Chapter 7.5, Problem 41HP

To determine

To write an exponential function which when graphed passes through the given points.

Answer to Problem 41HP

  $y=3e^{\ln \left(\frac{1}{2}\right)(x)}$

Explanation of Solution

Given:

Points: $(0,3)$ and $(1,6)$

Concept used:

General equation of exponential function:

  $y=a{e}^{\alpha (x)}$

Calculation:

Given that the points are $(0,3)$ and $(1,6).$

Using the equation of exponential function,

  $y=a{e}^{\alpha (x)}$...........(1)

Substitute 0 for x and 3 for yin the equation (1).

  $\begin{array}{l}3=a{e}^{\alpha (0)}\\ a=3\end{array}$

Substitute 3 for a in the equation (1).

  $y=3e^{\alpha (x)}$

Substitute 1 for x and 6 for y in the above equation.

  $\begin{array}{l}\Rightarrow 6=3e^{\alpha (1)}\\ \Rightarrow 2=e^{\alpha }\\ \Rightarrow \ln (2)=\ln (e^{\alpha })\text{ [Taking natural log on both sides]}\\ \Rightarrow \ln (2)=\alpha \end{array}$

Substitute $\alpha =\ln (2)$ in the above equation.

  $y=3e^{\ln (2)(x)}$

Conclusion:

Therefore, $y=3e^{\ln (2)(x)}$ is the exponential function for which the graph passes through $(0,3)$ and $(1,6).$