Question

If a and b are different positive primes such that

(i) ${\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)=a^x{b}^y$, find x and y.

(ii) ${\left(a+b\right)}^{-1}\left(a^{-1}+b^{-1}\right)=a^x{b}^y$, find x + y + 2.

Answer 1

(i)
$$\begin{gathered} {\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)=a^x{b}^y \\ \Rightarrow \left(\frac{a^{-7}{b}^{14}}{a^{14}{b}^{-28}}\right)÷\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)=a^x{b}^y \\ \Rightarrow \left(a^{-7-14}{b}^{14+28}\right)÷\left(a^{3+2}{b}^{-5-3}\right)=a^x{b}^y \\ \Rightarrow \left(a^{-21}{b}^{42}\right)÷\left(a^5{b}^{-8}\right)=a^x{b}^y \\ \Rightarrow a^{-21-5}{b}^{42+8}=a^x{b}^y \\ \Rightarrow a^{-26}{b}^{50}=a^x{b}^y \\ \Rightarrow x=-26\mathrm{and}y=50 \end{gathered}$$

(ii)
$$\begin{gathered} {\left(a+b\right)}^{-1}\left(a^{-1}+b^{-1}\right)=a^x{b}^y \\ \Rightarrow \frac{1}{\left(a+b\right)}\left(\frac{1}{a}+\frac{1}{b}\right)=a^x{b}^y \\ \Rightarrow \frac{1}{\left(a+b\right)}\left(\frac{a+b}{ab}\right)=a^x{b}^y \\ \Rightarrow \left(\frac{1}{ab}\right)=a^x{b}^y \\ \Rightarrow {\left(ab\right)}^{-1}=a^x{b}^y \\ \Rightarrow a^{-1}{b}^{\mathbf{-}\mathbf{1}}=a^x{b}^y \\ \Rightarrow x=-1\mathrm{and}y=-1 \end{gathered}$$
Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.