HELLO, I'm Mr. Tarrou. Let's finish part two
of this lesson about solving equations with
logarithms. It is the fall and near Halloween
so I wore this shirt, pumpkin pie my favorite
desert. We need to solve this trinomial, solve
it for x and it looks quite a bit different
than maybe we are used to. Let me tie this
in with an equation that maybe you are more
comfortable with and that is simply x squared
minus x minus six. Now I gave you this equation
to solve I highly doubt that anyone would
have a problem with it. When you factor a
trinomial, in this case a quadratic, you look
for factors of the last term that go into
the last two blanks and factors of the first
term go into the first two blanks. We are
looking for factors of one and so there is
no choice here it is one and one. And because
the leading coefficient is one, we are simply
going to be looking for factors of negative
six that subtract to be negative one. That
is going to be, in blue, negative three and
positive two. They multiply to give me negative
six and they also add up to the middle term
which has a coefficient of negative one. Now
normally when we have quadratics we can write
x and x without really thinking about it.
But why is it x and not x squared. Well because
we want to multiply these binomials back together.
We are doing x times x is x squared. So any
time we have a number times itself it is going
to be equal to x squared. Well when you take
a number and break it down into something
times itself equals the square root. The square
root of 16 is four. Why, because four times
four is sixteen. So when you just write x
and x without thinking from x squared, you
are doing the square root of x squared. Now
let's review fractional exponents for a minute.
Fractional exponents are when you have a power
over a root. So if you are taking the square
root that written as a fraction is a power
of one half. And so we have x squared in the
middle and round it being applied a power
of one half or a square root. When you have
stacked exponents you are multiplying. So,
one half of two is equal to one. When you
take the square root as far as exponents are
concerned, that is the same as dividing the
exponents by two or multiplying them by a
half. So when you have a trinomial it is going
to be factorable as long as, well as long
as it is factorable in the first place, as
far as the exponents are concerned they need
to count down by a half. Whatever this first
exponent is then half of that, because that
is what you do when you square root exponents
and then you just have a constant at the end.
So if this were a power of four, then the
middle term would have to have a power of
two. And x to the fourth minus x squared minus
six is equal to zero would factor down to...
The square root of x to the fourth is x squared.
So just remember with you square root symbol
written as a fractional exponent that is a
power of one half. As far as exponents are
concerned you are dividing them by two. So
in this harder looking example we have an
exponent of 4x. Well as long as the middle
exponent is half of the first exponent, the
could be a factorable trinomial. It will be
factorable as long as you can get your leading
coefficient and constant to come together
to some how give you that middle coefficient.
I don't know why I changed the sign on that
one. Let's factor this one over here. We have
a leading coefficient of one and we have a
constant of six. We want the factors of negative
six that add to the middle term of positive
one not the negative one that I put in this
example I made up. We are going to be looking
for e, now if you are going to square root
e to the 4x that is 4x divided by two, so
this is going to be e to the 2x. Now we have
plus three times e to the 2x minus two. Then
after we factor you set each factor equal
to zero to finish solving the problem. So
we are going to say that we have e to the
2x plus 3 equals zero and also e to the 2x
minus 2 equals zero. Now we are going to try
and solve these for x. We are going to subtract
both sides by three on this first equation.
And we are going to add two to both sides
here. And when you have a variable in the
exponent that you are trying to solve for,
a lot of times it will not come out of the
exponent without using logarithms. Excuse
me:) But I am not even going to bother with
this one. You are never going to get a positive
base, remember that e is approximately 2.718...
It is like pi and is a decimal that runs on
forever. It shows up a lot in nature problems,
electrical engineering, sound engineering,
continuous compound growth. Anyway, you are
never going to get a positive base raised
to any power give you a negative answer. So
this one is going to be undefined and have
no solution. Now over here we have e to the
2x equals positive two. We are going to apply
the natural log function to both sides to
get rid of the base of e allowing us to get
the variable down out of the exponent. We
are going to have the natural log of e to
the 2x equals the natural log of two. If the
natural log and the e did not cancel out we
would pull the two x out front as a coefficient
and then divide both sides by the natural
log of e. But, when you log a number that
has the same base as your logarithm they will
cancel out. This gives us 2x equals the natural
log of two. Then we will divide both sides
by two to finish solving this problem and
get x is equal to either the natural log of
two, that is not going to be two divided by
two and then the natural log of one...it is
not the natural log of one... The natural
log of two divided by two is going to be approximately,
if I am reading my work out right, positive
.347 if you get that typed into your calculator
correctly. Let's move on to the next one.
