Hi, let's take a look at solving some logarithmic equations.
The technique used to solve a specific log
equation really depends on what the original
problem looks like, so some analysis is
always in order before you begin.
Let's take a look at an example: log base 3 of 2 x minus 1 equals 2.
That's what we want to do, we want to solve
for x, so let's analyze what we have.
Well, we have a log expression: we have a
log, a base, we have this term, which is
called the argument, and then
we have this number out here, 2,
which is actually the logarithm.
Now think about the definition of logarithm.
So, we have log base a of x equals y,
and this means that a to the y equals x.
Now there are some restrictions on these variables.
First of all, a cannot be 1. What else?
a has to be greater than 0, and then there
is one other one, and this is probably the
most missed part of a problem like this.
What sits in this position, whether it be
just x like it is here in the definition or
2 x minus 1 as it is in our sample one, has to be greater than 0.
So, the first thing to do is to look at this
problem and realize the 2 x minus 1 has to
be greater than 0, and so this is the domain of this expression.
Alright next, let's take a look at the form of what we have.
We have a log, a base, an argument, and a
number, and this exactly matches log, base,
argument, number, which means all we have to
do in order to rewrite this is to follow the definition, and then
of course we have to remember at
the end that we're going to have to check to
be certain that the value we get for x,
when put back into here, makes this expression greater than 0.
Okay, so we rewrite 3, which is the base a,
to the second power equals 2 x minus 1, and
now we have a linear equation, and we solve for x.
So, we have 9 equals 2 x minus 1. 2 x equals 10, and so x is 5.
I switched the sides of the equation.
You don't have to do that, but it's just
traditional to have the variable on the left.
So, we got x is 5, are we done? No, we have to check.
We want to check to be certain that this
value, when placed into here, makes this
expression positive. So, what would we have?
We have 2 times 5 minus 1, which is 9, which
is definitely greater than 0, and so this is
a valid solution to this first log example.
Okay, now let's take a look at another example.
In this example we have the expression 2 log
base 4 of x equals log base 4 of 25.
First thing I notice when I analyze this is
that there's a coefficient, 2, in front of
this expression, and so I recall a property
that's going to allow me to rewrite this.
Okay, can you think of what the property is?
Well, here it is, it looks like this: if you
have log base a of some expression, x,
raised to a real number power, you can
rewrite this as r log base a of x.
Okay, in this case, we have the right-hand
side, we're going to rewrite it as the left-hand side.
The same restrictions hold here: a cannot be
1, a has to be greater than 0, and again
what sits in this position has to be positive.
Okay, using that property, the first thing
I'm going to do is rewrite this as log base 4 of x squared.
The 2 is written here, and then I just copy over the right-hand side.
Alright, now what do we have? Go back and look at example one.
Does this look like the same format as example one?
Not really, we have a log expression on the
left, but in example one, we had a constant
here, and in this example, we have another log expression.
So, instead of using the definition to
rewrite this and solve, we have to use the
one to one property of logarithms. So, let's go back over that.
Well it looks like this: if you have the log
base a of some expression, and we'll just
call it capital M and in this case that's
x squared, and it equals the log base a of
capital N, we're going to put an if in front of this.
If this is true, then what do you think is true?
Well, you're right: M equals N.
Now of course, there are restrictions here
again: M, N, and a all have to be positive and again a cannot be 1.
Alright, so how do we use this property?
Well we look at this and we see that the
bases are the same: a, a, in this case 4 and 4, and so what do we do?
We realize that the one to one property,
which is what this is called, allows us to
very simply reduce this expression to x squared equals 25.
Okay let's see, we know that we can use the
square root property now and get plus or minus 5,
and it's tempting to stop here, but don't.
Always remember that when you use or work
with a log expression at all, anything you
do, this has to be greater than 0, and so does this.
Okay, well the 25 is already greater than 0,
and so now all we have to do is check this. So, what happens?
