Chapter 4.7, Problem 17E

Solution

To find: A Power function which passes through the given points.

  $y=1.251x^{1\cdot 262}$

Given: Two points are given as $\left(2,3\right)$ and $\left(6,12\right)$.

Concept used:

If a function $y=f\left(x\right)$ passes through a point $\left(x_0,y_0\right)$ , then the point $\left(x_0,y_0\right)$ satisfies the equation of the function, that is, $y_0=f\left(x_0\right)$.

Calculation:

As the equation $y=a{\left(x\right)}^b$ passes through the point $\left(2,3\right)$ ; therefore,

  $3=a{(2)}^b$ …… (1)

Again, as the equation passes through the point $\left(6,12\right)$ ; therefore,

  $12=a{(6)}^b$ …… (2)

Now, find the value of $a$ from the equation (1):

  $a=\frac{3}{{\left(2\right)}^b}$ …… (3)

Further, substitute the value of $a$ in equation (2);

  $\begin{array}{c}12=3\left(\frac{6^b}{2^b}\right)\\ \frac{12}{3}={\left(\frac{6}{2}\right)}^b\\ 4={\left(3\right)}^b\end{array}$

On both sides, take the logarithm with base 2:

  $\begin{array}{c}lo{g}_{2}4=b({\log }_{2}3)\left[{\log }_e{m}^n=n{\log }_{e}m\right]\\ lo{g}_2{\left(2\right)}^2=b(1\cdot 584)\left[{\log }_{2}3=1\cdot 584\right]\\ 2lo{g}_2\left(2\right)=b(1\cdot 584)\left[{\log }_{n}n=1\right]\end{array}$

  $\begin{array}{c}2=b(1\cdot 584)\\ \frac{2}{1\cdot 584}=b\end{array}$

Thus,

  $b=1\cdot 262$

Now, substitute the value of $b$ in above equation (3);

  $\begin{array}{c}a=\frac{3}{{(2)}^{1\cdot 262}}\\ a=\frac{3}{2\cdot 398}\left[2^{1\cdot 262}=2\cdot 398\right]\\ a=1\cdot 251\end{array}$

Substitute the values of $a$ and $b$ in general equation of power function $y=a{\left(x\right)}^b$ :

  $y=1.251x^{1\cdot 262}$

Conclusion:

Hence, the power function that satisfies the given points is $y=1.251x^{1\cdot 262}$.