Chapter 6.1, Problem 65E

To determine

To determine the values of x and y .

Answer to Problem 65E

  $x=8,y=-1.$

Explanation of Solution

Given information :

  $\frac{b^x}{b^y}=b^9$

  $\frac{b^x\cdot b^2}{b^{3y}}=b^{13}$

We know that when the base of the power is same, then we can compare the powers of the equation.

Now we know $\frac{b^x}{b^y}=b^9$ or $b^{x-y}=b^9$

Hence we get,

  $\begin{array}{l}x-y=9\\ x=9+y\end{array}$

Similarly we know $\frac{b^x\cdot b^2}{b^{3y}}=b^{13}$

Hence we get,

  $\begin{array}{l}\frac{b^x\cdot b^2}{b^{3y}}=b^{13}\\ \frac{b^{x+2}}{b^{3y}}=b^{13}\\ b^{x+2-3y}=b^{13}\\ x+2-3y=13\end{array}$

Now put the value of x we calculated earlier, we get,

  $\begin{array}{l}x+2-3y=13\\ \left(9+y\right)+2-3y=13\\ 11-2y=13\\ y=-1\end{array}$

Put value of y in any equation we get, $x=8$

Therefore we get $x=8,y=-1.$