For positive integers \(m\) and \(n\text{,}\) \(\sqrt[n]{x^m}=x^{\sfrac{m}{n}}\text{.}\) This is universally true when \(x\) is positive. When \(x\) is negative, things are more complicated. It boils down to a choice. Some computer algebra systems choose to define an expression like \((-8)^{\sfrac{1}{3}}\) as a certain negative real number (namely, \(-2\)). Others choose to define an expression like \((-8)^{\sfrac{1}{3}}\) as a complex number in the upper half plane of the complex plane (\(1+i\sqrt{3}\)). Because of the ambiguity, some choose to simply declare expressions like \((-8)^{\sfrac{1}{3}}\text{,}\) with a negative base, to be undefined. The point is, if your computer tells you that say, \((-8)^{\sfrac{1}{3}}\) is undefined, you should realize that you may be expected to interpret \((-8)^{\sfrac{1}{3}}\) as \(-2\text{.}\)

For real numbers \(k\) and \(n\text{,}\) and \(x\neq0\text{,}\) \(\frac{k}{x^n}=kx^{-n}\text{.}\)

For a real number \(k\neq0\text{,}\) \(\frac{\fe{f}{x}}{k}=\frac{1}{k}\fe{f}{x}\text{.}\)

For positive real numbers \(A\) and \(B\) and all real numbers \(n\text{,}\)

• \(\fe{\ln}{AB}=\fe{\ln}{A}+\fe{\ln}{B}\)

• \(\fe{\ln}{\frac{A}{B}}=\fe{\ln}{A}-\fe{\ln}{B}\)

• \(\fe{\ln}{A^n}=n\fe{\ln}{A}\)