Solving for unknowns
Posted: Tue Feb 03, 2015 5:11 pm
Solve for $x$ and $y$ for which $(0.2)^{x}(0.5)^{y} = 200$.
Solution
$(0.2)^{x}(0.5)^{y} = 200$
$\left(\frac{2}{10} \right)^{x} \left(\frac{5}{10} \right)^{y} = (2^{3})(5^{2})$
$\left(\frac{1}{5} \right)^{x} \left(\frac{1}{2} \right)^{y} = (2^{3})(5^{2})$
$(5^{-x})(2^{-y}) = (2^{3})(5^{2})$
$(5)^{-x}(2)^{-y} = (5^{2}) (2^{3})$
Comparing powers of like terms for both sides:
$-x = 2$
$x = -2;$
$-y = 3$
$y = -3.$
Therefore $x = -2 \ \text{and} \ y = -3$.
Any Alternative approach?
Solution
$(0.2)^{x}(0.5)^{y} = 200$
$\left(\frac{2}{10} \right)^{x} \left(\frac{5}{10} \right)^{y} = (2^{3})(5^{2})$
$\left(\frac{1}{5} \right)^{x} \left(\frac{1}{2} \right)^{y} = (2^{3})(5^{2})$
$(5^{-x})(2^{-y}) = (2^{3})(5^{2})$
$(5)^{-x}(2)^{-y} = (5^{2}) (2^{3})$
Comparing powers of like terms for both sides:
$-x = 2$
$x = -2;$
$-y = 3$
$y = -3.$
Therefore $x = -2 \ \text{and} \ y = -3$.
Any Alternative approach?