# How do you simplify (a^4c^2e^0)/(b^-1d^-3) using only positive exponents?

Apr 11, 2016

$\frac{{a}^{4} {c}^{2} {e}^{0}}{{b}^{-} 1 {d}^{-} 3} = 1 {a}^{4} b {c}^{2} {d}^{3}$

#### Explanation:

$\frac{{a}^{4} {c}^{2} {e}^{0}}{{b}^{-} 1 {d}^{-} 3}$ to simplify this to all positive exponents there are three aspects of the expression that can be changed

any value to the zero exponent = 1
${e}^{0} = 1$

a negative exponent in the denominator can be made positive by moving the value to the numerator $\frac{1}{b} ^ - 1 = b$ and $\frac{1}{d} ^ - 3 = {d}^{3}$

Therefore
$\frac{{a}^{4} {c}^{2} {e}^{0}}{{b}^{-} 1 {d}^{-} 3} = 1 {a}^{4} b {c}^{2} {d}^{3}$