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The function f(x) = ax^r satisfies f(2) = 1 and f(32) = 4. Find r.

 Aug 13, 2017
 #1
avatar+9488 
+2
f(x) = a * xr    
         
f(2) = a * 2r    
                            Since  f(2) = 1  , we can replace  f(2)  with  1  .
1 = a * 2r    
        Divide both sides of the equation by  2r  .
12r = a    

 

 

f(32) = a * 32r    
                         Since  f(32) = 4  , we can replace  f(32)  with  4  .
4 = a * 32r    
        Divide both sides of the equation by  32r  .
432r = a    

 

 

And since  a = a .....

 

12r = 432r                  
                 Since  20 = 1  ,  22 = 4  , and  25 = 32  , we can say...
202r = 22(25)r    
        And  (xa)b  =  xab
202r = 2225r    
        And  xaxb=xab
20 - r = 22 - 5r    
        Now the bases are equal, so the exponents are equal.
0 - r = 2 - 5r    
         
0 = 2 - 4r    
         
-2 = -4r    
         
12 = r    
 Aug 13, 2017
 #2
avatar+26396 
+1

The function f(x) = axr satisfies

f(2) = 1 and

f(32) = 4.
Find r.

 

f(x)=axrx=2:f(2)=a2r=1(1)x=32:f(32)=a32r=4(2)(2)÷(1):a32ra2r=4132r2r=4(322)r=416r=4|16=424=4142r=412r=1|:2r=12

 

laugh

 Aug 14, 2017

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