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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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hectictar · Answer 1 · 2017-08-13T11:48:24

f(x)	=	a * x^r

f(2)	=	a * 2^r
			Since f(2) = 1 , we can replace f(2) with 1 .
1	=	a * 2^r
			Divide both sides of the equation by 2^r .
$\frac{1}{2^r}$	=	a

f(32)	=	a * 32^r
			Since f(32) = 4 , we can replace f(32) with 4 .
4	=	a * 32^r
			Divide both sides of the equation by 32^r .
$\frac{4}{32^r}$	=	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 (x^a)^b = x^ab
$\frac{2^0}{2^r}$	=	$\frac{2^2}{2^{5r}}$
			And $\frac{x^a}{x^b}=x^{a-b}$
2^{0 - r}	=	2^{2 - 5r}
			Now the bases are equal, so the exponents are equal.
0 - r	=	2 - 5r

0	=	2 - 4r

-2	=	-4r

$\frac12$	=	r

heureka · Answer 2 · 2017-08-14T09:26:04

The function f(x) = ax^r satisfies

f(2) = 1 and

f(32) = 4.
Find r.

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

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