>> Hello, this is Professor Jen-Mei Chang
at California State University Long Beach.
In this video lesson, we're going to look
at laws of logarithms. We're going assume
that a is a number that is greater than zero
and not equal to 1. Also assume that capital
A, B, and C are real numbers with A and B
strictly greater than zero. Then we have our
first rule which says that the logarithm of
base a of capital A times capital B is equal
to the log of base A, capital A, plus the
log of base A, capital B. This is to say that
the log of a product is equal to the sum of
2 logs. I'm going to prove that this rule
is true by first letting X equals the log
of base A, capital A. And let Y be the log
of base A, capital B. And by the rules of
the logarithm, I can rewrite that first equation
into little A raised to the X equals capital
A. Similarly, I can rewrite the second equation
here into little A raised to Y is equal to
capital B. This means now that I have a new
way to write log of base A, A times B. I can
now write it as log base A of A raised to
the X plus Y. But also we know this from the
rule of the log, this turns into just simply
X plus Y. But remember X is log base A, A
and Y is log base A, B. So we basically have
what we needed. A log of a product is equal
to the sum of the logs. Mathematicians likes
to use a little square at the end of a proof
to indicate that the proof is now finished.
The second rule is similar. It says that a
log of a quotient, A over B, is the same as
the log of A minus the log of B. We're assuming
that the send base applies to everybody. And
notice that this has to be the same base in
every single expression. This is also because
we have to assume that the base A is greater
than zero, and it's not equal to 1. Because
we're translating all of that logarithm language
into the exponential language, and we require
that the base in exponential function to be
nonzero, not equal to 1, and strictly greater
than zero. Again, I'll prove this quickly
because we know that capital A can be written
as capital A over B times B. I'm going to
use the previous rule in the following deduction.
I can write log A as log A over B times B.
That's just what we have from this statement.
For right now, use the rule number 1 on this
expression here. I would then obtain log A
over B plus log B. We now simply subtract
log B both sides. We have our desired expression
for the log A over B. And that is the end
of proof for this one here. We have another
rule on the next slide which says a log of
A raised to a power C of base A is equal to
or equivalent to C times log A with the same
base, little A. This rule is very useful because
that allows you simply bring that power down
in the front to make it into a simpler expression.
To prove that, I'm going to let log base A,
capital A, equal to this expression called
U. Then rewriting this expression, I get A
to the U is equal to capital A. This allows
us to write the left hand side of the rule
so that I have log A to the C now is equal
to log A to the U and the entire expression
raised to the power C. But I also know that
I can simplify this by multiplying the U and
the C together to get A to the C, U. Now I
can simplify this expression simply into C
times U because of the rule of the logarithm.
Remember U is logarithm of A. And we have
the rule as desired. There are 4 types of
problems that one can ask you about logarithms.
First let's start with simple evaluations
using the rules. For example, evaluate log
base 4, 2 plus log base 4, 32. If you were
to evaluate this directly, you have to use
a calculator because to answer the question
of 4 raised to some power equals 2 is not
so easy. And similarly to answer the question
about 4 raised to a power equals just 32 is
probably not going to be as easy as well.
However, you can actually combine them by
using the rules. The rule number 1 says that
I can combine the sum of 2 logs into a single
log of the product. So I can write this as
log of base 4 of 2 times 32. This turns into
a log of base 4 of 64. And then I ask myself
the question of 4 raised to what power is
equal to 64, and that's simple, it's just
3. Or you can think of 64 as 4 raised to the
third. So they used the rule number 3, I can
write this as base 4 and then 4 raised to
the power of 3. And as I can bring down this
power of 3 and write this as a log base 4
and 4. But log base 4, 4 is simply just equal
to 1. This, I get 3 times 1 which is equal
to 3. I do this, I don't really have to use
a calculator at all. Let's look at another
example next. For the same reason, similarly,
if you have to answer question of 2 raised
to something is equal to 80 and that's not
so easy. But I can now combine these 2 logs
by using rule number 2 because the difference
of 2 logs can be combined into a log of the
quotient. So I can write this as log of base
2, 80 divided by 5. And that is equal to the
log of base 2, 16. But then I can write the
16 as 2 raised to the fourth power and by
bringing the 4 down, I have log of base 2,
2. But that is, again, equal to 1. So I have
all together 4. One last example here, I have
an 8 to 1/3 log base 10, 8. Remember that
if I don't put anything down here for the
base, it's assumed it's the common log which
is a base 10. This one I can write it as a
negative 1/3 log of 2 to the 3. And I can
bring this 3 down to the front, combine that
with the negative 1/3. So I have negative
the log of base 10, 2. Now for log of 10,
2, it's really not something that I know.
So I would have to use a calculator to get
an estimate for this. Or you can think of
this as the log of 2 raised to the negative
1 which can be thought of as a log of 1/2.
