13. x 2 = x 2 = x 2 = x 2 = x 3 = x 3 = x 4 = x 4 = x 5 = x 5 =
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1 Section 8. Eponents and Roots Eercises In Eercises -, compute the eact value (/) 4. (/) (/) 8. (/) (/) 4. (/6). In Eercises -4, perform each of the following tasks for the given equation. i. Load the left- and right-hand sides of the given equation into Y and Y, respectivel. Adjust the WINDOW parameters until all points of intersection (if an) are visible in our viewing window. Use the intersect utilit in the CALC menu to determine the coordinates of an points of intersection. ii. Make a cop of the image in our viewing window on our homework paper. Label and scale each ais with min, ma, min, and ma. Label each graph with its equation. Drop dashed vertical lines from each point of intersection to the -ais, then shade and label each solution of the given equation on the -ais. Remember to draw all lines with a ruler. iii. Solve each problem algebraicall. Use a calculator to approimate an radicals and compare these solutions with those found in parts (i) and (ii).. = 4. = 7. = 7 6. = 7. = 6 8. = = = 7. = 8. = 4. 6 = 4. 6 = 9 In Eercises -40, simplif the given radical epression Coprighted material. See: Version: Fall 007
2 766 Chapter 8 Eponential and Logarithmic Functions In Eercises 7-64, simplif the product, and write our answer in the form r Compare and contrast ( ) and ( ). 4. Compare and contrast 4 ( ) 4 and ( 4 ) Compare and contrast ( ) and ( ). 44. Compare and contrast ( ) and ( ). In Eercises 4-6, compute the eact value In Eercises 6-7, simplif the quotient, and write our answer in the form r Version: Fall 007
3 Section 8. Eponents and Roots In Eercises 7-80, simplif the epression, and write our answer in the form r Version: Fall 007
4 Chapter 8 Eponential and Logarithmic Functions 8. Solutions. = = 4. = = 6 = 6 7. = ( = ) 8 7 = = 7. = 6 6. Using a graphing calculator, note that the graph of = intersects the graph of = in two places. The intersect was used to determine the -values of the points of intersection. These are labeled on the -ais on the image that follows. 0 = = To solve the problem algebraicall, the solutions of = are called square roots of and are denoted = = ± ± Version: Fall 007
5 Section 8. Eponents and Roots A calculator was used to find the approimation. This result agrees nicel with the approimations found using the intersect utilit above.. In the image that follows, note that the graph of = does not intersect the graph of = 7. Therefore, the equation = 7 has no real solutions. 0 = 0 0 = 7 0 To solve the problem algebraicall, note that it is not possible to square a real number and obtain 7. Therefore, the equation has no real solutions. = 7 7. In the image that follows, note that the graph of = intersects the graph of = 6 in one location. The intersect utilit on the graphing calculator was used to find the -value of this point of intersection and this approimation is placed on the -ais. 0 = = 6 To solve the problem algebraicall, note that the solution of = 6 is called the cube root of 6 and is denoted b 0 Version: Fall 007
6 Chapter 8 Eponential and Logarithmic Functions = 6 = A calculator was used to obtain the last approimation. Note that this approimation agrees nicel with the approimation found above with the intersect utilit. 9. Note that the graph of = 4 intersects the graph of = 4 in two places. Therefore, the equation 4 = 4 has two real solutions, which are found with the calculator s intersect utilit and labeled on the -ais of the image that follows. 0 = 4 = To solve the equation algebraicall, note that the solutions of 4 = 4 are called fourth roots of 4 and are denoted b 4 = 4 = ± 4 4 ±.446 A calculator was used to find the last approimation. Note that these agree nicel with the results found with the intersect utilit above.. Note that the graph of = intersects the graph of = 8 in one location. Hence, the equation = 8 has one real solution, which is found using the intersect utilit and labeled on the -ais in the image that follows. Version: Fall 007
7 Section 8. Eponents and Roots =8 0 = To solve the equation algebraicall, note that the solution of = 8 is called the fifth root of 8 and is denoted b = 8 = The last approimation was found b a calculator and agrees nicel with the approimation found with the intersect utilit above.. Note that the graph of = 6 does not intersect the graph of =. Hence, the equation 6 = has no real solutions. 0 = = 0 To solve the equation algebraicall, note that it is not possible to raise a real number to the sith power and get. Hence, the equation has no real solutions. 6 =. The notation 49 calls for the positive square root of 49. Note that 7 = 49. Thus, 49 = 7. Version: Fall 007
8 Chapter 8 Eponential and Logarithmic Functions 7. The notation 6 calls for the positive square root of 6. It is not possible to square a real number and get 6. Therefore, 6 is not a real number. 9. The notation 7 calls for the cube root of 7. Note that = 7. Hence, 7 =.. The notation calls for the cube root of. Note that ( ) =. Hence, =.. The notation 4 6 calls for the positive fourth root of 6. Note that it is not possible to raise a real number to the fourth power and get 6. Thus, 4 6 is not a real number.. The notation 4 6 calls for the positive fourth root of 6. Note that 4 = 6. Hence, 4 6 =. 7. The notation calls for the fifth root of. Note that ( ) =. Hence, =. 9. The notation 04 calls for the fifth root of 04. Note that 4 = 04. Thus, 04 = ( ) = 4 =. However, is not defined, so ( ) is not a real number. 4. ( ) = = and ( ) =. Therefore, the two epressions are equal. 4. = = ( ) = = = (8 ) 4 = 4 = = (6 ) = 4 = = (7 ) = = = (6 4 ) = 4 = = 6 4 = (6 4 ) = 4 = = = 9. = + = = 4 4 = 8 Version: Fall 007
9 Section 8. Eponents and Roots 6. = = = 4 = = + = = + 4 = = 4 = = ( )( 4 ) = = ( 4)( ) = 8 = ( )( ) = 4 = ( )( ) = 0 Version: Fall 007
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