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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.
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Letβs look at a few ways to solve this.
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The first way we can solve this is by taking our powers to a power.
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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.
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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.
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We need to follow the same procedure for π₯ to the negative six to the fourth power.
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It becomes equal to π₯ to the negative six times four, π₯ to the negative 24th power.
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And weβll bring down our denominator.
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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.
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All of our exponents here have a base of π₯.
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In the numerator, weβre multiplying π₯ to the negative 16th times π₯ to the negative 24th.
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Thatβs the same thing as saying π₯ to the negative 16 plus the negative 24.
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We add those together.
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And we have a numerator of π₯ to the negative 40th.
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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.
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Negative eight plus negative three equals negative 11.
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We also know that one over π₯ to the π power equals π₯ to the negative π power.
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And by extension, that means that one over π₯ to the negative π power is equal to π₯ to the π power.
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In practice, what we usually say is that negative exponents in the numerator move to the denominator.
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And the sign becomes positive.
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And exponents that are negative in the denominator move to the numerator and become positive.
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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.
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And now we need to do some canceling.
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We know the π₯ to the 11th power is equal to π₯ times itself 11 times.
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And π₯ to the 40th power is π₯ times itself 40 times.
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But instead of writing this all the way out, instead of expanding, we can think about it like this.
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40 is equal to 11 plus 29.
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So π₯ to the 11th power times π₯ to the 29th power equals π₯ to the 40th power.
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If we bring our numerator across, we realize that the π₯ to the 11th power in the numerator and the denominator cancel out.
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And that leaves us with one over π₯ to the 29th power.
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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.
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The simplified form of the given expression is π₯ to the 29th power.
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But I said I wanted to try a few different ways.
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So Iβm gonna clear this off and solve this again with a different method.
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This time, when we solve, I want to switch the exponents first.
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We have the exponent π₯ to the negative eight.
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That needs to move to the numerator and become positive.
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And π₯ to the eight to the negative two needs to move to the denominator and become positive.
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π₯ to the negative three moves to the numerator and becomes positive.
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π₯ to the negative six to the fourth power moves to the denominator and becomes positive.
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And then I want to consider what π₯ to the eighth power squared means.
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We can rewrite that as π₯ to the eighth power times π₯ to the eighth power.
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We have π₯ to the eighth power times itself.
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From there, I wanna rewrite π₯ to the six power to the fourth power.
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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.
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And then weβre taking the fourth power.
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But remember, a power to a power means multiply.
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And two times four equals eight.
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This is what weβre saying.
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π₯ to the sixth power to the fourth power is the same thing as saying π₯ cubed to the eighth power.
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Both of them are versions of π₯ to the 24th power.
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Six times four equals 24.
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And eight times three equals 24.
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And hopefully itβll be really clear why we did this.
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Because we have an π₯ cubed in the numerator.
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Imagine that our π₯ cubed is our π₯ cubed to the first power.
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So we have one π₯ cubed in the numerator.
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And we have eight π₯ cubed in the denominator.
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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.
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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.
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When we do this, we get one over π₯ to the eighth power times one over π₯ cubed to the seventh power.
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π₯ cubed to the seventh power is the same thing as π₯ to the 21st power.
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Three times seven equals 21.
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And we have these positive exponents with a numerator of one.
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So we rewrite them as π₯ to the negative eight times π₯ to the negative 21.
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And then because they have the same base, we can add the exponents together.
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π₯ to the negative eight plus negative 21 equals π₯ to the negative 29th.
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These are very different methods that give us the same final result, π₯ to the negative 29th power.