### Video Transcript

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.