Recall the
exponential function
y = bx.
The graph of this function for b > 1 passes through the Y axis at y = 1, is asymptotic to
the negative X axis, and is always increasing. And for 0 < b < 1 it passes through the
Y axis at y = 1, is asymptotic to the positive X axis, and is always decreasing.
The domain for y in either case is y > 0. No exponent exists that can make y ≤ 0.
The exponential function y = bx has the inverse function: x = logb(y).
The inverse function for the exponential function is called the logarithm function.
The exponent x in the exponential function is called the
logarithm base b of y in the logarithm funciton.
For the base, b, the logarithm function maps all
positive numbers to the exponents of this base so that when b is raised to these exponents
the results are the positive numbers that are mapped.
Example, say 10 is the base of the
exponential. Then the logarithm function base 10 maps the numbers
10, 100, 1000, 0.1 to the exponents 1, 2, 3, and -1, because 101
= 10, 102 = 100, 103 = 1000 and
10-1 = 0.1.
As another example, say 2 is the base
of the exponential. Then the logarithm function base 2 maps 2, 16,
64, 0.25 to 1, 4, 6, and -2 respectively since
21 =
2, 24 = 16, 26 = 64 and 2-2
= 0.25.
We usually shorten logarithm
to log.
Definition:
logb(x) = y iff
by = x
b
is the base.
y is the power of b that gives the number x.
For now, we restrict b and x to be
positive Real numbers; y can take on all values.
In the examples above :
log1010 = 1
log101000 = 3 log100.1 = -1
and
log22 = 1
log264 = 6 log20.25 = -2
Simply put, a logarithm is an exponent.
There are two primary motivations for
logarithms.
The first and oldest is: they
make multiplication of large numbers easier
to do since they turn the operation of multiplication into addition.
This was important prior to the calculator. In those days
tables of logarithms were used to do this multiplication. Later the
slide rule which used logarithmic scales was used to do these
multiplications. I was one of those who used a slide rule in High
School for math, chemistry, physics, etc., and I was also one of
those who was elated with my first Texas Instruments TI55 calculator. It was
mesmerizing. The glow of the red led display was transfixing for
sure! The decimal places it could carry in its calculations! (More
on the table and the slide rule will be shown later.) As a side
note, the first computer, the ENIAC, spent most of its time
calculating logarithms to determine trajectories of projectiles.
The second and most used today is:
logarithmic scales allow
plotting data that spans powers of a base rather than multiples of
the base. For example noise levels are measured in powers of 10. A
linear scale could handle from 1 to 100, but not 1 to 1000
simultaneously with the same accuracy. (More on this later.)
Examples, find the missing parameter
b, x, or y:
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1) log5(25) = y
5y = 25
y = 2
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Rewrite this equation into its exponential form.
By inspection 5 squared equals 25, so y = 2.
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2) logb(100) = 2
b2 = 100
b = 10
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Again, rewrite this equation into its exponential form.
By inspection, the only number squared that equals 100 is 10.
So b must be 10.
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3) log10x = -2
10-2 = x
x = 0.01
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Rewrite this equation into its exponential form.
10-2 = 0.01
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4) log10(10) = y
10y = 10
y = 1
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Rewrite this equation into its exponential form.
By inspection, 101 = 10.
In general,
logb(b) = 1.
We'll call this the Identity Rule for logarithms.
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5) ln(ea) = y
ea = ey
a = y
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Rewrite the equation into its exponential form.
By inspection y must equal a.
In general,
logb(ba)
= a
We'll call this the Identity Power Rule for logarithms.
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6) eln(a) = y
ln(y) = ln(a)
a = y
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Rewrite the equation into logarithmic form.
ln(y) is the exponent of e for the number represented by
eln(a) .
The logarithm function is a one to one function so the
arguments of the logarithms must be equal.
In general,
We'll call this the Log Power Rule for logarithms
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As you can see with all of these
example we used equivalent exponential or logarithmic equations to solve for the
unknown.
Base 10 and base e are the two most common logarithm bases.
