### Video Transcript

Solve log base five of π₯ squared
minus six log base five of π₯ plus eight equals zero, where π₯ is an element of the
real numbers.

Now, the fact that we have two
terms in log base five of π₯ shouldnβt throw us off here. What we need to spot is we have a
special type of quadratic equation. And so what weβre going to do is
make a substitution. Now, we donβt necessarily need to
do this, but it can make the whole process a little less complicated. Weβre going to let π¦ be equal to
log base five of π₯. Then log base five of π₯ all
squared becomes π¦ squared. Six log base five of π₯ is six
π¦. And so our full equation becomes π¦
squared minus six π¦ plus eight equals zero.

And now, we have quite a
straightforward, quadratic equation. We can solve for π¦ by first
factoring the expression on the left-hand side. It will be the product of two
binomials whose first term is π¦. And thatβs because when we
redistribute these, we get π¦ times π¦ which gives us the π¦ squared. And so next, weβre looking for two
terms whose product is eight and whose sum is negative six. Well, those two numbers are
negative four and negative two. Remember, negative four times
negative two is positive eight. But their sum is negative six as
required.

So our equation is π¦ minus four
times π¦ minus two equals zero. Well, π¦ minus four and π¦ minus
two are just numbers. And the only way for the product of
these two numbers to be equal to zero is if either π¦ minus four is equal to zero or
π¦ minus two is equal to zero. And so to solve the first equation,
we add four to both sides. And we find π¦ is equal to
four. Similarly, for our second equation,
we add two to both sides. And we get π¦ is equal to two. But weβre not finished. Our original equation is in terms
of π₯. And weβre looking to solve for
π₯. So we go back to our earlier
substitution. We let π¦ be equal to log base five
of π₯. And so we replace π¦ with log base
five of π₯. And we find our two equations are
now log base five of π₯ is equal to four and log base five of π₯ is equal to
two.

So how do we solve these? Well, essentially, the logarithm
answers the question for us. It answers what exponent do we need
for one number to become another number. Here, five is the base. Four is the power or the
exponent. And then π₯ is the number that
becomes. So we can actually rewrite
this. And we can say that π₯ must be
equal to five to the fourth power, which is 625. Similarly, in our second equation,
five is the base, two is the exponent, and, this time, π₯ is the number that
becomes. So π₯ is equal to five to the power
of two or five squared, which is equal to 25. And so the solution set to the
equation log base five of π₯ squared minus six log base five of π₯ plus eight equals
zero, where π₯ is an element of the real number set, is 25 and 625.