### Video Transcript

In this video, we will learn how to
solve logarithmic equations with like bases using laws of exponents and
logarithms. Before we do that, letβs review how
logs work. If this is a general statement for
a logarithmic equation, itβs made up of a few pieces. We have logarithmic, abbreviated
log. And we often say this is the log
with a base π of some value π₯ is equal to π. The log base π of π₯ equals π is
mathematically equivalent to the statement π₯ is equal to π to the π power. The log base π of π₯ is asking the
question what power of π equals π₯, which we see here is π. If π to the π power equals π₯,
then the log base π of π₯ equals π.

In order to solve problems where
weβre dealing with more than one logarithm, but where these logs have the same
bases, weβll need to remember some properties of logs. We can start with the product
rule. If we have some log base π of π₯
times π¦, we can rewrite that as log base π of π₯ plus log base π of π¦.

The quotient rule is a bit
similar. In this case, if we have log base
π of π₯ divided by π¦, we can rewrite that as log base π of π₯ minus log base π
of π¦. From there, we have the power
rule. If we have log base π of π₯ to the
π power, we can rewrite that as π times the log base π of π₯. And then we have the log of
one. This is the log base π of π, and
that equals one.

Remember, the log is trying to
answer the question what power do we take π to to get π. π to the first power is equal to
π. So the log base π of π equals
one. And the final principle that will
be really helpful when working with logarithms with like bases is this. If the log base π of π₯ is equal
to the log base π of π¦, then π₯ equals π¦.

Letβs look at an example where
weβre dealing with logs with the same base.

Find π₯ such that log base six of
π₯ plus log base six of three is equal to three.

The first thing we notice here is
that we have two log values both with a base of six. When weβre working with problems
like this, the first step is usually to simplify the equation using laws of
logarithms. In this equation, weβre adding two
different logs with the same base, which should remind us of the product rule that
tells us log base π of π₯ plus log base π of π¦ is equal to log base π of π₯
times π¦. And that means we can combine these
two logarithms by multiplying the values in the log. Weβll then have log base six of
three times π₯, of three π₯, so that log base six of three π₯ equals three.

At this point, we can no longer do
any simplification. So weβll want to rewrite this log
in exponential form. We remember if log base π of π₯
equals π, then π₯ equals π to the π power. The base π becomes the base of our
exponential equation. The three is the exponent, which is
set equal to the value we were taking the log of. Since six cubed equals 216, we have
three π₯ equals 216. And from there, we divide both
sides of the equation by three, to see that π₯ equals 72.

The process here is to simplify and
then rewrite the equation in exponential form. This was a very simple example. But very often when we do problems
like this, the simplification and the rearranging will be what we spend most of our
time doing.

Letβs look at some examples where
we have to do that.

Given that the log base 12 of π£
plus three equals one, find the value of π£.

Letβs consider this in two
different ways. The first way is to rewrite this
log in exponential form. If we know that log base π of π₯
equals π, then we can rewrite that as π to the π power equals π₯. For us, that will be 12 to the
first power equals π£ plus three. We know that 12 to the first power
equals 12. So we subtract three from both
sides of the equation. And we see that nine equals π£ or
more commonly π£ equals nine.

But is there another way we can
think about this question? Well, if we know that log base π
of π equals one, we could rewrite one in the form log base 12 of 12, since log base
12 of 12 does equal one. And then since weβre dealing with
two logs with the same base, set equal to each other, we can say that π£ plus three
is equal to 12. And again, weβll subtract three
from both sides, to see that π£ equals nine. Both methods are valid ways to
rearrange the equation to solve for π£.

Letβs consider another example.

Determine the solution set of the
equation log base eight of π₯ minus six plus log base eight of π₯ plus six is equal
to log base eight of 64 in all real numbers.

If we look at our equation, we see
that we are adding two log values that have the same base, which means we can use
the product rule log base π of π₯ plus log base π of π¦ equals log base π of π₯
times π¦. However, we need to be careful
here. In these logarithms, the π₯-value,
the factor of our first log, will be π₯ minus six and the π¦-value, the factor of
our second log, will be π₯ plus six. And this means that weβll be
dealing with the log base eight of π₯ minus six times π₯ plus six, which will be
equal to log base eight of 64.

We wanna make sure weβre correctly
dealing with whatβs up inside the log, which means we can FOIL. Or we can recognize that this is
the difference of two squares. When we have π₯ plus π times π₯
minus π, itβs equal to π₯ squared minus π squared. That would mean for us π₯ squared
minus six squared, which equals 36. Now we have log base eight of π₯
squared minus 36 is equal to log base eight of 64. And once we have only one log on
either side of the equal sign and theyβre taken to the same base, we can say that π₯
squared minus 36 must be equal to 64.

To solve, we add 36 to both sides
of the equation. And we see that π₯ squared equals
100. If we take the square root of both
sides of the equation, we end up with π₯ is equal to plus or minus 10. But again, weβll need to do some
checks. Since our π₯-value goes inside a
log, and we know that we canβt take the log of a negative value, if we plugged in
negative 10 for π₯, we would be trying to find the value of a negative log, which is
not possible. And so we want to say that π₯ in
fact does not equal negative 10. And if we check positive 10, we are
taking the log of positive values. And so we can say the only value
that π₯ can be is 10. The solution set is then just
10.

In our next example, weβll consider
logs where we also have a square root.

Find the solution set of log base
eight of the square root of nine π₯ minus 26 plus log base eight of the square root
of π₯ plus one is equal to log base eight of 128 minus two in all real numbers.

If we copy down our equation, we
see that weβre dealing with three terms that are log base eight. But one of our terms is a
constant. So letβs focus on the right side of
the equation, the side with the constant first. We have a few options, but itβs
going to require us to think creatively.

