Solving ${\left( {\frac{{9.1}}{{10 - x}}} \right)^{0.1}} = {\left( {\frac{{9 + y}}{{9.1}}} \right)^{0.9}}$

Question

how to find real $x$ and $y$ such as ${\left( {\frac{{9.1}}{{10 - x}}} \right)^{0.1}} = {\left( {\frac{{9 + y}}{{9.1}}} \right)^{0.9}}$ ?

Answer 1

Well solving for $y$ you get $y=9.1\times(\frac{9.1}{10-x})^\frac{1}{9}-9$. So just pick any $x$ not equal to $10$ and then find $y$ using the above formula.

Answer 2

Notice that if each side is $1$, then the equation is verified. To make those fractions $1$, take $x = 0.9$ and $y = 0.1$.