Question

How do you re-write this using only positive exponents?

3x^-4/3 (1+2x^5/3)

Answer 1

`\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\ a^{-{ n}} \implies \cfrac{1}{a^( n)} \qquad \qquad \cfrac{1}{a^( n)}\implies a^{-{ n}} \qquad \qquad a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}} \\\\\\ and\qquad \quad a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^( n)} \qquad \qquad \sqrt[{ m}]{a^( n)}\implies a^{\frac{{ n}}{{ m}}}\\\\ -------------------------------\\\\`


`\bf \cfrac{3x^{-(4)/(3)}}{3(1+2x^{(5)/(3)})}\implies \cfrac{3}{3}\cdot \cfrac{1}{x^{(4)/(3)}}\cdot \cfrac{1}{(1+2x^{(5)/(3)})}\implies \cfrac{1}{x^{(4)/(3)}(1+2x^{(5)/(3)})}`


`\bf \cfrac{1}{x^{(4)/(3)}+2x^{(5)/(3)}x^{(4)/(3)}}\implies \cfrac{1}{x^{(4)/(3)}+2x^{(5)/(3)+(4)/(3)}}\implies \cfrac{1}{x^{(4)/(3)}2x^{(9)/(3)}}\implies \cfrac{1}{x^{(4)/(3)}+2x^3} \\\\\\ \cfrac{1}{\sqrt[3]{x^4}+2x^3}\implies \cfrac{1}{x\sqrt[3]{x}+2x^3}\implies \cfrac{1}{x(\sqrt[3]{x}+2x^2)}`