how do you work out$${\left({\frac{{\mathtt{4}}}{{\mathtt{9}}}}\right)}^{{\mathtt{1.5}}}$$? I know the answers $${\frac{{\mathtt{8}}}{{\mathtt{27}}}}$$ but not why?
Why don't you join up? You are asking some good questions today :)
$$\\\left(\frac{4}{9}\right)^{1.5}\\\\
=\left(\frac{4}{9}\right)^{3/2}\\\\
=\left(\left(\frac{4}{9}\right)^{1/2}\right)^3\\\\
=\left(\frac{\sqrt4}{\sqrt9}\right)^3\\\\
=\left(\frac{2}{3}\right)^3\\\\
=\frac{2^3}{3^3}\\\\
=\frac{8}{27}\\\\$$
Why don't you join up? You are asking some good questions today :)
$$\\\left(\frac{4}{9}\right)^{1.5}\\\\
=\left(\frac{4}{9}\right)^{3/2}\\\\
=\left(\left(\frac{4}{9}\right)^{1/2}\right)^3\\\\
=\left(\frac{\sqrt4}{\sqrt9}\right)^3\\\\
=\left(\frac{2}{3}\right)^3\\\\
=\frac{2^3}{3^3}\\\\
=\frac{8}{27}\\\\$$