Chapter 4.7, Problem 6E

Solution

To find: An exponential function which passes through the given points.

  $y=0\cdot 125{\left(2\right)}^x$ And $y=-0\cdot 125{\left(-2\right)}^x$

Given: Two points are given as $\left(3,27\right)$ and $\left(5,243\right)$.

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 $y=a{\left(b\right)}^x$ passes through the point $\left(3,27\right)$ ; therefore,

  $27=a{b}^3$ …… (1)

Again, as the equation passes through the point $\left(5,243\right)$ ; therefore,

  $243=a{b}^5$ …… (2)

Therefore by equation (2) and (1);

  $\frac{a{b}^5}{a{b}^3}=\frac{243}{27}$

  $b^2=9$

Thus;

  $b=3$ And $b=-3$

Now, substitute the value of $b=3$ in the equation (1);

  $\begin{array}{l}a=\frac{27}{27}\\ a=1\end{array}$

Now, substitute the value of $b=-3$ in the equation (1);

  $a=-1$

Thus,

Two set of value of $a$ and $b$ is possible one is $\left(1,3\right)$ and another is $\left(-1,-3\right)$.

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

  $y={\left(3\right)}^x$ And $y=-{\left(-3\right)}^x$

Conclusion:

Hence, the exponential function that satisfies the given points is $y={\left(3\right)}^x$ and $y=-{\left(-3\right)}^x$.