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

MIRB16  Aug 13, 2017
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 #1
avatar+4488 
+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  .
\(\frac{1}{2^r}\) = 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  .
\(\frac{4}{32^r}\) = a    

 

 

And since  a = a .....

 

\( \frac1{2^r}\) = \(\frac4{32^r}\)                  
                 Since  20 = 1  ,  22 = 4  , and  25 = 32  , we can say...
\( \frac{2^0}{2^r}\) = \( \frac{2^2}{(2^5)^r}\)    
        And  (xa)b  =  xab
\( \frac{2^0}{2^r}\) = \( \frac{2^2}{2^{5r}}\)    
        And  \(\frac{x^a}{x^b}=x^{a-b}\)
20 - r = 22 - 5r    
        Now the bases are equal, so the exponents are equal.
0 - r = 2 - 5r    
         
0 = 2 - 4r    
         
-2 = -4r    
         
\(\frac12\) = r    
hectictar  Aug 13, 2017
 #2
avatar+18573 
+1

The function f(x) = axr satisfies

f(2) = 1 and

f(32) = 4.
Find r.

 

\(\begin{array}{|lrclclr|} \hline & \mathbf{f(x)} & \mathbf{=} & \mathbf{ax^r} \\\\ x=2 : & f(2) &=& a\cdot 2^r &=& 1 & (1) \\ x=32: & f(32) &=& a\cdot 32^r &=& 4 & (2) \\\\ \hline (2) \div (1) : & \frac{a\cdot 32^r} { a\cdot 2^r } &=& \frac{4}{1} \\ & \frac{ 32^r} { 2^r } &=& 4 \\ & \left( \frac{ 32 } { 2 } \right)^r &=& 4 \\ & 16^r &=& 4 \quad &&& | \quad 16 = 4^2 \qquad 4 = 4^1 \\ & 4^{2r} &=& 4^1 \\ & 2r &=& 1 \quad &&& | \quad : 2 \\ & \mathbf{r} & \mathbf{=} & \mathbf{ \frac12 } \\ \hline \end{array}\)

 

laugh

heureka  Aug 14, 2017

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