Welcome ladies and gentlemen.
So what I'd like
to do is show you
how to solve
logarithmic equations.
Now when solving logarithmic
equation, previously what
we did was solve a logarithmic
equation just by rewriting them
in exponential form.
But you can see in each
and every one of these,
we have more than one logarithm.
So we just can't quickly and
easily reach switch it over
to exponential form because
exponential form-- to go
from logarithmic to
exponential form,
we had to have a
logarithm isolated.
Now if we have logarithms
on the same side,
we can still use the
rules of logarithms
to condense it down
to one logarithm.
And if we have a
logarithm on both sides,
we can also use the
one to one property
that we use in exponential form
by taking the log of base x
equals log base b to y.
I'm about to sneeze.
Since the bases
are the same, you
could just set the
values of each logarithm
equal to one another.
The next thing that
we also want to do
is, since we have
two logarithms,
we also want to make sure we can
check for extraneous solutions
or solutions that are not work.
Because if you
remember, when we're
looking at the solution
of a logarithm,
here's a logarithm
with no transformation.
Now all of these are going
to have transformations.
So it is possible to
have negative solutions.
However, in a logarithm
with no solution,
we can't take an x value and
plug it into log base b of x.
There is no negative
x that we can plug.
So whenever we
take our solutions,
when we plug them back
into the original format,
we can't take the log
of a negative answer.
And we'll get to this
as we work along.
But I wanted to kind of
show you just the graph
with no transformations.
You can't take the log
of a negative value.
So in the first case here,
I have two logarithms
on the same side.
Now I don't want to add
them to the other side
to do my one to one property,
because if I did that, I still
have this two here.
And the one to one
property only works
when you have a logarithm
equal to a logarithm.
Since there is that two here,
we can't complete the one
to one property.
So what I'm going to have to
do is keep the logarithms here
and condense them
down to one logarithm.
And I can do that using
my rules of exponents,
which is the product rule.
And remember the
product rule states
if you have two logarithms, as
long as you have the same base
and you're adding them,
you can rewrite them
as the product of
those two values.
Now I have a logarithm that's
been isolated equal to a value.
I can rewrite this
in exponential form.
So 4 squared equals
negative x squared--
I'll apply distributive
property here-- minus 10x.
Now I can rewrite this as 16
equals-- I'm going to factor.
Let's keep that in there first.
Actually, let's leave this out.
I'll do one more.
16 equals negative
x squared minus 10x.
Now you can see that
I have a quadratic.
My x that I need to
solve for is being
raised to the second power.
So therefore I'm going to treat
this as a quadratic equation
where I can either
solve by factoring,
I can solve by the quadratic
formula, completing the square,
and so forth.
So the first thing
I want to do is
try to see if I can
solve by factoring.
So to do that, I need to get
all the terms on the one side
and set it equal to 0.
So I have 0 equals negative
x squared minus 10x minus 16.
I don't like factoring
when my a is negative,
so I'm going to factor
out a negative 1.
That's not going to
affect my solution
because I can divide out
that negative 1 at any time.
So I have 0 equals x
squared plus 10x plus 16.
Now to go ahead and
factor that, I'm
just going to go ahead and
rewrite this in factored form.
So 0 equals-- let's see, what
two numbers multiply to give me
16, add to give me 10.
That's going to be x plus
8 as well as x plus 2.
Running out of space.
So therefore now
what I can do is
use the zero product property,
say x plus 8 equals 0
and x plus 2 equals 0.
To solve, I get x
equals negative 8
and x equals negative 2.
Now going back to what I
was describing over here,
I have to go back and make
sure my solutions work.
Because you can see
they're negative, right?
And a lot of times when
students see negative,
they're like oh,
it's extraneous.
Well not necessarily.
This graph, if you were to graph
this, has some transformations.
It's shifting left,
right, up or down, crazy.
This has no transformations.
So yes, if no transformation,
you can't have a negative x.
However, when you
have transformations,
we just can't take the value of
the logarithm to be negative.
Now watch what happens.
When I take negative
8 and plug it
in for x for both
of these logarithms,
let's see if it works.
Negative 8 in for
negative x is positive 8.
That works.
Negative 8 in for x
here is positive 2.
So that works.
So negative 8 checks out.
So I just do a nice little check
mark, which me as a student
would tell my teacher that
I've checked both solutions.
Obviously if we had a
little bit more time,
if I wasn't doing
six examples, you
could also just plug it
in and show your work.
But I'm going to kind of move
ahead a little bit quicker
than that.
Negative 2, plug in for
negative x is positive 2.
That works.
Negative 2 plugged in for
negative x is positive 8.
So that works.
So both my solutions work out.
And the next example is very
similar to the other one.
So you have the logarithms
on the same side.
You have another number.
You have the same bases.
So let's just go ahead
and do the same work
that we did here last time.
Again, I'm going to rewrite
this in exponential form
to kind of save a little work.
I'm just going 2 to the
cubed, third power, is 8.
And then I'll distribute
this, x squared minus 2x.
I'll subtract an
8 on both sides.
So I get 0 equals x
squared minus 2x minus 8.
Now I need to do is
determine which values
multiply to give me negative 8.
So now I can factor again.
Which values multiplied
give me negative 8
but then add to
give me negative 4?
So it's 0 equals x
minus 4 times x plus 2.
You apply the zero
product property.
So now I've solved
for my two solutions.
Now what I want
to do is plug them
back in to see if either
of them are extraneous.
I take 4.
I can plug 4 directly in for x.
That works.
I can plug 4 in for x here.
4 minus 2 is 2.
That works.
Then I take negative 2.
Negative 2 in for x, you
can't take a log base 2
of negative 2, right?
So therefore that is extraneous.
