Question

If a and b are different primes such that $\left(i\right){\left[\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right]}^7÷\left[\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right]=a^x{b}^y,findxandy.$

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

Answer 1

$\left(i\right){\left[\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right]}^7÷\left[\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right]=a^x{b}^y\Rightarrow \frac{a^{-7}{b}^{14}}{a^{14}{b}^{-28}}÷\frac{a^3{b}^{-5}}{a^{-2}{b}^3}=a^x{b}^y\Rightarrow \frac{a^{-7}{b}^{14}}{a^{14}{b}^{-28}}\times \frac{a^{-2}{b}^3}{a^3{b}^{-5}}=a^x{b}^y\Rightarrow a^{-7-14-2-3}.b^{14+28+3+5}=a^x{b}^y\Rightarrow a^{-26}\times b^{50}=a^x{b}^{y}Comparing,weget,x=-26,y=50$

$\left(ii\right)(a+b{)}^{-1}(a^{-1}+b^{-1})=a^x{b}^y\Rightarrow (a+b{)}^{-1}(\frac{1}{a}+\frac{1}{b})=a^x{b}^y\Rightarrow \frac{1}{a+b}\times \frac{b+a}{ab}=a^x{b}^y\Rightarrow \frac{1}{ab}=a^x{b}^y\Rightarrow a^{-1}{b}^{-1}=a^x.b^{y}Comparing,x=-1,y=-1Nowputthevalueofxandyinx+y+2,x+y+2=-1-1+2=0$