Well, when we check positive 5, of course
positive 5 put into here is greater than 0, and so that's fine.
But negative 5 put into here is not greater
than 0, and so that root does not check.
So, what we have here is actually one root:
positive 5, and that is the solution to this example.
Okay now let's look at what happens when the
log expressions get a little more complicated.
In this third example, we have the sum of
two log expressions and then a constant, so
this example looks a little different in
terms of the format of the first two that we did.
So, let's analyze this. What could we possibly do to simplify this
to get it into a form where we could easily
use either the process we did for example
one, where we use the definition, or the
process we did for example two, where we use
the one to one property? Well, you're right.
What we can do is we can add these two
expressions, so let's review how this is done.
If you have the log base a of an expression
added to the log base a of another
expression, then yes, you can rewrite this
as the log base a of M times N.
Okay, same restrictions: a, M, N have to be
greater than 0, and what's the other one, a cannot equal 1.
So, let's check: a is 5, this is M, this is N, and so what can we do?
We can combine this expression, so we get
log base 5 of M times N, so that would be
x plus 6 times x plus 2 equals 1.
Alright, did we improve the situation?
Yes, we did because we have to multiply
these two terms in order to eventually
solve, but what's important about what we
did is we got this in standard form.
So, we have log, base, argument, and then
the constant here, which means we can go
back to the definition now, like we did in
example one, where we have log base a of x
equals y means a to the y equals x.
Same restrictions: x of course has to be greater than 0.
Okay, so let's do that.
As I copy this over, I'm going to multiply M
times N so I'm going to get x squared plus
8 x plus 12, that's this expression,
and then I'm going to rewrite the logarithm
in exponential form, so I would then have 5 to the 1.
Now in this case, I have these two sides
reversed, but this is an equality, and
that's fine, and we usually find it more
convenient to have the x on the left.
Alright, so let's solve this quadratic
equation: x squared plus 8 x, subtract 5
from both sides, and then let's see
if this factors easily. I believe it does.
Okay, we set each factor equal to 0, and we
get x equals negative 1 and x equals negative 7.
Are we finished? No, we have to check.
We have to check both roots, and not only
both roots but both roots in this expression
and in this expression. If we find one
situation where negative 1 does not produce
a positive argument here, then that is not a root.
Alright, so let's check the negative 1 first.
Okay, when I put negative 1 into here,
negative 1 substituted here would be 5, so
that's greater than 0. That's good.
Negative 1 here would be negative 1 plus 2, which is 1.
That means this is a valid root.
Even though it's a negative number,
it's not x that has to be positive, it's the entire expression.
Okay, let's check negative 7.
Did we just assume that it works or not works?
No, you always have to check it.
Alright, so negative 7 into here, what happens?
Right away, you see that you have negative 7
plus 6 which is negative 1, and that is
clearly not greater than 0. So this root does not check.
Let's go back and recap what we did here.
First, we used this law property to combine the two log terms.
Then, when we had this in standard form, we
used the definition of logarithm, just like
we did in example one, to rewrite this in
exponential form, solve for x, and then we always check.
Okay, let's look at one more example.
In our fourth example, let's look at what we have.
First of all, try to remember what does this mean?
This is log base e or natural log, so we
have four natural log expressions.
What can we do to solve for x?
Well, let's see if we can condense some of this.
Remember the property that looks like this:
if you have log base a of M minus
log base a of N. How can you combine these two expressions?
Rewrite it as this, exactly, same
restrictions as always in a, M, and N.
Okay, we're going to be able to use this
property on the left side and the right side.
Okay, we're going to rewrite this.
Bases are the same, so I'm going to rewrite
this as natural log of x which is M divided
by x plus 1, and on the right-hand side the
natural log of the quotient of x plus 3 and x plus 5.
Alright, we've simplified it, what do we do now?
Well, we generally have two things we can
use: we can use the definition, which is not
really going to be that helpful here, or we
can go back to that example two and the one to one property.
So, we have log base e of an expression
equals log base e of an expression.