We will use special notation:
log(x)
for log10(x)
ln(x) for loge(x).
log(x) is spoken “the common
logarithm of x” and
ln(x) is spoken “the natural
logarithm of x” and
logb(x)
is spoken “the log base b of x.”
BE CAREFUL HERE: Many texts will
simply use log(x) to mean the base e natural logarithm and specify
other bases b, by naming them, even base 10. Usually the base is assumed
to be understood by the context of the material involved. I will not use that convention here
since ln(x) and log(x) are more commonly used and the intended base is made clear
by the notation itself.
Graphs of ln(x) and log(x)
The graphs are asymptotic to the Y
axis, they cross at x=1 and are always increasing for positive x.
Notice log(x) is flatter than ln(x). For two exponentials of different
bases the larger base requires a smaller
exponent while the smaller base requires a larger exponent for both
exponentials to equal the same number.
From the graph we read the common log of 7
is about 0.8. The natural log of 7 is about 2.0. So as the base of
the logarithm increases its graph lies closer to the X axis. Note,
however, that the graph never becomes parallel to the X axis, it
steadily increases.
Both graphs cross at x = 1. This is because every nonzero base
raised to the exponent 0 equals 1. At x = 10, log(x) = 1 and at
x = 2.74, ln(x) = 1. This makes sense since any base with the exponent 1 is that base.
Next is a closer look at the interval
0< x < 1.
Here the larger base 10 requires a
smaller exponent (in magnitude) for a given number than the smaller
base does. For example, for x = 0.4, 10 needs the power
-0.4 while
2.72 requires -0.9. Comparing the magnitudes these exponents we see
that 10 has smaller exponent. As before, as the base of the
logarithm increases the graph of the logarithm flattens toward the X
axis.
Notice how the graphs become asymptotic to the negative Y axis
rather quickly. A very small change in x near zero results in a very large
change in y. This means from a practical point of
view a large number of significant digits in the logarithm are required to
distinguish one x value form another from another in this range.
These two graphs cannot cross the Y axis. There is no exponent of a base that will
cause that exponential to be zero.
Comments on Notation:
Grouping symbols aren't necessary for
the exponent in an exponential. For example, ex+1
has the exponent x + 1. However, for lnx+1 the argument for ln is
x, NOT x + 1. What we have is: add 1 to the lnx. Parenthesis
must be used for logarithms. We will write ln(x+1).
Now what does logAB mean? The best
answer is to be consistent; use parenthesis. If we mean the
common log of the product AB then remove all doubt and write log(AB). “logAB” is ambiguous, however you will see it used to
mean the same as log(AB) elsewhere. Here I will be explicit by using parenthesis. Use parenthesis.
These parenthesis are not grouping symbols. We know that A·(x+2) equals (Ax + 2A).
The logarithm function is a function not an algebraic expression.
As in any other function, f(x), A · f(x) does not equal f(Ax).
Likewise, A · log(x) does not equal log(Ax).
It's a good idea to review the product
rule for exponents at this time.
Let's have a closer look at
logarithms and how they can be used to manipulate and solve
equations.
Product Rule for Logarithms:
Motivation: when you multiply
exponentials with the same base you add their exponents. Since
logarithms are exponents of the exponentials that equal these numbers, when these numbers are
multiplied you add their logarithms to find the logarithm of their
product.
Here's why.
Start with the numbers L and M.
Let bf
= L and bg = M
Now multiply L and M: LM = bf⋅ bg
and bf⋅ bg = bf + g
(product rule for exponentials)
so: LM = bf + g
In logarithmic form: logb(LM) = f + g
Back to the first two equations: bf = L and bg
= M.
In logarithmic form they become: logb(L) = f and logb(M)
= g.
Substituting these back in for f and g into logb(LM) = f + g, we get
the Product Rule for Logarithms:
logb(LM) = logb(L) + logb(M)
This is the rule that turns multiplication into addition.
Examples:
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7) log(1000)
= log(10 ·
10 · 10)
= log(10) + log(10) + log(10)
= 1+1+1 = 3
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Write 1000 into factors of log base b=10.
Use identity rule (example 4.)
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8) ln(abc) =
ln(a) + ln(b) + ln(c)
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Direct application of the product rule.
