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# Help!!!!!

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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
+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
+18573
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The function f(x) = axr satisfies

f(2) = 1 and

f(32) = 4.
Find r.

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

heureka  Aug 14, 2017

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