You actually put this into your calculator
to get an estimate which is about negative
0.301. The next type of question I can ask
you about log is to expand. Starting from
a single expression in the log, I wanted to
rewrite it using the rules into multiple terms.
For example, I have a natural log of the quantity
A times B divided by cube root of C. I can
first using the second rule by writing this
as difference. So I have natural log of A,
B minus the natural log of cube root of C.
Then I use the first rule to write the product
of natural log into the sum of the 2 logs.
So I have natural log of A plus natural log
of B. I will then use the third rule to write
the last 1 here by realizing that Q root of
C is really C raised to the 1/3 power. This
last expression becomes 1/3 natural log of
C. Everything else stays the same. And that
will be my final answer when I expanded everything.
Another example would be a log base 5, quantity
X times Y to the sixth power. A lot of
the times, I would suggest you use parentheses
to indicate what's multiplying by what. In
this case, all of this together is the input
and the natural log function. If you don't
use the parentheses around it, someone might
interpret this as log of 5 X and then multiply
this entire thing times Y to the sixth. So
wherever needed to avoid confusion, I would
recommend using parentheses. To expand this,
I will use the rule number 1 to write this
into a sum. So I have log base 5 of X plus
log of base 5 Y to the sixth power. And then
use the rule number 3 on the second expression
only to write it into 6 times log of base
5 Y. The next type of example is to go the
other way given some expanded expression I
wanted to simplify into a single logarithm
expression. For example, combined 3 log X
plus 1/2 log of X plus 1 into a single logarithm.
If we compare the operations here with what
we did before, notice that before we applied
the rule number 3 last. Here, we're actually
going to apply rule number 3 first because
we're basically doing the opposite operations
here. So rule number 3 allows us to rewrite
that first expression into a log of X cubed.
Rule number 3 allows us to write the second
expression into log of X plus 1 raised 1/2.
And then we applied the rule number 1 which
turns the sum of logs into a log of a single
product. So I have log of X cubed times X
plus 1 to the 1/2. Now you can definitely
turn this 1/2 into a square root if you want,
but you don't have to. So you can write this
as X cubed times the square root of X plus
1. Again, this is assuming of a common log
of base 10. Let's look at another example.
Here we have 3 terms. Each terms has a coefficient
in front of them. Okay, we going to apply
rule number 3 again first into natural log
of S cubed plus natural log of T to the 1/2,
minus natural log of the quantity T squared
plus 1 raised to the fourth power. And when
I combine terms, I do 2 at a time. So if I
use rule number 1 applied to the first 2 terms,
I will get natural log of S cubed times T
to the 1/2. And then I have the leftover of
natural log of T squared plus 1 raised to
the fourth power. Here those 2 terms are separated
by a subtraction sign which means I'm going
to apply rule number 2 to combine them into
a fraction. So I have natural log of S cubed
times T to the 1/2. But you can actually write
that as a square root of T if you want and
divide it by the quantity T squared plus 1,
the whole thing raised to the fourth power.
This is now a single logarithm. You may wonder
when you do what. A lot of the times, we want
to simplify our problem by solving the variable
involved. We wanted to start with some complicated
combinations of logs and it just turned into
a single log that allows us to solve problem
easily. So this type of operation is actually
more useful than the other ones. Now that
we're familiar with how to manipulate logarithms
from a single to many and for many to a single,
we can solve. For example, let's solve the
log P equals log of P zero minus C times log
quantity T plus 1 for the variable P. Here,
we're assuming that P subzero and C are some
known constant values. T is really the only
variable in this equation. So when I ask you
to solve this equation for P, what I'm really
asking you is to write P in terms of T or
write P as a function of T. This problem really
is where I combine multiple logarithms into
a single logarithm and then go from there.
I'm going to try to simplify the right hand
side as much as I can using the rules we have
learned. But first we're going to turn it
into a power using the rule number 3. So I
have log P subzero minus log quantity T plus
1 raised to the C. Then I use the rule number
2 to combine the 2 logs into a single log
of quotient. So I have log of P subzero divided
by T plus 1, raised to the C power. Going
from this equation, if I want to isolate P,
I need to undo the process of log which is
applying its inverse both sides. And we know
that the inverse of a log function is exponential
function. And because we're dealing with the
common log here, we're assuming that base
10. I need to apply the exponential function
of base 10 both sides. So I have 10 raised
to the power of log P is equal to 10 raised
to the log of this entire thing. That's P
over T plus 1 raised to the C's power. And
because the exponential function of base 10,
the logarithm function of base 10 are inverses
of each other, I simply just get the input.
Which in this case, would be P on the left.
And on the right, the 10 in log kind of cancel.
I simply get the P subzero divided by the
T plus 1 raised to the C's power. So this
expression at the end is a function in terms
of T and that gives rise to expression for
P.