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9) ln(3) + ln(4) + ln(5)
= ln(3 ·
4 · 5)
= ln(30)
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Direct application of the product rule in reverse.
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Notice how the multiplication signs exchanged places with the
plus signs.
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10) log(3x + 5)
The product rule does NOT apply.
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CAREFUL HERE
The argument must be a single term comprised of
factors. We have a sum of two terms. Don't let that first
term fool you. Yes, it is a product of the two factors 3 and x,
but the overall argument of the logarithm is a SUM not a
PRODUCT.
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11) log(3x) + 5
log(3) + log(x) + 5
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Compare this example to the
previous example. 5 is not part of the argument of the logarithm.
3x is now the single product with factors 3 and x and the product
rule for logarithms can now be applied.
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12) log(x + y)
The product rule does NOT apply.
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Same as example 10.
Logarithms with arguments that involve multiple terms can
not be simplified.
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13) log(x3)
= log(x · x
· x)
= log(x) + log(x) + log(x)
= 3log(x)
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Write the exponential as a repeated
product.
Apply the product rule.
In general,
logb(a)k = k·logb(a)
We'll call this the Power Rule
for Logarithms.
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14) log(x2y2)
= log((xy)2)
= 2 log(xy)
= 2( log(x) + log(y) )
= 2log(x) + 2log(y)
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Use the exponent rule for the
distribution of a power over a product of exponential factors.
Then use the Power Rule.
Split using the product rule of
logarithms and simplify.
Simplify.
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15) log(w2z3)
= log(w2) + log(z3)
= 2log(w) + 3log(z)
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This time use the product rule for
logarithms first since the exponents are unequal.
Use the Power Rule.
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16) log(x2 + y2)
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No rule applies. The argument of
the logarithm must be a single term.
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17) log(x2 −
y2)
= log( (x + y)(x −
y) )
= log(x + y) + log(x −
y)
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This looks like example 16 but we
can factor this argument into a single term.
Now apply the product rule for
logarithms.
Study this example. Make sure you
see the difference between example 16 and this example.
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18)
= log(xy-1)
= log(x) + log(y-1)
= log(x) + (-1)log(y)
= log(x) − log(y)
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Write
the ratio as a product.
Use the Product Rule.
Use the Power Rule.
Simplify.
In
general for y ≠ 0:
We'll call
this the Quotient Rule for
Logarithms.
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19)
= log(1) − log(x)
= 0 −
log(x)
= − log(x)
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Use the Quotient Rule for
Logarithms.
Regardless of the base the exponent
0 is the logarithm of 1.
In general:
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20)
= log(x) −
log(3)
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Directly apply the Quotient Rule.
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21)
log(x + 1) −
log(y)
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Directly apply the Quotient Rule.
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22)
log(x) −
log(y + 1)
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Directly apply the Quotient Rule.
The minus sign DOES NOT
distribute across the argument of the logarithm.
You cannot write
−log(y + 1)
as
log( −y − 1)
nor, even worse, as
− log(y) − log(1)!
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23)
= log(xz) −
log(y)
= log(x) + log(z) −
log(y)
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Directly apply the Quotient Rule.
Now use the product rule for
logarithms.
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24)
= log(x) −
log(yz)
= log(x) − (
log(y) + log(z) )
= log(x) −
log(y) − log(z)
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Directly apply the Quotient Rule.
Now apply the product rule being
careful with the minus sign. Use parenthesis so you do not make a
mistake.
Clear the parenthesis. Watch the
minus sign!
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Do you see a pattern here? All factors in the numerator result
in positive signs and all factors in the denominator have negative
signs.
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25) Write as a single logarithm:
log(x) − log(y)
+ log(w) − log(z)
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The positive logarithms involve x
and w so they are the factors in the numerator. The negative
logarithms involve y and z so they are the factors in the
denominator.
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The Power Rule for
Logarithms (example 13) leads directly to the Change of Base Rule
for Logarithms.
Change of Base Rule for Logarithms
Given two bases b
and c then
Here's why:
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Start with logarithm base b, we're looking for the equivalent
logarithm base c.
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Rewrite equation into exponential form so we can use the Power
Rule for Logarithms with the base c.
