+9479 | 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 |
+26393 The function f(x) = axr satisfies
f(2) = 1 and
f(32) = 4.
Find r.
\(\begin{array}{|lrclclr|} \hline & \mathbf{f(x)} & \mathbf{=} & \mathbf{ax^r} \\\\ x=2 : & f(2) &=& a\cdot 2^r &=& 1 & (1) \\ x=32: & f(32) &=& a\cdot 32^r &=& 4 & (2) \\\\ \hline (2) \div (1) : & \frac{a\cdot 32^r} { a\cdot 2^r } &=& \frac{4}{1} \\ & \frac{ 32^r} { 2^r } &=& 4 \\ & \left( \frac{ 32 } { 2 } \right)^r &=& 4 \\ & 16^r &=& 4 \quad &&& | \quad 16 = 4^2 \qquad 4 = 4^1 \\ & 4^{2r} &=& 4^1 \\ & 2r &=& 1 \quad &&& | \quad : 2 \\ & \mathbf{r} & \mathbf{=} & \mathbf{ \frac12 } \\ \hline \end{array}\)
