Hi, this is Tracy Kight with Georgia Highlands,
and today's video is on properties of logarithms.
This is from Unit 4 of College Algebra – and
we're also going to look at expanding and
condensing logarithms.
The first thing we have to look at is the
properties of logarithms.
The properties of logarithms –just like
the graphs – link very closely to the rules
for exponents.
So let's look at the first rule.
We have X to the 5th times X to the 3rd.
What is the rule when we are multiplying like
bases?
Well we simply add the exponents.
So that would be 5 plus 3 – or X to the
8th.
Well the same thing is true for logarithms.
When we are multiplying two values, we can
separate them to simplify by changing it to
addition.
Okay, when we were dividing exponents we subtracted
the exponents as long as the bases were the
same.
When we are dividing with a logarithm, we
change it to subtraction.
And when we're raising a power to a power,
we multiply – 5 times 3 or X to the 15 – and
here, if I had log base A of X to the 3rd
power, this 3 would come out front to be multiplied
times our logarithm.
Now these logarithm properties are true no
matter what the base is, including natural
logs.
It also applies if I'm multiplying I can change
them to addition; if it is division I expand
it by subtracting; and if we're raising a
power – we're raising it to a power we multiply
it.
So let's look at some examples.
We're going to use the properties of logarithms
to expand each of the following expressions.
So on this first one, the first thing I'm
going to do is take this 2 and multiply it
out front.
Since it goes with both of the terms, I'm
going to put it in parentheses so that the
2 goes with both the A and B.
Now the A and B are being multiplied, so I'm
going to change them to addition – so log
base 3 of A plus log base 3 of B.
We can also distribute this 2 to both terms
and write it a log base 3 of A plus log base
3 of B.
For the next one, we have to remember some
properties of exponents.
We have to remember that taking the cube root
of an expression is the exact same thing as
raising it to the 1/3 power.
So just like before, I'm going to bring the
1/3 out front and it goes with each term so
I'm going to put this in parentheses.
X, Y, and Z are each being multiplied; so
I'm going to change that to addition.
Do 1/3 log X plus log Y plus log Z – and
just as before, we can put the 1/3 with each
of the terms and have 1/3 log of X plus 1/3
log Y plus 1/3 log Z.
Okay, and now let's look at this last one.
The first thing I'm going to do is see that
these two values – X cubed and the square
root of Z divided by Y cubed – are being
multiplied.
So I'm going to write LNX cubed plus LN, the
square root of Z over Y cubed.
Next I'm going to bring the 3 out front, and
I'm also going to write each of these to the
1/2.
So 3 LNX plus LNZ to the 1/2 over Y to the
3/2.
The reason is to the 3/2 is because it was
to the 3rd power and 3 times 1/2 gives me
3/2.
Now we're going to rewrite it – 3 LNX plus
LNZ to the 1/2 minus LNY to the 3/2, which
gives me 3 LNX plus 1/2 LNZ minus 3/2 LNY.
All right, let's look at some more examples.
In this next example, we are asked to condense
each of them to a single logarithm.
The first thing I'm going to do is take care
of the 1/2, 1/3, and 1/4; I'm going to move
them from the front of the expression to the
exponent.
So I've got log base 3 with 64 to the 1/2,
log base 3 of 125 to the 1/3 minus log base
3 of 16 to the 1/4.
Now if you'll remember, 64 to the 1/2 is the
same as saying the square root of 64.
The square root of 64 is 8 – 125 to the
1/3 is the same as asking for the key root
of 125, which is 5.
And on the last one, 16 to the 1/4 is asking
for the 4th root of 16, which is 2.
Next I'm going to change these first two that
are being combined with addition to multiplication.
This would give us log base 3 of 8 times 5
– or 40.
Now I'm going to combine the log base 3 of
40 minus log base 3 of 2 to get log base 3
of 40 divided by 2, which simplifies to log
base 3 of 20.
Now, we took that long original expression
and condensed it to a small expression that
can be evaluated.
Let's condense this one.
On this one, the first term that I see is
the 17; so I'm going to move it to the exponent.
I've got log base B of M to the 17 plus log
base B of 2 minus log base B of X.
Next I'm going to combine these two terms
that are being added and change them to multiply
– log base B of 2 times M to the 17th minus
log base B of X.
And then the last step is to change the subtraction
to division –
log base B of 2M to the 17 over X.
And that is our single logarithm.
And for our last one, I'm going to take care
of the 3 and move it here to the exponent,
and I'm also going to take care of my parentheses
as well.
This is going to give us log base B of X cubed
minus – and in the parenthesis it's being
added, so we change it to multiplication.
Now the subtraction sign tells us to change
it to division.
So log base B of X cubed over 4X.
I can reduce these two X's and this is going
to give us log base B of X squared over 4
for our answer.
And now we've condensed it to a single logarithm.
