Chapter 4.7, Problem 20E

Solution

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

  $y=0.214x^{2\cdot 387}$

  $a=$

Given: Two points are given as $\left(5,10\right)$ and $\left(12,81\right)$.

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 $\left(5,10\right)$ ; therefore,

  $10=a{(5)}^b$ …… (1)

Again, as the equation passes through the point $\left(12,81\right)$ ; therefore,

  $81=a{(12)}^b$ …… (2)

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

  $a=\frac{10}{5^b}$ …… (3)

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

  $\begin{array}{c}\frac{81}{10}=\frac{{12}^b}{5^b}\\ \frac{81}{10}={\left(\frac{12}{5}\right)}^b\end{array}$

On both sides, take the logarithm with base 2:

  $\begin{array}{c}{\log }_{2}81-{\log }_{2}10=b({\log }_{2}12-{\log }_{2}5)\\ $ {\log }_2{(3)}^4-{\log }_{2}10=b({\log }_{2}12-{\log }_{2}5)\left[{\log }_e{m}^n=n{\log }_{e}m\right]$4{\log }_{2}3-{\log }_{2}10=b({\log }_{2}12-{\log }_{2}5)\\ 4(1\cdot 584)-3\cdot 321=b(3\cdot 584-2\cdot 321)\end{array}$

Thus;

  $\begin{array}{c}6\cdot 336-3\cdot 321=b(1\cdot 263)\\ 3\cdot 015=(1\cdot 263)b\\ b=\frac{3\cdot 015}{1\cdot 263}\\ b=2\cdot 387\end{array}$

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

  $\begin{array}{c}a=\frac{10}{{(5)}^{2\cdot 387}}\\ a=\frac{10}{46\cdot 605}\\ a=0\cdot 214\end{array}$

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

  $y=0.214x^{2\cdot 387}$

Conclusion:

Hence, the power function that satisfies the given points is $y=0.214x^{2\cdot 387}$.