Chapter 3.5, Problem 1E

To determine

To match:

The given models with their equations.

Answer to Problem 1E

  $\begin{array}{l}\left(a\right)\text{ matches with }\left(iv\right)\\ \left(b\right)\text{ matches with }\left(i\right)\\ \left(c\right)\text{ matches with }\left(iii\right)\\ \left(d\right)\text{ matches with }\left(vi\right)\\ \left(e\right)\text{ matches with }\left(ii\right)\\ \left(f\right)\text{ matches with }\left(v\right)\end{array}$

Explanation of Solution

Given:

Some random models and some equations

  $\begin{array}{l}\left(a\right)\text{Exonential growth model}\left(i\right)y=a{e}^{-bx},b>0\\ \left(b\right)\text{Exponential decay model}\left(ii\right)y=a+b\ln x\\ \left(c\right)\text{Logistic growth model}\left(iii\right)y=\frac{a}{1+b{e}^{-rx}}\\ \left(d\right)\text{Gaussian model}\left(iv\right)y=a{e}^{bx},b>0\\ \left(e\right)\text{Natural logarithmic model}\left(v\right)y=a+b{\log }_{10}x\\ \left(f\right)\text{Common logarithmic model}\left(vi\right)y=a{e}^{-\frac{{\left(x-b\right)}^2}{c}}\end{array}$

The exponential growth model is given by $y=a{e}^{bx},b>0$ and the exponential decay model is given by $y=a{e}^{-bx},b>0$ . Therefore, (a) matches with (iv) and (b) matches with (i).

Logistic growth model is given by $y=\frac{a}{1+b{e}^{-rx}}$ . Therefore, (c) matches with (iii).

Gaussian model is given by $y=a{e}^{-\frac{{\left(x-b\right)}^2}{c}}$ . Therefore, (d) matches with (vi).

Natural logarithmic model is given by $y=a+b\ln x$ , and common logarithmic model is given by $y=a+b{\log }_{10}x$ . Therefore, (e) matches with (ii), and (f) matches with (v).