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Which of the following is equal to two to the negative two power times 27 to the negative one-third power times 25 to the negative one-half power over four to the negative three power.
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A) One over 240, B) 15 over 16, C) 16 over 15, D) 48 over five, or E) 240
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One thing that we notice almost immediately is that all four of these exponent values are negative.
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And we know that π₯ to the negative π power is equal to one over π₯ to the π power.
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That is, when you have a negative exponent in the numerator, you can move that exponent to the denominator and make it positive.
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And by extension, we can say one over π₯ to the negative π power equals π₯ to the positive π power.
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In our case, it means all the values with negative exponents in the numerator should be moved to the denominator.
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And they will be positive.
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Remember, this is only possible because we are multiplying all three of these values together.
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We now have in the denominator two to the positive two power times 27 to the positive one-third power times 25 to the positive one-half power.
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And four to the negative three power can be moved to the numerator as four to the positive three power.
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If we keep our numerator the same, we can calculate two squared, which is four.
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And then, we have to remember that 27 to the one-third power is equal to the cube root of 27.
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Itβs saying that some value cubed equals 27.
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And if π₯ cubed equals 27, then π₯ equals 27 to the one-third power, or the cube root of 27.
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We know that three times three times three equals 27; three cubed equals 27.
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And that means the cube root of 27 is three.
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And so, in place of 27 to the one-third power, we can write three.
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After that, we have the square root of 25.
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If π₯ squared equals 25, then π₯ equals 25 to the one-half power, which is the same thing as the square root of 25.
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We know the square root of 25 is five because five times itself is 25.
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There is one more thing we can simplify.
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If we look at our four in the denominator, we could say that that is four to the first power.
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And since we have four cubed in the numerator and four to the first power in the denominator, we can simplify that to have four squared in the numerator.
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This is because when the exponents have the same base, π₯ to the π power over π₯ to the π power equals π₯ to the π minus π power.
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In our case, we had four cubed over four to the first power.
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And that is four to the three minus one power, four squared.
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Three times five is 15.
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Four squared is 16.
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So, this simplification is 16 over 15, which is option C.