The log base three of x minus one equals three.
Now if you have a log function in your equation
the first thing that you want to do is isolate
that log function. Mine already is. This is
a pretty straight forward question. I can
actually go right into exponential form by
remembering out of a logarithm you get an
exponent. Remember the log of 100 it 2, because
then squared is a hundred. You always get
exponents from log functions. So because this
is a very basic equation I can just write
3 to the third power equals x minus one. But
for teaching purposes and not all equations
come out that easily I want to remind you
that to get rid of log functions you use the
exponential function. Again let's think about
it. The log of a 100 equals 2. So I logged
a hundred and I got an answer of two. If I
would like to get back to a hundred, I do
ten squared equals one hundred. That exponential
function will undo the log. They are inverse
functions. So I am going to make both sides
of my equation an exponent of three and when
you have a base of three and a log with that
same base value, of three in this case, those
are inverse functions and they will cancel
out. This allows the x minus one to simply
drop down. And we get x minus one equals three
to the third. That is x minus one is equals
three to the third is twenty seven. Add both
sides one and you get a final answer of twenty
eight. And if you want to make sure that this
works, simply go back and plug it into your
original question. And, my original video,
this is the second part, when you have an
original function that can become undefined
like you cannot divide by zero. So if x is
in the denominator you may have a problem.
You cannot square root number and here have
x inside of a log function. You cannot take
the log of a negative number. You cannot square
root a negative number, you cannot divide
by zero, so when you are working with an equation
that has some major restrictions on the domain,
you had better plug that back into the original
problem and make sure that your answer does
not make it undefined. I forgot that step
and therefore I messed up on my last two examples
in my video. So, here I am doing my second
take on Halloween. Plug that in. Twenty eight
minus one is twenty seven and at least you
are logging a positive number. Ok, I try not
to make mistakes in these videos but every
once awhile when I am trying to talk and write
at the same time that happens:) If you ever
notice a mistake in my videos please send
me a notice and I will correct it or in this
case reshoot that video. One of my students
actually found the mistake so I thought not
only do watch my videos but he knew enough
to find a mistake in my work so he got a bonus
point for that:) Here we go. The natural log
of the square root of x plus one equals five.
If you have a log function in your equation
and you would like to get rid of it, the exponential
function will undo the log because they are
inverses. I want to get rid of this natural
log, it is already isolated. Natural log is
log base e and if I would like to undo that
log base e I am going to make both sides of
the equation an exponent of e. That is going
to cancel out the log function and give me
x plus one, now it is the square root of x
plus one but I am going to for just teaching
purposes and it is usually a good idea...
The more into advanced math you get, the more
comfortable you want to get with fractional
exponents. They are much nicer to deal with
than radical notation. My square root I am
writing as the power of one half. Power over
root, it was the second root or square root.
That is equal to e to the fifth power. Now
if you have a fractional exponent you are
going to raise both sides of your equation
by the reciprocal of that power. So if I have
an expression raised to the one half, and
I want to get rid of that power of one over
two, I am going to raise both sides by the
reciprocal of one half which is just two.
You probably would have done that anyway had
you left the radical in your equation. You
undo a square root by squaring both sides.
So two times one half is one. Excuse me I
am still getting over a cold. We have x plus
one equals, this is again power to power and
we get e to the tenth. And now we are going
to get the plus one away from the x with subtraction.
We get x equals e to the tenth power minus
one. If you put that into your calculator
the answer to this question is unbelievable
it is so large. You get x equals 22,025.5
This is rounded off a bit and is the approximate
answer, this is the exact form of the answer,
to this equation. Again you do need to check
this answer because our variable is in a log
function and you cannot log a negative number.
So make sure that you plug that back in to
see if it works. Moving on to our next example.
We have log base three of x minus five plus
log base three of x plus three is equal two.
I am going to combine these log functions.
I have two minutes left. That is going to
give me log base three of x squared minus
two x minus fifteen equals two. If I apply
a power of three to both sides, I am going
to get x squared minus two x minus fifteen
is equal to nine. Subtract both sides by nine
and I get x squared minus two x minus twenty
four equals zero. That is factorable. I am
going to get x minus six times x plus four
equals zero. My solutions are x equals 6 and
x equals negative four. Now if I plug these
into the original equation, negative four
minus five and negative four plus three, both
of those are negative and you cannot log a
negative number. So the only answer is six.
I am Mr. Tarrou. BAM!!! Thank you for watching,
thank you allowing me to help you out. Go
Do Your Homework!
