In this example we are asked to write each
expression as a single logarithm.
In example 2, you may recall we were going
from a single logarithm to a multiple logarithm.
This is the reverse.
So we are going from multiple logarithm and
we want to combine into a single logarithm.
So these are the reverse.
But the same rules will allow us to do that.
If we have a difference like in part a, a
difference of two logarithms, we know if we
can combine them together to one logarithm
by taking the quotient of their argument.
Before we jump to that rule we may observe
there is a 2 in front of the logarithm.
In order to use the rule to combine them,
we have to remove the 2 in the front, so we
are going to use the reverse of the rule that
allows us to bring the exponent in the front.
We are going to move it back up.
We are going to move the 2 back up to the
exponent of the argument.
So we can rewrite this as the log base 3 of
u squared.
Now we have a logarithm and an argument, a
logarithm and an argument.
They both have a base 3.
I can subtract which allows us to combine
these into one logarithm.
We take the quotient u squared over v.
We went from two separate logarithms and combined
into one.
I would like you to try part b. on your own.
I am going to jump to part c here next.
In part c we have three logarithms, three
natural logarithms, base e.
We want to combine all of these.
So we have a sum, these are being added and
then we are subtracting.
Let's go to two at a time.
I don't recommend going from three directly
to one.
I recommend paring up two of them.
And then pairing up 2 more.
Let's look at the first two.
Natural log of x over x - 1 plus natural log
of x + 1 over x.
If I want to add two logarithms, I can combine
them into 1 by taking the product of the argument.
I can rewrite this as natural log of x over
x - 1 times x + 1 over x.
I am taking the natural log of that product.
Let's just leave the third logarithm here
alone for a while.
Now I want to multiply x over x - 1 times
x + 1 over x.
We can see the x will divide out.
That makes this a little simpler.
We divide an x into both of these and we are
left with 1 in the numerator here and denominator
here.
Now if we simplify our argument this becomes
the natural log of x + 1 over x - 1.
I will bring down my natural log of x squared
minus 1.
Once we join together the first two now we
can combine the resulting logarithm and the
natural log of x squared - 1.
Even if the help me solve this or the e textbook
suggests that you combine these all in one
step, I would recommend you do these two at
a time.
Now we have a difference so we can use division
to combine these together.
Now I want to take the natural log of the
quotient x + 1 over x - 1 and I want to divide
by x squared - 1.
When I divide, that is the same as multiplication
by the reciprocal.
Rather than doing this division of a fraction,
let's change it to multiplication by the reciprocal.
Now we can see this x squared - 1, this can
factor because this is the difference of two
squares.
I can rewrite this as x + 1 over x - 1 times
1 over x + 1 times x -1.
We have a common factor in the numerator and
denominator.
So we can divide that out.
We can divide x+ 1 by x + 1 and get 1 and
divide this x + 1 by x + 1 and get 1.
So now that argument will simplify into 1
times 1 or 1 in the numerator.
In the denominator we have x -1 times x - 1.
We can rewrite that x - 1 squared.
Now that is useful because we want to write
this argument using an exponent if possible.
I have an exponent in the denominator or the
argument.
We know if we have a reciprocal, we can rewrite
that 1 over this using a negative exponent.
I can rewrite this as the natural log of x
- 1 to the negative 2 power.
Now I have the power up on the argument and
I can move this out into the front and multiply.
I will take my negative 2 move it out into
the front and then finally we have -2 times
natural log of x - 1.
When you have three or more logarithms, you
probably won't have any more than 3, I would
definitely pair up two at a time.
Combine two into 1 first and then combine
the remaining two.
I think that easier way to do this.
I want you to try part d on your own.
