#### Answer

$b^{x-y}=\frac{b^{x}}{b^{y}}$
This is known as second Law of Exponents; where x and y are real numbers.

#### Work Step by Step

We already know that $b^{x}=e^{xlnb}$
Now,
$b^{x-y}=e^{(x-y)lnb}$
This implies
$b^{x-y}=e^{xlnb-ylnb}=\frac{e^{xlnb}}{e^{ylnb}}$
Hence, $b^{x-y}=\frac{b^{x}}{b^{y}}$
This is known as second Law of Exponents; where x and y are real numbers.