Question

how does a to the 4 b to -5 over c to -3 d to the 6th get simplified?

Answer 1

Given the expression

$\frac{a^4b^{-5}}{c^{-3}d^6}$

To simplify the expression above, we convert all negative indices to positive indices

Applying the rule of indices

$a^{-x}=\frac{1}{a^x}$

Applying the rule to the given expression gives

$\begin{gathered} \text{Where b}^{-5}=\frac{1}{b^5}\text{ and} \\ c^{-3}=\frac{1}{c^3} \end{gathered}$

Substitute the above deduction into the given expression

$\begin{gathered} \frac{a^4b^{-5}}{c^{-3}d^6}=a^4\times\frac{1}{b^5}\times\frac{1}{\frac{1}{c^3}}\times\frac{1}{d^6} \\ \text{Where }\frac{1}{\frac{1}{c^3}}=c^3 \\ =a^4\times\frac{1}{b^5}\times\frac{1}{\frac{1}{c^3}}\times\frac{1}{d^6}=a^4\times\frac{1}{b^5}\times c^3\times\frac{1}{d^6} \\ =\frac{a^4c^3}{b^5d^6} \\ \frac{a^4b^{-5}}{c^{-3}d^6}=\frac{a^4c^3}{b^5d^6} \end{gathered}$

Hence, the simplified form is

$\frac{a^4c^3}{b^5d^6}$