>> Here we are going to
use the Laws of Logarithms
to combine expressions.
Whereas the Exponentiation
Law is used last
when we expand logarithms, it is
used first when we combine them.
And so whenever we see a
coefficient in front of a term
that is not one, we are
going to change that to log
of a power using that
coefficient as an exponent.
So, in the first case, we
have a coefficient of a half.
And so I'm going to write the
given expression as common log
of 14, plus common log of 3
to 1/2, minus common log of 7.
Then I'm going to
use the Product Rule
on the first two terms.
I'm also going to
change 3 to the 1/2
to the square root of a 3.
So instead of two
common logs we have log
of the product of the arguments..
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So 14 times the square root
of 3, minus common log of 7.
And so we're going to have
the Quotient Rule in play now.
So we can do log of the quotient
of 14 square root
of 3, divided by 7.
The 7 cancels into
the 14 to get the log
of the quantity 2
square root 3.
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In the second example, we
don't have any coefficients
that are not one, but we do have
a product that can be factored
in the first argument.
It's log base 5 of the quantity,
x squared minus 16, minus log
base 5 of the quantity,
x minus 4.
So we will, first of all,
factor the x squared minus 16
as x plus 4, times, x minus 4.
We have a second term: minus
log base 5 of, x minus 4.
Then we'll use the Product
Rule, which will mean
that we're changing the log
of the product to the sum
of two logs base 5, where 1
factor is in each of the terms.
So log base 5 of the quantity,
x plus 4, plus, log base 5
of the quantity, x minus 4,
minus log base 5 of, x minus 4.
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We can drop the brackets.
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And see that we've
got log base 5
of the quantity, x
plus 4, plus, log base 5
of the quantity, x minus
4, minus, log base 5
of the quantity, x minus 4.
So the log base 5's of the, x
minus 4's, neutralize each other,
and we're left with log base
5 of the quantity, x plus 4.
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In the third problem, we have
2 times the entire quantity,
log base 6 of x, plus, 2 log base
6 of y, minus, 4 log base 6 of z.
And so first of all, I'm
going to distribute that 2.
So it'll be 2 log base
6 of x, plus, 4 log base 6
of y, minus, 8 log base
6 of z. Now I'm going
to use the Power Rule because
we have coefficients of 2, 4,
and 8, that we can
write as exponents.
So, log base 6 of, x squared,
plus log base 6 of y
to the 4th, minus, log
base 6 of z to the 8th.
We use the Sum Rule to write
the log of the product.
So we get log base
6 of the product,
x squared y to the 4th, minus
log base 6 of z to the 8th.
Now since we have a
difference, we can write this
as the log base 6 of the
quotient of the arguments.
So it will be the
quantity x squared, y
to the 4th, divided
by z to the 8th --
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-- as the argument
of log base 6.
And one last example
where we have log base a
of b, plus, c log base a of
d, minus, v log base a of w.
And so two coefficients
that are not one.
So we will rewrite this as log
base a of b, plus, log base a of d
to c power, minus, log base
a of w to the v power.
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Then we'll be able to combine
again, starting with the sum.
We have the sum of
two log base a's.
We'll be able to write that
as log base a of the product
of the arguments,
which will be b 
d to the c power, minus, log base
a of w to the v. And so lastly,
we've got the difference
of two log base a's.
We'll be able to write
that as log base a
of the quotient of
their arguments.
So the numerator will
be b d to the c power --
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-- divided by w to the v power.
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So log base a of the quantity, b
d to the c power, divided
by w to the v power.
