Chapter 7.4, Problem 81PPE

To determine

To find: suppose $\frac{a^x}{a^y}=a^3$ and $\frac{a^x}{a^{3y}}=a^{-5}$ . Find the values of x and y.

Answer to Problem 81PPE

The required answer is: $x=7$ and $y=4$

Explanation of Solution

Given information:

The given information: $\frac{a^x}{a^y}=a^3$ and $\frac{a^x}{a^{3y}}=a^{-5}$

Calculation:

  $\begin{array}{l}\frac{a^x}{a^y}=a^3\\ a^{x-y}=a^3\end{array}$

(Subtract exponents when dividing powers with the same base)

So, $x-y=3$ (equation 1)

  $\begin{array}{l}\frac{a^x}{a^{3y}}=a^{-5}\\ a^{x-3y}=a^{-5}\end{array}$

(Subtract exponents when dividing powers with the same base)

So, $x-3y=-5$ (equation 2)

Now, subtract equation 2 from equation 1

  $\begin{array}{l}\underset{\_}{\begin{array}{l}x-y=3\\ (-)x-3y=-5\end{array}}\\ x-x-y-(-3y)=3-(-5)\end{array}$

  $\begin{array}{l}\underset{\_}{\begin{array}{l}x-y=3\\ (-)x-3y=-5\end{array}}\\ -y+3y=3+5\end{array}$

  $\begin{array}{l}\underset{\_}{\begin{array}{l}x-y=3\\ (-)x-3y=-5\end{array}}\\ 2y=8\end{array}$

  $\frac{2y}{2}=\frac{8}{2}$

(Divide 2 on both sides)

  $y=4$

(Simplify)

Place the value of $y=4$ in equation 1

  $\begin{array}{l}x-y=3\\ x-4=3\\ x-4+4=3+4\end{array}$

(Add 4 on both sides)

  $x=7$

(Simplify)