Okay, so let's go back and look at that
property: log base a of M equals log base a
of N, and if this is true then M equals N,
which means that what I can do is I can
rewrite these two expressions in this form.
Do a little algebra, solve for x, and remember to check.
Okay, let's cross multiply.
So, x times x plus 5 would be x squared plus 5 x.
Multiplying across this way, I would have x squared plus 4 x plus 3.
Okay, well look what happens: x squared and
x squared, and so I have 5 x equals 4 x plus 3,
so I subtract 4 x from both sides, and I get x is 3.
Alright, we still have to check.
Once again, what are we checking for?
We are checking to be certain that this value is in the domain of
all four of these expressions: this one, this one, this one, and this one.
Well, let's see what happens. Three is positive, so 3 here checks
positive, greater than 0; 3 plus 1 is 4,
that's also greater than 0 so that checks;
it's going to check here, 6, and it's also going to check here, 8.
So, what does that mean? We have the solution, and we now have a
solution that has been checked.
These four examples are examples that you should go back and study.
You should rework them on your own, then you should go back and do
similar exercises for practice.
That's what it will take to build skill solving logarithmic equations.
Let's take a look at solving some exponential equations.
The technique that is used to solve a
specific exponential equation really depends
on what the original problem looks like, so
some analysis is always in order before you begin.
For example, in this exponential equation,
we have 3 to the x plus 1 equals 1 over 81.
The x is in the exponential position, and so
what we have to do is figure out how we should approach this problem?
The simplest thing to do here is to realize
that 81 can be rewritten as 9 times 9 or
3 to the 4th, which means 1 over 81 can be
rewritten as 1 over 3 to the 4th, which can
then further be rewritten as 3 to the negative 4.
This means that with very little effort
I can rewrite the original problem so that I have like bases.
When this happens, I can use a one-to-one property. Let's refresh.
If I have a to the M equals a to the N, then what is also true?
Right, M equals N. Here, as always,
the base a cannot be 1 and has to be greater than 0.
So, what can we do now?
Well, we can set the two exponents, the M
and the N, equal. Solve for x, and we get x is negative 5.
We're not required to check this as we are
when we're solving logarithmic equations to
check for whether or not the value is in the
domain, but it's always a good idea. Let's just check.
Negative 5 into here would be negative 5
plus 1, so this would be 3 to the negative 4,
which we know is 3 to the negative 4.
Working with a problem of this type is
called solving an exponential equation with like bases.
This only applies for certain problems that can be rewritten like this.
So, let's take a look at what happens when
you can't rewrite an expression with like bases.
In this example, 2 to the x equals 5.
We will not be able to rewrite each side in
terms of the same base, so we need to look for another method.
Recall the definition of logarithms that
allows you to change from exponential form
to log form: a to the y equals x means log base a of x equals y.
Be careful when you do this: these are
positional notations, and so this x in this
position is not the same as this x here.
So, what we actually have is: this is the 2,
this is the x, and this is the 5. So, let's rewrite in log form.
So, it would be log base 2 of 5 equals x.
Now this is an exact answer, but if you
needed to evaluate this on a scientific
calculator, you would have to have it in
either base e or base 10, so let's convert this.
Right now, it's in base 2. We have the change of base formula: a log
base a of b can be rewritten as the natural
log of b divided by the natural log of a.
That's using base e, or you can use base 10: log of b divided by log of a.
So, let's use natural log for this.
I would rewrite this expression as the
natural log of 5 divided by the natural log
of 2, and this is an exact value for x that
could be easily evaluated on a scientific calculator.
Let's take a look at another example.
In this example, 4 times 3 to the x plus 1 equals 6.
We want to solve for x. The first thing to notice is that this
exponential equation is not in what we consider standard form.
This 4 is not part of the base, the base is 3,
so the first step is going to be to rewrite this in standard form.
I'm going to divide both sides by 4, and so
I get 3 to the x plus 1 equals 6 over 4.
