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Which of the following is equal to three-fifths to the negative sixth power times three-fifths to the negative three power over three-fifths to the eighth power?
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A) Three-fifths to the 11th power, B) Three-fifths to the negative one power, C) Three-fifths to the negative 11 power, D) Three-fifths to the negative 17 power, or E) Three-fifths to the negative 25th power.
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First, let’s write down what we’re starting with.
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And before we do anything else, let’s try and remember some laws of exponents.
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We know the product rule which says that 𝑥 to the 𝑚 power times 𝑥 to the 𝑛 power is equal to the 𝑥 to the 𝑚 plus 𝑛 power.
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When we’re multiplying exponents with the same base, we add the two values together.
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The quotient rule when we’re dividing 𝑥 to the 𝑚 power by 𝑥 to the 𝑛 power, it will be equal to 𝑥 to the 𝑚 minus 𝑛 power.
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When we’re dividing two exponents with the same base, we subtract their values.
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And finally, negative exponents: If we have 𝑥 to the negative 𝑚 power, that’s the same thing as one over 𝑥 to the 𝑚 power.
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So back to our problem.
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It’s really tempting to try and distribute this to the negative six power and to the negative three power, so that we get something like this.
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But in the end, that doesn’t help us simplify very much.
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We need a different strategy.
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Imagine that we had 𝑥 to the negative six power times 𝑥 to the negative three power divided by 𝑥 to the eighth power.
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First, we would use our product rule and add the two exponents in the numerator.
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The base stays the same 𝑥.
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And negative six plus negative three equals negative nine.
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Now, we have a statement that says 𝑥 to the negative nine over 𝑥 to the eighth power.
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And so, we move on to the quotient rule.
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The base will stay the same.
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And we’ll subtract the exponent in the denominator from the exponent in the numerator.
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𝑥 to the negative nine minus eight equals 𝑥 to the negative 17.
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We can say let 𝑥 equal three-fifths.
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And if 𝑥 equals three-fifths, we can plug that in to have three-fifths to the negative 17 power, which is option D.
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Using this method prevents us from making calculation mistakes.
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Let’s look at a longer method that doesn’t use variable substitution.
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We start at the same place.
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And we recognize that the two numerators have the same base.
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So we can add their exponents, negative six plus negative three.
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And then, we copy down our denominator.
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You could then distribute the exponent across the fractions.
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Our numerator, three to the negative ninth power, over five to the negative ninth power is being divided by three to the eighth power over five to the eighth power.
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Division is the same thing as multiplying by the reciprocal.
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We have to multiply the numerator and multiply the denominator.
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But we switched the order because we can multiply in any order that we want.
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Which means we could say three to the negative ninth over three to the eighth times five to the eighth over five to the negative ninth.
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And then, we’ll have to use our quotient rule and use subtraction of exponents.
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Three to the negative ninth minus eight and five to the eighth minus negative nine gives us three to the negative 17 power times five to the 17th power.
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If we move that three to the negative 17th into the denominator, it becomes three to the positive 17th, which is five-thirds to the 17th power.
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But that’s not an answer choice.
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If we change three to the 17th power into the numerator, it needs to be three to the negative 17th power.
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And five to the negative 17th power can be moved into the denominator.
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Which shows that three-fifths to the negative 17th power is equal to what we started with.
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Taking so many steps to simplify could result very easily in us missing a negative sign or adjusting something incorrectly.
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At this stage, three over five to the negative ninth power over three-fifths to the eighth power.
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We could’ve simply said three over five to the negative nine minus eight power is equal to three over five to the negative 17th power.
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What matters here is recognizing that when we’re dealing with the same base, we can treat it as a single unit instead of trying to break it up into smaller pieces.
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And doing that saves us a lot of trouble.
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But both ways give us the final answer of D, three-fifths to the negative 17th power.