In Exercises 117–124, simplify each exponential expression.(2x²y⁴...

Question

In Exercises 117–124, simplify each exponential expression.(2x²y⁴)⁻¹(4xy³)⁻³ / (x²y)⁻⁵(x³y²)⁴

Accepted Answer

Hello, today we're gonna be simplifying the following expression. So what we are given is seven X cubed Y to the fifth race to the power of negative two times two X squared Y to the fourth race to the power of negative four all over X to the fourth Y squared race to the power of negative four times X to the fifth Y cubed race to the power of five. Now we have some negative exponents that we need to rewrite and we can turn the na negative exponents positive by using the exponential rule for negative exponents. And that states that if we have a to the power of negative M, if we take the reciprocal, we can rewrite this as a to the power of M. And the same also applies for the opposite. If we have one over A to the power of negative M, this can be rewritten as A to the power of M because that would be the reciprocal. So for every value in the numerator that has a negative exponent, we can rewrite that as a positive exponent by moving these quantities to the denominator. Furthermore X to the power of four times Y squared race to the power of negative four can be written with a positive exponent if we move it to the numerator, so we can rewrite this expression to give us X to the power of four Y squared race. The power of positive four all over seven X cubed Y to the fifth race to the power of two times two X squared Y to the fourth race to the power of four times X to the fifth Y cubed race to the power of five. Now using our exponential multiplication, we have a property that states that if we have a race to the power of M race to the power of N, we can take the product of the exponents and rewrite this as a to the power of M times N. So we're going to use this property to simplify the outermost exponents. We're going to distribute the exponent of four to X to the fourth and Y squared, the exponent of 2 to 7 X cubed and Y to the fifth, the exponent of 4 to 2 X squared and Y to the fourth and the exponent of five to X to the fifth and Y cubed. So in the numerator X to the power of four times four is going to give us X to the power of 16 and Y to the power of two times four is going to give us Y to the power of eight. And in the denominator, we have seven squared times X to the power of three times two, which is going to give us X to the power of six times Y to the power of five times two, which is going to give us Y to the power of 10 times 2 to the power of four times X to the power of two times four, which is going to give us X to the eighth times Y to the power of four times four, which is going to give us Y to the power of 16. And finally X to the power of five times five, which is going to give us X to the power of 25 times Y to the power of three times five, which is going to give us Y to the power of 15. So we have quite a bit to simplify in the denominator. Let's go ahead and group up the constant values together, group up all the X values together and group up all the Y values together. So for the constant, we have seven squared times 2 to the power of four, seven squared is just going to be seven times itself two times which is going to give us 49 and two to the power of four is just two times itself four times which is going to give us 16 And 49 times is going to simplify to give us 784. So that's what the constants in the nme denominator are going to simplify too. And for the X terms and Y terms in the denominator, we can use the exponential property for addition to simplify those values that states that if we have a, to the power of M times A to the power of N, you can simplify these two values by taking the sum of the exponents. And that is going to give us a A to the power of M plus N. So that's going to be the next property we're going to use in the denominator. So in the denominator, we have X to the power of six times X to the power of eight times X to the power of 25 we can simplify these three values by taking the sum of the exponents that is going to give us X to the power of six plus eight plus 25 which is going to simplify to give us X to the power of 39. And for the Y values, we have Y to the power of 10 times Y to the power of 16 times y to the power of 15. If we add these values together, this is going to give us Y to the power of 10 plus 16 plus 15, which is going to simplify to give us why to the power of 41. So we can rewrite the expression as X to the power of 16 times y to the power of eight, all over 784 times x to the power of 39 times y to the power of 41. And we can simplify this even further by using the division property for exponential values. The division property states that if we have a to the power of M over A to the power of N, this can be simplified by rewriting this as A to the power of M minus N. So for X to the 16th over X to the 39 we can simplify this quantity to give us X to the power of 16 minus 39 X to the power of 16 - is going to give us X to the power of -23. And if we take the reciprocal of this exponential value, this can be simplified as one over X to the power of 23. And the same thing applies for the Y terms we have Y to the power of eight over Y to the power of 41. This can be simplified by use by Y to the power of eight minus 41 which itself can be rewritten as Y to the power Of -33. And taking the reciprocal of this value gives us one over Y to the power of 33. So we can rewrite the expression as one over 784 times x to the power of times Y to the power of 33. And this is going to be the simplified form of the given expression. And with that being said, the answer to this problem is going to be B so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 117–124, simplify each exponential expression.(2x²y⁴)⁻¹(4xy³)⁻³ / (x²y)⁻⁵(x³y²)⁴

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