- [Instructor] Contract using
the rules of logarithms,
and express your answer
as a single logarithm.
Well, let's first of all
begin by reviewing the rules
of logarithms.
The first rule says that the sum
of two logarithms is a single
logarithm of a product.
The second rule of logarithms
says that the difference
of two logarithms is a single
logarithm of a quotient.
The third rule of logarithms
says a constant multiplied
by a single logarithm is
equal to the single logarithm
of a power.
For part A, we have three
log of K minus two log
of K plus three.
At first glance, it might appear
that we have the difference
of two logarithms,
but each of those logarithms
has a coefficient,
so let's first of all
use rule number three
to write each of those as
a logarithm of a power.
So the first expression
will become the logarithm
of K cubed,
and the second one will
become the logarithm
of the square of K plus three.
Now that we have the
difference of two logarithms,
we can apply rule number two.
The difference of two
logarithms is a logarithm
of a quotient,
so I end up with the logarithm
of K cubed over the square
of K plus three.
For part B, we have
minus the logarithm of M
plus five times the
logarithm of three plus N.
Well, first of all, let's
use the commutative property
to rewrite this in a different order.
Instead of writing it
as a sum, we'll write it
as a difference.
So we have five times the
logarithm of three plus N
minus the logarithm of M.
Now we can use the same
strategy that we used in part A.
First of all, we'll use rule number three
to write five logarithm of three plus N
as the logarithm of the
fifth power of three plus N.
Now that we have a
difference of two logarithms,
we'll write a single
logarithm of a quotient.
So we end up with the
logarithm of the fifth power
of three plus N divided by M.
For part C, we have four
log of T plus 1/2 log of T.
So first of all, let's
apply rule number three.
Will have the logarithm of T to the fourth
plus the logarithm of T to the 1/2 power.
Now that we have the logarithm of a sum,
we can rewrite that as a
single logarithm of a product,
so that will give it the logarithm
of T to the fourth times T to the 1/2.
We have a single logarithm, but
we can simplify the argument
of the logarithm using the
properties of exponents.
We have the same base, and the
operation is multiplication,
so we can add the exponents.
Four plus 1/2 is 4 and 1/2,
but we can right that as 9/2,
so we have the logarithm of T to the 9/2,
and the meaning of the 9/2
exponents is the square root
of T to the ninth power,
so our final answer is
logarithm of the square root
of T to the ninth power.
For part D, we have 1/3, the quantity,
the log of X plus two times the log of Y.
Let's start inside the parentheses,
and we'll first of all
use rule number three
on the second term.
We'll rewrite two log of Y as
the logarithm of Y squared,
and let's continue working
inside the parentheses.
We'll rewrite the sum
of the two logarithms
as a single logarithm of a product,
so the log of X plus the log
of Y squared will become the
logarithm of X times Y squared,
and now we can use rule number three.
We'll take the 1/3 coefficient,
and we'll write it as an
exponent inside of the logarithm,
so we have the logarithm
of the 1/3 power of X times Y squared,
but the 1/3 exponent means
to take the cube root
of an expression,
and so our final answer
will be the logarithm
of the cube root of X times Y squared.
