Chapter 10.7, Problem 59E

To determine

Tofind:the correct statement which express the given equation.

Answer to Problem 59E

Option $B$ is correct.

Explanation of Solution

Given:

  $\frac{5a^8{b}^5}{180a^6{b}^2}$ .

Concept used:

Equation can be taking power to all term:

  $\begin{array}{l}x<y<z.\\ x^n<y^n<z^n.\end{array}$

Equation can be simplified also by:

  $\begin{array}{l}x<y<z.\\ a<b<c.\\ xa<yb<cz.\end{array}$

Calculation:

Suppose $1<b<2\text{ and }2<a<3.$

Consider the expression:

  $\frac{5a^8{b}^5}{180a^6{b}^2}$ .

Simplify as:

  $\begin{array}{l}\frac{5\times a^6\times a^2\times b^2\times b^3}{5\times 36\times a^6\times b^2}.\\ \Rightarrow \frac{a^2{b}^3}{36}.\end{array}$

Cancel out equal terms in the numerator and the denominator.

  $\begin{array}{l}1<b<2{\text{ and 1}}^3<a<2^3.\\ \Rightarrow 1<b^3<8.\\ 2<a<3{\text{ and 2}}^2<a^2<3^2\text{.}\\ \Rightarrow {\text{4<a}}^2<9.\end{array}$

Consider the products respectively.

  $\begin{array}{l}1\times 4<a^2{b}^3<8\times 9.\\ \Rightarrow 5<a^2{b}^3<72.\end{array}$

Divide throughout with $36$ :

  $\begin{array}{l}\frac{5}{36}<\frac{a^2{b}^3}{36}<\frac{72}{36}.\\ \frac{5}{36}<\frac{a^2{b}^3}{36}<\mathrm{2........}\left(i\right)\end{array}$

Also, I tis clear that $4<5$ and so, $\frac{4}{36}<\frac{5}{36}$ .

That is $\frac{1}{9}<\frac{5}{36}\mathrm{.....}\left(ii\right)$

Putting g $\left(i\right)\text{ and }\left(ii\right)$ together;

  $\frac{1}{9}<\frac{5}{36}<\frac{a^2{b}^3}{36}<2$ .

In otherwords:

  $\frac{1}{9}<\frac{a^2{b}^3}{36}<2$ .

Hence, option $B$ is correct.