Logarithms and their lawsEdexcel A-Level Maths: Revision notes

The logarithm $\log_ax$ (with $a>0$, $a\ne1$, $x>0$) is the power to which $a$ must be raised to give $x$: $y=\log_ax\iff a^y=x.$ It is the inverse function of $a^x$, so $\log_a(a^x)=x$ and $a^{\log_ax}=x$. Two results to know: $\log_aa=1$ and $\log_a1=0$. Examples: $\log_5125=3$, $\log_2\frac18=-3$ and $\log_48=\frac32$ (because $4^{3/2}=8$). The graph of $y=\log_ax$ is the reflection of $y=a^x$ in the line $y=x$: it passes through $(1,0)$, has the $y$-axis as an asymptote and exists only for $x>0$.

$\ln x$ means $\log_ex$. It is the inverse of $e^x$, so $\ln(e^x)=x$ and $e^{\ln x}=x$. Its graph is the reflection of $y=e^x$ in $y=x$: it passes through $(1,0)$, has the $y$-axis as an asymptote, and increases slowly. Useful values: $\ln e=1$, $\ln1=0$. To solve equations with $e$ or $\ln$:

• $e^{ax+b}=p\Rightarrow ax+b=\ln p$ (needs $p>0$)
• $\ln(ax+b)=q\Rightarrow ax+b=e^q$ Example: $2000e^{0.04t}=3000$ gives $0.04t=\ln1.5$ and $t=10.1$.

For $x,y>0$ and any base $a$: $\log_ax+\log_ay=\log_a(xy)$ $\log_ax-\log_ay=\log_a\left(\frac xy\right)$ $k\log_ax=\log_a(x^k)$ The power law includes $k=-1$ (so $-\log_ax=\log_a\frac1x$) and $k=-\frac12$ (so $-\frac12\log_ax=\log_a\frac1{\sqrt x}$), and $\log_a\sqrt x=\frac12\log_ax$. They come from the laws of indices, since logarithms are powers. Example: with $p=\log_a2$, $q=\log_a5$: $\log_a10=p+q$, $\log_a0.4=p-q$, $\log_a\sqrt{20}=p+\frac12q$.

• Unknown index: take logarithms of both sides, e.g. $1500\times1.05^t=2000e^{0.04t}$ gives $\ln1500+t\ln1.05=\ln2000+0.04t$, so $t=\frac{\ln(4/3)}{\ln1.05-0.04}=32.7$.
• Equations with logs: combine into one logarithm, then convert, e.g. $\log_3x+\log_3y=3$ gives $xy=27$.
• Simultaneous log equations: use the laws to get $xy$ and $\frac xy$, then solve. Always check that arguments are positive, rejecting any solution that would give the logarithm of a negative number.

Solve $\log_3x+\log_3y=3$ and $\log_3x-\log_3y=1$. Add the logs: $\log_3(xy)=3\Rightarrow xy=27$. Subtract: $\log_3\frac xy=1\Rightarrow\frac xy=3$. So $x=3y$, $3y^2=27$, $y=3$ and $x=9$. Then $\log_3(x^2y)=2\log_3x+\log_3y=2(2)+1=5$, and $\log_3\frac1{\sqrt y}=-\frac12$. Alternatively adding the equations gives $2\log_3x=4$, so $x=9$ directly.