Laws of indicesAQA A-Level Maths: Revision notes

A power $a^{n}$ has base $a$ and index (exponent) $n$. For any base and any indices $m$ and $n$: $a^m\times a^n=a^{m+n},\qquad a^m\div a^n=a^{m-n},\qquad (a^m)^n=a^{mn}.$ Also $(ab)^n=a^nb^n$ and $\left(\frac ab\right)^n=\frac{a^n}{b^n}$. The laws only apply when the bases are the same (or you make them the same first, for example writing $8=2^3$). With numbers in front, deal with the coefficients and the powers separately: $3x^4\times2x^5=6x^9$.

$a^0=1\ (a\neq0),\qquad a^{-n}=\frac{1}{a^n},\qquad \left(\frac ab\right)^{-n}=\left(\frac ba\right)^n.$ A negative index means a reciprocal. It does not make the answer negative: $2^{-3}=\frac18$. For example $\left(\frac23\right)^{-2}=\left(\frac32\right)^2=\frac94$, and $\dfrac{1}{x^{-4}}=x^4$. Only the base immediately beneath the index is affected: $3x^{-2}=\frac{3}{x^2}$, but $(3x)^{-2}=\frac1{9x^2}$.

A fractional index is a root: $a^{\frac1n}=\sqrt[n]{a},\qquad a^{\frac mn}=\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m}.$ The denominator is the root and the numerator is the power. The laws of indices hold for all rational exponents, so $x^{\frac12}\times x^{\frac12}=x$ and $\sqrt{x}=x^{\frac12}$. Worked example. $27^{\frac23}=\left(\sqrt[3]{27}\right)^2=3^2=9$. With a negative sign, $27^{-\frac23}=\frac19$, and $\left(\frac{27}{8}\right)^{-\frac23}=\left(\frac{8}{27}\right)^{\frac23}=\left(\frac23\right)^2=\frac49$.

Apply the power to every factor inside a bracket, numbers included: $\left(8x^6\right)^{\frac23}=8^{\frac23}\left(x^6\right)^{\frac23}=4x^4.$ To simplify a fraction with a single term on the bottom, split it and subtract indices: $\dfrac{x^2+3x}{\sqrt x}=x^{2-\frac12}+3x^{1-\frac12}=x^{\frac32}+3x^{\frac12}$. Rewrite roots and reciprocals as powers of $x$ first, such as $\frac{1}{\sqrt x}=x^{-\frac12}$, then combine. For example $\dfrac{16x^{\frac32}\times x^{-\frac12}}{(8x^6)^{\frac13}}=\dfrac{16x}{2x^2}=\dfrac8x$.

To solve $x^{\frac mn}=c$, raise both sides to the reciprocal power $\frac nm$: $x^{\frac32}=8\Rightarrow x=8^{\frac23}=4,\qquad x^{-\frac32}=\frac18\Rightarrow x^{\frac32}=8\Rightarrow x=4.$ If a common factor appears, factorise instead of dividing, so that no solution is lost: $x^{\frac32}-x^{\frac12}=0\Rightarrow x^{\frac12}(x-1)=0$. Since $x^{\frac12}=0$ gives $x=0$, check whether the question's condition (such as $x>0$) allows it.