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If 3^(x + y) = 81 and 81^(x - y) = 9 then what is the value of the product xy? Express your answer as a common fraction.

 Dec 10, 2020
 #1
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If 3^(x + y) = 81 and 81^(x - y) = 9 then what is the value of the product xy? Express your answer as a common fraction.     

 

34 = 81 so (x+y) must equal 4  

 

81–2 = 9 so (x–y) must equal –2  

 

                                                                x + y  =  4  

                                                                x – y  =  –2  

 

Add those two equations and obtain       2x       =  2     therefore x =1  

 

Sub 2 back into one of the equations       1 + y  =  4     therefore y=3  

 

                                                                 x • y  =  1 • 3  = 3   

 

I suppose, to comply with the terms of the                                3  

problem, the answer should be expressed                     xy  =  ––  

as a fraction.                                                                             1  

 Dec 10, 2020
 #2
avatar+117546 
+1

3^(x +y)   = 81

81^(x - y)  =  9

 

Write everything in terms of base 3

 

3^(x +y)  = 3^(4)

3^[4(x -y)]  = 3^(2)

 

Equate  the exponents  ......we have  these  equations

 

x + y  = 4

4 [ x - y] =  2   →   2x - 2y  = 1  →  x - y  =1/2

 

x + y =  4

x - y  =1/2           add these equations

 

2x  = 9/2

x = 9/4

 

And

x + y =  4

9/4 + y  =16/4

y = 7/4

 

xy  =  (9/4)(7/4)   =   63/16

 

cool cool cool

 Dec 11, 2020
edited by CPhill  Dec 11, 2020

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