Question

Simplify: $\frac{\left(2^5\times 5^2\times t^8\right)}{\left({10}^3\times t^4\right)}$

Answer 1

Simplify $\frac{\left(2^5\times 5^2\times t^8\right)}{\left({10}^3\times t^4\right)}$ using the laws of exponents.

We know that, according to the laws of exponents:

$\begin{array}{l}{\mathbf{a}}^{\mathbf{m}\mathbf{}}\mathbf{÷}{\mathbf{a}}^{\mathbf{n}}\mathbf{}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{-}\mathbf{n}}\\ {\mathbf{a}}^{\mathbf{m}\mathbf{}}\mathbf{\times }{\mathbf{b}}^{\mathbf{m}}\mathbf{}\mathbf{=}{\left(\mathrm{ab}\right)}^{\mathbf{m}}\end{array}$

where $a$ and $b$ are non-zero integers and $m,n$ are any integers.

Therefore,

$\begin{array}{c}\frac{\left(2^5\times 5^2\times t^8\right)}{\left({10}^3\times t^4\right)}=\frac{2^5\times 5^2\times t^8}{{10}^3\times t^4}\\ =\frac{2^5\times 5^2\times t^8}{{\left(2\times 5\right)}^3\times t^4}\\ =\frac{2^5\times 5^2\times t^8}{2^3\times 5^3\times t^4}\mathbf{}\left[\because {\left(ab\right)}^m=a^m\times b^m\right]\\ =\frac{2^{5-3}\times t^{8-4}}{5^{3-2}}\left[\because a^m÷a^n=a^{m-n}\right]\\ =\frac{2^2\times t^4}{5^1}\\ =\frac{4t^4}{5}\end{array}$

Hence the simplified form of $\frac{\left(2^5\times 5^2\times t^8\right)}{\left({10}^3\times t^4\right)}$ is $\frac{4t^4}{5}$.