.
3.5 to 5.4
Often felt, but rarely causes damage.
Under 6.0
At most slight damage to well-designed buildings.
6.1 to 6.9
Can be destructive in areas up to about 100 kilometers
7.0 to 7.9
Major earthquake. Can cause serious damage over larger areas.
>8
Great earthquake. Can cause serious damage in areas
several hundred kilometers across.
If the amplitude of an earthquake is Amm (as measured by a Wood-Anderson seismograph), then
Richter Magnitude = log10 ÝAÞ + ÝDistance Correction FactorÞ
where the distance factor comes from the distance of the seismometer from the earthquake and is tabulated in
Richter’s (1958) book “Elementary Seismology ”.
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