### Video Transcript

Simplify four to the three π plus
three power times 25 to the one minus three π power all divided by two to the nine
π plus three power times 50 to the one minus three π power.

Here, we are working with bases of
four, 25, two, and 50. And if we can get these bases to be
alike these properties, π₯ to the π times π₯ to the π is equal to π₯ to the π
plus π power and π₯ to the π divided by π₯ to the π is equal to π₯ to the π
minus π power. And if we would have π₯ to the
negative π power, we could move that to the denominator and make the exponent
positive. So it would be one over π₯ to the
π.

So we need to somehow figure out
how to rewrite these bases so they are all alike. So here are the bases that weβre
working with. We can rewrite four as two
squared. So it would be a base of two. Two is already a base of two. Itβs two to the first power. 25 we can rewrite as five
squared.

However, 50 isnβt a perfect
square. But since weβve been using bases of
two and five, could we somehow rewrite 50 using them? Well, 50 is 25 times two and 25 is
equal to five squared. So, weβve written 50 using a base
of five and two. So, letβs begin to simplify.

So beginning with our numerator, we
will replace the four with two squared and 25 with five squared. And now to simplify, we must
distribute the exponents. And we have two to the six π plus
six power times five to the two minus six π power. We will now use this property, but
backwards. So notice π₯ to the π times π₯ to
the π equals π₯ to the π plus π. And what we have so far? We have two to the six π plus six
power β so we need to separate those β and five to the two minus six π power and we
can separate those. So, we have two to the six π plus
six power.

So we could rewrite that as two to
the six π power times two to the six power and then five to the second power times
five to the negative six π power. And since we have a negative
exponent, we will use this property. So we need to move this since itβs
already on the numerator to the denominator and then we can change it to be a
positive six π power, which we would have here.

Now we also could have used the
fact that when we subtract our exponents, itβs like dividing. So, we would have five to the
second power divided by five to the six π power. Thatβs how we have gotten the
subtraction right here. Okay, so weβve simplified the
numerator. Now, letβs simplify the
denominator.

So weβll put a one on top to
represent the fact that these numbers are on the denominator. So we have our bases of two and
50. So we will keep our base of
two. And 50 we decided to rewrite it as
five squared times two, which weβve done here. So now, we need to simplify the
five squared times two to the one minus three π power. And now, weβve separated them. Before moving on, letβs just
distribute the two to the one and the negative three π. So we have five to the two minus
six π power.

Now, letβs use this property again
to separate them. So we have first separated the nine
π and the three. So weβve two to the nine π power
times two to the third power times five squared times five to the negative six π
power times two to the first power times two to the negative three π power. So weβve two negative powers. And we can make them positive by
moving them up to the numerator. So now, weβve completely simplified
the denominator.

So here we had simplified our
numerator. But we had moved five to the six π
to the denominator. And here, we simplified our
denominator, but moved these up to the numerator. So we need to put them
together. And we can put them together by
multiplying because everything on top was supposed to go to the numerator and
everything on the bottom was supposed to go to denominator.

Beginning with the numerator, we
have two to the six π and two to the three π and weβre multiplying them. So, we can rewrite that as two to
the six π plus three π power. And six π plus three π is nine
π. Then, we had two to the six power
times five to the second power times five to the six π power. So, we didnβt put any of the fives
together because we had a two and a six π. And those do not simplify
together. We only put the two to the six π
and two to the three π together because they were both πs.

On the denominator, weβve five to
the six π times two to the nine π. And now, we have two to the third
power and then two, just two to the first power. So, that would be two to the three
plus one power, which is two to the fourth power, times five squared. So now a few things can cancel. The two to the nine πs cancel. The five squares cancel. The five to the six π powers
cancel. And we are left with two to the six
power divided by two to the fourth power.

And when dividing like bases with
exponents, we subtract their exponents and six minus four is two. And two squared is equal to
four. So, after simplifying, our final
answer will be four.