### Video Transcript

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.