How do i put (5x)-5/4 in radical form

assuming you mean \((5x)^ \frac{-5}{4}\)

remember
\(x^{-m}= \frac{1}{x^m}\) and
\(x^ \frac{m}{n}= \sqrt[n]{x^m}\)

combining we get
\(x^ \frac{-m}{n}= \frac{1}{x^\frac{m}{n}} = \frac{1}{ \sqrt[n]{x^m} }\)
so

\((5x)^ \frac{-5}{6}= \frac{1}{(5x)^\frac{5}{4}} = \frac{1}{ \sqrt[4]{(5x)^5} } =\frac{1}{ 5x\sqrt[4]{5x} }\)

Answer Link

jdoe0001 jdoe0001 · Answer 2 · 2017-02-21T18:26:32-05:00

\(\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad \sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}} \\\quad \\a^{-\frac{{ n}}{{ m}}} = \cfrac{1}{a^{\frac{{ n}}{{ m}}}} \implies \cfrac{1}{\sqrt[{ m}]{a^{ n}}}\qquad\qquad \cfrac{1}{\sqrt[{ m}]{a^{ n}}}= \cfrac{1}{a^{\frac{{ n}}{{ m}}}}\implies a^{-\frac{{ n}}{{ m}}} \\\\ -----------------------------\\\\\)

\(\bf thus \\\\ (5x)^{-\frac{5}{4}}\implies \cfrac{1}{(5x)^{\frac{5}{4}}}\implies \cfrac{1}{\sqrt[4]{(5x)^5}}\implies \cfrac{1}{\sqrt[4]{5^5x^5}} \\\\ \cfrac{1}{5x\sqrt[4]{5x}}\)