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The logarithm base c of both sides of the equation must be
equal as well.
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Now use the Power Rule.
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Solve for K. (We started with K defined in the first equation
as the logarithm base b of x.)
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Use the first equation to rewrite K with logb(x).
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Examples:
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26) base 10 to base 5.
log(x) = log5(x)/log5(10)
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Direct application of the Change of Base Rule.
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27) base e to base 10
ln(x) = log(x)/log(e)
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Direct application of the Change of Base Rule.
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28) base 10 to base e
log(x) = ln(x)/ln(10)
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Direct application of the Change of Base Rule.
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29) log2(16)
log2(16) = log(16)/log(2)
1.20412/0.30103 = 4
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Direct substitution into the Change of Base Rule.
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30) log3(30)
log3(30) = log(30)/log(3)
1.4771/0.4771 = 3.0959
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Direct substitution into the Change of Base Rule.
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The utility of the Change of Base Rule is to find the exponent for some number
whose base is anything other than ten. A significant amount of effort has
been spent over the years to create tables of logarithms to many decimal
points of precision to improve the accuracy of calculations.
It was significantly faster for early computers to use look-up tables of
logarithms to perform multiplication.
We're almost there.
Recall in our
study of exponential equations I stated that the
exponentials must have the same base in order to solve the equations.
Now we can remove that restriction by “taking the logarithm
of both sides of the equation” to solve for x.
Example:
solve for x
52x
= 6x-1
Take
the log of both sides.
log(
52x ) = log( 6x-1
)
It
doesn't matter which logarithm base you use. Since The natural and
common logarithms are readily available on a calculator, use one or the other.
Use
the power rule for logarithms.
2x
log(5) = (x – 1) log(6)
Now
solve this linear equation for x.
2x
log(5) = x log(6) – log(6)
First
move the log(6) x term from the right to the left to isolate the x
terms.
2x
log(5) – x log(6) = – log(6)
Now
factor out x from these two terms on the left.
x
( 2log(5) – log(6) ) = –
log(6)
Divide
the equation by the coefficient of x.
Use a
calculator now. You could do this with a log table as well. In my
day, the day without a calculator, I had to memorize the common
logarithms of 2, 3, 5, 7 and that was all that was needed. A slide
rule also helped. We'll talk about this later. Anyway, carefully
using a calculator the result is:
x
= -1.26
If
you substitute this value back in for x to check you'll find 0.017 =
0.017.
(Use
the yx or the xy
key.)
It is
most important to check your result! The calculation for x can be
quite error prone.
Knowing
the logarithms and using them in this case we would have:
-
(3010+4771) /( 2(6989) – (3010+4771))
(Decimal
numbers are unnecessary, we kept track of them in our head.)
Now
if I want an approximate answer I'd round to the first digit to make
the math simple. (or simpler)
-
(8000) / (14000 - 8000) = -8000 /(6000) -4/3 = - 1.33
Compare
the answer: 1.33 – 1.26 it is 0.07 off. Not bad for an
estimate.
Another
example:
solve
for x:
42x
= 7x-3
ln(
42x ) = ln( 7 x-3
)
2x
ln(4) = (x-3) ln(7)
2x
ln(4) = x ln(7) – 3 ln(7)
2x
ln(4) - x ln(7) = -3ln(7)
x
(2ln(4) - ln(7)) = -3ln(7)
x
= -7.06
Checking
this value you'll find 3x10-9 = 3x10-9
Round
off in my calculations will be seen in your results with further
digits differing
If you go back and
take a look at the graph for the interval 0 to 1 you will see that y = -7
is where all the graphs for all logarithm bases merge. This
means that quite a few significant digits in the calculation would be
required to distinguish one point on the curves from another. And this is why
the digits beyond the 3 in this example differ.
Table of Rules
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Identity Rule
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logb(b) = 1
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Identity Power Rule
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logb(ba)
= a
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Log Power Rule
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Product Rule
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logb(LM)
= logbL + logbM
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Power Rule
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logb(ak)
= k ·logb(a)
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Quotient Rule
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y ≠
0
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Reciprocal Rule
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Change of Base Rule
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