Now that it is in standard form, what method should we use?
Because we have a base, an exponent, and
a number, we can simply rewrite this as a log expression.
Remember that, if you have a to the y equals
x, that this means log base a of x equals y.
So, from this example, the a is the 3, the y
is the x plus 1 term, not to be confused
with this x term, and then the 6/4 is x.
So, when we rewrite in log form, we would
have a 3 here, we would have 6/4 here,
and x plus 1 out here. So, let's do that.
Log base 3 of 6/4 equals x plus 1.
We want to solve for x, so the next thing to
do is to subtract 1 from both sides.
We would have log base 3 of 6/4 minus 1, and
you can see that we're going to need
parentheses here because we want it to be
perfectly clear that the 6/4 is part of the
log, but the 1 is not, and we get x.
This is an exact value, but it is not in
either base e or base 10, which is where we
would want this if we had to evaluate.
So, we look at the change of base formula:
log base a of b can be rewritten as the
natural log of b divided by the natural log
of a or log base 10, so it would be log base 10 of b divided by log base 10 of a.
I'm going to use natural log, and so I could
rewrite this expression in parentheses as
the natural log of 6/4 divided by the
natural log of 3 and then minus 1 equals x. This is the exact value.
If you needed to evaluate this, you could do
this on a scientific calculator.
Now let's take a look at a more complicated example.
In this fourth example, we have 5 to the
x minus 2 equals 4 to the 3 x plus 1.
Right away, you can see that because you
have variables on both sides in the
exponential position, you're not going to be
able to use the definition immediately to simplify this.
So, instead what we're going to do is use
the following property: if M equals N, now
let's look at what an M is, M is this entire
expression, N is this expression, the two
expressions are equal, then we can rewrite
this as a log base a of M equals the log base a of N.
I would recommend either using natural log
or log base 10 because you may, at some
point, want to get an approximate answer,
and that's what your scientific calculator has.
So, applying this property and using natural
log, on the left side I would get the
natural log of 5 to the x minus 2, and on
the right side the natural log of 4 to the 3 x plus 1.
Okay, what have I done? Have I made this any easier to solve?
Well, it looks longer, but as it turns out,
we can do something else at this point that
will make it much simpler to solve.
We can recognize here that if we have the
log base a of x to the r, so that's log base
e of 5 to the x minus 2, where the x minus 2
is the r, we can rewrite this as r times log base a of x.
That's one of the properties.
So, what we're going to do is take this
expression and rewrite it with the r in the
front, in this case x minus 2 is the r.
So, we have x minus 2 times the natural log
of 5 and 3 x plus 1 times the natural log of 4.
The natural log of 5 is a constant.
Let's distribute it, so we get x ln 5, and then minus 2 ln 5.
On the right-hand side, we distribute ln 4
here and here, so we get 3 x ln 4 plus 1 ln 4.
Gathering the x terms on the left,
I subtract 3 x ln 4 from both sides.
Copy this expression, and add 2 ln 5.
Alright, let's look for a minute.
So, 3 x ln 4, because it has an x, has been subtracted from both sides.
This term does not have an x: it has been
added to both sides and is now here.
We want to factor out the x because that is what we are trying to solve.
So, when we factor out the x, what remains is ln 5 minus 3 ln 4.
One times ln 4 can be rewritten as ln 4, and then here we have 2 ln 5.
Dividing both sides by what is in
parentheses, I get x is the natural log of 4
plus 2 times the natural log of 5,
all divided by the natural log of 5 minus 3 natural log of 4.
This is the exact value of x using all of the
original numbers from the original problem.
There are things you could do to simplify
this, including using your calculator to get an approximate answer.
The best thing to do at this point is to
read the directions and decide what you are
trying to do in terms of the form of your answer.
You can see from these four examples that
there are varying techniques used to solve exponential equations.
Study these examples, practice them on your
own, and then practice similar examples
until you have achieved a high level of skill solving these equations.