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Rational powers in practice
Solving exercises about basic properties of rational powers.
Hello. Let us try to solve the exercise 1 of the rational powers in practice step. We want to simplify the following three expressions. Let us divide our light board in three parts to solve the three little exercises. First, consider 2 squared times 2 to the minus 1 over 2. Then you see you have the same base. And therefore, to get the product, it is sufficient just to sum the two exponents.
2 plus minus 1 over 2, which is equal to 2 with exponent 3 over 2. Now, I can rewrite this one as 2 to 1 plus 1 over 2, which is equal 2 to 1 2 with the exponent 1 over 2. And this is, of course, equal to 2 times square root of 2.
And now, consider the second expression. Then we have 3 cubed minus 5 squared over 2 to the 5 times 2 to minus cube. OK, now, regarding the numerator we just computed, these two numbers. And we get 27 minus 25 over– here, again, you have the same base and then you can just add the two exponents– 2 to 5 minus 3, which is equal to 2 over 2 squared. Again, this can be written like 2 times 2 to minus 2, which is equal, because here we have a product between powers with the same base. And therefore, we can just add the two exponents. We have 2 minus 1, which is equal to 1 over 2. And finally, consider the last expression.
2 cubed with the exponent 2 cubed over 2 3 squared with exponent 3. Now, when you have the power of a power, you have to multiply the exponents. Therefore, this is equal to 2 cube times 2 cubed over 2 3 squared times 3. And this is exactly like to write 2 3 times 2 cubed times 2 to minus 3 cubed times 3. Now, again, we have the same base. And therefore, we can just add the two exponents. OK, maybe we can just collect one copy of 3. And then we have 2 with the exponent 3 times 2 cubed minus 3 squared, which is equal 2 cubed times– OK, 2 cubed is equal to 8. 3 squared is equal to 9.
Therefore, 8 minus 9, which is equal to 2 minus 3, which is 1 over 2 to the 3rd, which is 1 over 8. Thank you for your attention. Hello. Let us consider the exercise number 2 of the rational powers in practice step. OK, we have two real numbers, y and x, with y greater or equal than x, which is greater or equal than 0. And we want to compare these two quantities, 2 times x to 3 over 5, and this expression. OK, let us try to solve this exercise.
We have to ask ourselves, if 2 times x with the exponent 3 over 5, is less or equal or greater than y times y with exponent minus 2 over 5 plus y with exponent of minus 3 over 5 times the 5th root of y. OK, this second expression is equal to y times y minus 2 over 5 plus y y minus 3 over 5 times the 5th root of y. How can we rewrite this?
OK, y to 1 is exactly like to write y to 5 over 5. And I repeat, y to minus 2 over 5 plus, again, y to 5 over 5 y minus 3 over 5. And now, the 5th root of y is exactly y to 1 over 5.
Here, we have a product of two powers with the same base. Therefore, it is sufficient just to add the two exponents, and we get y to 3 over 5. Again here, we have the y to 2 over 5 times y 1 over 5. But again, here we have the same base. And then we get y to 3 over 5, which is 2 times y to 3 over 5. OK, now, you see, the function which sends z to z with the exponent of 3 over 5 is defined on positive real numbers, and is strictly increasing. Therefore, x to 3 over 5 is surely less or equal than y to 3 over 5 under our hypothesis.
But also, 2 times x to 3 over 5 is less or equal to 2 times y 3 over 5. Therefore, which is the disequality between the two initial quantities. Observe that, of course, for x equal y, these two quantities are equal. And otherwise, the quantity on the left is less than the quantity on the right. Therefore, we can conclude that the relation of it, between the two initial quantities, is the following one. 2 times x to 3 over 5 is less or equal than this expression.
The following exercises are solved in this step.
We invite you to try to solve them before watching the video.
In any case, you will find below a PDF file with the solutions.
Exercise 1.
Simplify the following expressions: [a) 2^2cdot 2^{-frac{1}{2}},quad b) frac{3^3-5^2}{2^5cdot 2^{-3}},quad c) frac{(2^3)^{(2^3)}}{(2^{(3^2)})^3}.]
Exercise 2.
Let (ygeq xgeq0) and (ynot=0). Compare the quantities [2x^{frac{3}{5}}quadtext{and} quad yleft(y^{-frac{2}{5}}+y^{-frac{3}{5}}sqrt[5]{y}right).]
Exercise 3. [Solved only in the PDF file]
Compare the quantities [left(frac{1}{7}right)^{frac{1}{5}}quadtext{and}quad sqrt[5]{left(frac{1}{7}right)^4}.]
This article is from the online course:
Precalculus: the Mathematics of Numbers, Functions and Equations
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Precalculus: the Mathematics of Numbers, Functions and Equations
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