Question

Simplify and write $\frac{{\left({20}^3\right)}^5}{{\left(4^3\right)}^5\times {\left(5^5\right)}^3}$ as a single exponent.

Answer 1

We know that when we are having same bases and different powers, then the powers get added in multiplication and subtracted in case of division.

That means: $$\begin{gathered} a^m\times a^n=a^{m+n} \\ \frac{a^m}{a^n}=a^{m-n} \end{gathered}$$

So, on solving, we get :

$\frac{{\left({20}^3\right)}^5}{{\left(4^3\right)}^5\times {\left(5^5\right)}^3}$$$\begin{gathered} =\frac{{(4\times 5)}^{3\times 5}}{{\left(4\right)}^{3\times 5}\times {\left(5\right)}^{5\times 3}} \\ =\frac{{(4\times 5)}^{15}}{{\left(4\right)}^{15}\times {\left(5\right)}^{15}} \end{gathered}$$

$$\begin{gathered} =\frac{4^{15}\times 5^{15}}{4^{15}\times 5^{15}} \\ =4^{15-15}\times {15}^{15-15} \\ =4^0\times 5^0 \\ =1\times 1 \\ =1 \end{gathered}$$

Hence, $\frac{{\left({20}^3\right)}^5}{{\left(4^3\right)}^5\times {\left(5^5\right)}^3}$ is equal to $1$.