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.
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.