Laws of logarithmsEdexcel International A Level Maths: Revision notes

A logarithm answers the question: what power of the base gives this number? For $a>0$, $a\neq1$ and $x>0$: $\log_a x=y\iff a^y=x.$ So $\log_28=3$ because $2^3=8$, and $\log_{10}1000=3$. A logarithm is only defined for positive $x$, because $a^y$ is always positive. Putting $y=1$ gives $a^1=a$, so $\log_aa=1$ for any valid base. All the laws below are the index laws written in logarithm form, and they need the same base throughout.

$\log_a(xy)=\log_ax+\log_ay.$ Proof: let $\log_ax=m$ and $\log_ay=n$, so $x=a^m$ and $y=a^n$. Then $xy=a^{m+n}$, so $\log_a(xy)=m+n$. Example: $\log_26+\log_22=\log_212$. If $\log_a2=p$ and $\log_a5=q$ then $\log_a10=p+q$, and $\log_a20=\log_a4+\log_a5=2p+q$.

$\log_a\left(\frac xy\right)=\log_ax-\log_ay,\qquad\log_a\left(\frac1x\right)=-\log_ax.$ These come from $\frac{a^m}{a^n}=a^{m-n}$ and $\frac1{a^m}=a^{-m}$. Order matters in the quotient law: $\log_a\left(\frac52\right)=\log_a5-\log_a2$, not $\log_a2-\log_a5$. Example: with $\log_ax=3$, $\log_a\left(\frac1x\right)=-3$.

$\log_a\left(x^k\right)=k\log_ax$ for any real $k$, including negative and fractional values. It follows from $\left(a^m\right)^k=a^{mk}$. Roots are fractional powers: $\log_a\sqrt y=\frac12\log_ay$. So $\log_a\left(\frac{x^2}{y}\right)=2\log_ax-\log_ay$, and with $\log_ax=3$ and $\log_ay=5$ this equals $6-5=1$.

Since $\log_aa=1$, we also have $\log_a\left(a^n\right)=n$. A lone $a$ inside a logarithm contributes 1: $\log_a\left(\frac a{x^2}\right)=1-2\log_ax$. To write several logarithms as a single logarithm, first use the power law to remove coefficients, then combine with the product and quotient laws: $3\log_a2+\log_a12-\tfrac12\log_a9=\log_a8+\log_a12-\log_a3=\log_a\frac{8\times12}{3}=\log_a32.$ To show an identity, work on one side only, stating each law you use, for example $\log_a48-\log_a3=\log_a16=4\log_a2$.

Three false rules appear regularly: $\log_a(x+y)\neq\log_ax+\log_ay$; $\log_a(x^2)\neq\left(\log_ax\right)^2$; and $\frac{\log_ax}{\log_ay}\neq\log_a\frac xy$. A quick counter-example with $a=2$ shows each is false: $x=y=4$ gives $\log_2(8)=3$ but $\log_24+\log_24=4$. Check a result by substituting numbers, and remember that $x$ and $y$ must be positive for every law to be valid. Some terms may cancel: in $3\log_ax-2\log_ay+\frac12\log_a\left(xy^4\right)$ the $\log_ay$ terms cancel completely.