### Video Transcript

Simplify π₯ to the eighth power to the negative two power times π₯ to the negative six power to the fourth power over π₯ to the negative eight power times π₯ to the negative three power, given that π₯ is not equal to zero.

Letβs look at a few ways to solve this. The first way we can solve this is by taking our powers to a power. The power to a power is the rule that says π₯ to the π times π power is equal to π₯ to the π power to the π power. And that tells us that π₯ to the eighth power to the negative two power is the same thing as π₯ to the eight times negative two power, π₯ to the negative 16th power. We need to follow the same procedure for π₯ to the negative six to the fourth power. It becomes equal to π₯ to the negative six times four, π₯ to the negative 24th power. And weβll bring down our denominator.

The next thing we need to remember is that when we deal with exponents of the same base and we multiply them together, we add their exponents. All of our exponents here have a base of π₯. In the numerator, weβre multiplying π₯ to the negative 16th times π₯ to the negative 24th. Thatβs the same thing as saying π₯ to the negative 16 plus the negative 24. We add those together. And we have a numerator of π₯ to the negative 40th. In the denominator, π₯ to the negative eighth times π₯ to the negative three is equal to π₯ to the negative eight plus the negative three. Negative eight plus negative three equals negative 11.

We also know that one over π₯ to the π power equals π₯ to the negative π power. And by extension, that means that one over π₯ to the negative π power is equal to π₯ to the π power. In practice, what we usually say is that negative exponents in the numerator move to the denominator. And the sign becomes positive. And exponents that are negative in the denominator move to the numerator and become positive. So π₯ to the negative 40th over π₯ to the negative 11 is the same thing as π₯ to the 11th power over π₯ to the 40th power.

And now we need to do some canceling. We know the π₯ to the 11th power is equal to π₯ times itself 11 times. And π₯ to the 40th power is π₯ times itself 40 times. But instead of writing this all the way out, instead of expanding, we can think about it like this. 40 is equal to 11 plus 29. So π₯ to the 11th power times π₯ to the 29th power equals π₯ to the 40th power. If we bring our numerator across, we realize that the π₯ to the 11th power in the numerator and the denominator cancel out. And that leaves us with one over π₯ to the 29th power. And if we have a one as the numerator and our exponents are in the denominator, we usually rewrite it to say π₯ to the negative 29th power.

The simplified form of the given expression is π₯ to the 29th power. But I said I wanted to try a few different ways. So Iβm gonna clear this off and solve this again with a different method. This time, when we solve, I want to switch the exponents first. We have the exponent π₯ to the negative eight. That needs to move to the numerator and become positive. And π₯ to the eight to the negative two needs to move to the denominator and become positive. π₯ to the negative three moves to the numerator and becomes positive. π₯ to the negative six to the fourth power moves to the denominator and becomes positive.

And then I want to consider what π₯ to the eighth power squared means. We can rewrite that as π₯ to the eighth power times π₯ to the eighth power. We have π₯ to the eighth power times itself. From there, I wanna rewrite π₯ to the six power to the fourth power. Because I see in our numerator we have π₯ cubed, I know that I can write π₯ to the sixth power as π₯ cubed squared, because three times two equals six. And then weβre taking the fourth power. But remember, a power to a power means multiply. And two times four equals eight. This is what weβre saying. π₯ to the sixth power to the fourth power is the same thing as saying π₯ cubed to the eighth power. Both of them are versions of π₯ to the 24th power. Six times four equals 24. And eight times three equals 24.

And hopefully itβll be really clear why we did this. Because we have an π₯ cubed in the numerator. Imagine that our π₯ cubed is our π₯ cubed to the first power. So we have one π₯ cubed in the numerator. And we have eight π₯ cubed in the denominator. And we can say that the π₯ cubed in the numerator cancels out and that there are seven π₯ cubed terms left in the denominator. We can also cancel eight to the π₯ power over eight to the π₯ power because we have that term in the numerator and the denominator.

When we do this, we get one over π₯ to the eighth power times one over π₯ cubed to the seventh power. π₯ cubed to the seventh power is the same thing as π₯ to the 21st power. Three times seven equals 21. And we have these positive exponents with a numerator of one. So we rewrite them as π₯ to the negative eight times π₯ to the negative 21. And then because they have the same base, we can add the exponents together. π₯ to the negative eight plus negative 21 equals π₯ to the negative 29th. These are very different methods that give us the same final result, π₯ to the negative 29th power.