One way of solving this is writing
two as a log base eight. We do that by remembering log base
π of π is one. And that means we could rewrite the
constant two as two times log base eight of eight. That would be two times one. The other thing we remember is that
if we have a constant multiplied by a log base π of π₯, thatβs the same thing as
log base π of π₯ taken to the π power. This means the constant two can be
rewritten as log base eight of eight squared, which in fact will work very well for
us. So weβll just bring down this log
base eight of 128.

Now, if we have log base π of π₯
minus log base π of π¦, we can rewrite that as log base π of π₯ divided by π¦. And this means we can rewrite log
base eight of 128 minus log base eight of eight squared as a log base eight of 128
divided by eight squared. Eight squared is 64. 128 divided by 64 is two, which
means weβve been able to simplify the log base eight of 128 minus two down to log
base eight of two.

From there, we can turn our
attention to the left side of the equation. Both of our terms are log base
eight. But weβre taking the log of the
square root of both of these. But this will be much more helpful
if we rewrite the square root as to the one-half power. Once we do that, weβre able to pull
out that one-half power so that we now have one-half times the log base eight of
nine π₯ minus 26 plus one-half times log base eight of π₯ plus one. Since one-half is a factor of both
of these terms, we can undistribute it.

Now, inside these brackets, we have
two logs with the same base that are being added together. And that means we can use the rule
log base π of π₯ plus log base π of π¦ is equal to log base π of π₯ times π¦. Weβll now have a log base eight of
nine π₯ minus 26 times π₯ plus one. Weβll need to distribute and expand
here so that we have nine π₯ squared plus nine π₯ minus 26π₯ minus 26. Nine π₯ minus 26π₯ equals negative
17π₯.

At this point, weβre very close to
something we can work with. But we donβt have exactly something
we can work with yet because of this one-half. And so what we want to do is
actually put this one-half back in exponent form. We want to instead call this the
log base eight of nine π₯ squared minus 17π₯ minus 26 to the one-half power. And hopefully, youβll see why in
just a second. Because if we do that, we now have
a log base π of π₯ equal to a log base π of π¦, which tells us π₯ is equal to π¦,
which means nine π₯ squared minus 17π₯ minus 26 to the one-half power is equal to
two.

We can get rid of that one-half
power by squaring both sides of our equation, which gives us nine π₯ squared minus
17π₯ minus 26 equals four. If we subtract four from both
sides, we get the quadratic nine π₯ squared minus 17π₯ minus 30 equals zero. And this is something that we can
factor to solve for π₯.

At this point, weβre not really
dealing with any log rules. Weβre simply factoring to solve a
quadratic. So I wanna find the terms that
multiply together. Positive 10 and negative three
multiply together to equal negative 30. And negative 27 plus 10 equals
negative 17. Setting both of our factors equal
to zero, we see that π₯ will either be negative ten-ninths or π₯ will be equal to
three.

However, because of our properties
of logs, we know that we canβt take the log of a negative value. If we try to plug in negative
ten-ninths into nine π₯ minus 26, we would end up with a negative value. And that tells us that negative
ten-ninths is not a valid solution for π₯. If we use the same method to check
for three, nine times three minus 26 is positive. And if we substitute three in for
π₯ plus one, again, we get a positive value. The solution set for π₯ here is
just positive three.

In our final example, weβll solve a
logarithmic equation where the base is unknown.

Find the solution set of log base
π₯ of five plus log base π₯ of 40 minus two times log base π₯ of four is equal to
two plus log base π₯ of eight in all real numbers.

First, weβll copy down our
equation. And though there are a few
strategies we could use, letβs just start by working from left to right. So we have log base π of π₯ plus
log base π of π¦, which will be equal to log base π of π₯ times π¦, which means we
can combine our first two terms and call them log base π₯ of five times 40, log base
π₯ of 200. And then weβll just bring
everything else down.

Weβre close to being able to
combine log base π₯ of 200 and two times log base π₯ of four. And we can use the rule that π
times log base π of π₯ is equal to log base π of π₯ to the π power, which means
two times log base π₯ of four is equal to log base π₯ of four squared. And then weβre subtracting two logs
that have the same base. And we know when that is the case,
we can divide each of their values. log base π₯ of 200 minus log base π₯ of four
squared will be equal to log base π₯ of 200 over 16.

It seems like thatβs all we can do
here on the left side of the equation. What we can do now is we can move
this log base π₯ of eight to the left side of the equation by subtracting log base
π₯ of eight from both sides. We now have log base π₯ of 200 over
16 minus log base π₯ of eight. And weβll use this subtraction rule
again, which will be log base π₯ of 200 over 16 divided by eight. We know that divided by eight is
the same thing as multiplied by one-eighth. And if we do that simplification,
we end up with log base π₯ of 25 over 16 is equal to two.

And now we want to take this out of
logarithm form and put it into exponent form. If we have log base π of π₯ equals
π, then we can rewrite that as π₯ equals π to the π power. So we have 25 over 16 is equal to
π₯ squared. By taking the square root of both
sides of the equation, we get that π₯ is equal to plus or minus five over four,
since the square root of 25 is five and the square root of 16 is four. Since weβre only looking for values
that are in the set of all reals, we want to say that π₯ cannot be negative
five-fourths. We donβt wanna deal with a negative
base here, as this would yield an answer that is imaginary. And so π₯ is only equal to
five-fourths, which makes the solution set just five-fourths.

Before we finish, letβs review the
key points. To solve logarithmic equations with
like bases, first simplify the equation using law of logarithms. This might be enough to solve. If not, we can proceed to rewrite
the logarithm in exponential form.