And I just circle it and I
just write EXT for extraneous.
It is a solution, but it's
an extraneous solution.
The next one, we have the
same thing, basically,
going on, except now
I have two binomials.
So I'm going to do log base
3 of x plus 6 times x plus 4
equals 1.
Raise it.
Rewrite it in exponential form.
3 to the first
power is equal to x
squared plus 6x plus 4x plus
24, just expanding that out.
I'll subtract a 3 on both sides.
So I get 0 equals x
squared plus 10x plus 21.
How does this happen again?
OK, well this one always works.
So now I need to go
ahead and factor this.
So I'll go ahead at 0 equals
x plus 7 times x plus 3.
So therefore I get x plus 7
equals 0. x plus 3 equals 0.
So therefore x equals negative
7. x equals negative 3.
Now when I plug in
negative 7 in here,
negative 7 plus 6 is negative 1.
That's extraneous you
can't take the logarithm
of a negative number.
However, when I try negative
3, negative 3 in for x
gives me positive 3.
Negative 3 in for x
gives me a positive 1.
So that works.
So just because an
answer is negative
does not mean it's extraneous.
You have to make sure
you plug it back in
to make sure it works.
A lot of students get
that kind of mixed up.
And the next one's going to take
us a little bit of extra work.
So I'm just going to kind
of short line a little bit.
So I have 5 squared
is equal to-- I'm
going to multiply these out.
It's going to be the
product of these two.
Oh wait, I guess I-- let's
simplify this to log base 5.
Let's multiply these out.
x squared plus 5x
plus 4 equals 2.
I'm going to need a
little bit of extra work,
so that's why I'm kind
of speeding this along.
So therefore that's going to
equal 5 squared is equal to x.
So that's going to
be 25 is equal to x
squared plus 5x plus 4.
So therefore I can subtract
a 25 on both sides.
And I get 0 equals x
squared plus 5x minus 21.
So in this problem, we have
a little bit of an issue
because I can't factor this.
I can't find two numbers that
multiply to give negative 21
and add to give me 5.
So if you remember when I said
hey, when we have quadratics,
if we can find the
solutions, then we
have to look for
factoring, then we
have to look to the
quadratic formula.
x equals opposite
of b plus or minus
the square root b squared minus
4 times a times c all over 2a.
Remember, that's
from your equation
ax squared plus bx plus c.
So now I'm going to just plug
this into the quadratic form.
So my solutions are going to
be x equals plus or minus--
or sorry, opposite of b, which
is negative 5, plus or minus
the square root of b squared
minus 4 times a times c.
And then all divided by 2a.
So let's go ahead
and simplify this.
I'll grab my calculator here.
So it's going to be
negative 5 plus or minus.
It's going to be 25.
Let's see, that's going
to be positive 21.
That's going to be 42.
That's going to be 84.
109.
Let me just-- yeah, 4 times
21, 84, right, plus 25-- 109.
So therefore I have
x equals negative 5
plus or minus the square
root of 109 divided by 2.
All right, now here's where it
kind of gets a little crazy.
Because you might
think all right,
well you could easily do
the decimal version of this.
Negative 5 plus square root
of 25-- square root of 109
is approximately 10.
So negative 5 plus-- let's just
round it to 10-- divided by 2
is going to give me
a positive number.
If I plug in a positive
number back into my equation,
am I still going to have
two positive logarithms?
Yes.
However, what if I did
negative 5 minus negative 10,
approximately, right?
It's 10.4, 10.44
repeating or irrational.
Negative 5 minus
10 is negative 15.
Divided by 2 is like
negative 7 and change.
If I plug in negative 7 into
both of these logarithms,
therefore it's not
going to work, right?
Both of these are
going to be negative.
So therefore my only solution
is x equals negative 5
plus the square root
of 109 divided by 2.
So I'm just going
to compute that.
Minus 5-- so I'll do
the square roots of 109
plus a negative, which is
same thing as minusing 5.
And then divide by 2.
And I get 2.7--
approximately 2.72.
I'll just round to
the nearest hundredth.
So in the next
example, you can see we
have logarithms on both sides.
So now we can't just isolate
a logarithm on one side.
What we have to do is get the
logarithm-- what we have to do,
what we want to do is use
the one to one property.
So in this case what--
so in this case basically
what I'll want to do here
is basically combine these.
I can combine these so
I get log of 3x minus 3
is equal to log of
x plus 1 times 4.
Well now what's important
about this is now
that I have a logarithm
equal to another logarithm,
their values are going
to be exactly the same.
3x minus 3 is
equal to 4x plus 4.
So now I'll just solve
for my logarithmic.
So I subtract a 3x.
Subtract a 3x.
Negative 3 equals x plus 4.
Subtract 4.
Subtract 4.
Negative 7 equals x.
So negative 7 is my solution.
However I want to make
sure I plug that back in.
And when I do, when I plug that
back in, I plug negative 7.
So it's only one solution.
When I plug negative
7 back in here,
I get negative 21 minus
23 is negative 24.
Doesn't work, so
that's extraneous.
Negative 7 plus 1 is negative 6.
So therefore it's extraneous.
And since that's
the only solution,
we're just going to
write no solution.
In the next example, this
is one of the easier ones
that we come up to.
Whenever you have a logarithm
with another logarithm,
you can just apply the
one to one property.
So therefore I have 2x
minus 3 equals x plus 4.
Yes ladies and gentlemen,
it is that simple.
So now x equals 7.
However when I plug x
equals 7 into this equation,
I get 14 minus 3, which is
11, and 7 plus 4, which is 11.
So it works for both of them.
So there you go
ladies and gentlemen.
That is how you solve
logarithmic equations.
Thanks.
