Logarithms and their lawsAQA A-Level Maths: Revision notes

The logarithm $\log_ax$ is the power to which $a$ must be raised to give $x$: $\log_ax=y\iff a^y=x,\qquad a>0,\ x>0.$ So $\log_381=4$ because $3^4=81$, and $\log_2\frac18=-3$ because $2^{-3}=\frac18$. Two results always hold: $\log_a1=0$ and $\log_aa=1$. Because $\log_ax$ and $a^x$ are inverse functions, $a^{\log_ax}=x$ and $\log_a(a^x)=x$, and 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 only exists for $x>0$.

The natural logarithm is the logarithm to base $e$: $\ln x=\log_ex$. It is the inverse of $e^x$: $e^{\ln x}=x\ (x>0),\qquad\ln(e^x)=x.$ The graph of $y=\ln x$ is the reflection of $y=e^x$ in $y=x$: it passes through $(1,0)$, increases slowly, and has the $y$-axis as an asymptote. So $\ln x=3$ gives $x=e^3$, and $e^x=5$ gives $x=\ln5$. Also $\ln e=1$ and $\ln1=0$.

For any base $a>0$ and $x,y>0$:

• Addition: $\log_ax+\log_ay=\log_a(xy)$
• Subtraction: $\log_ax-\log_ay=\log_a\left(\frac xy\right)$
• Power: $k\log_ax=\log_a(x^k)$ The power law holds for any $k$. Two special cases are used a lot:
• $k=-1$: $-\log_ax=\log_a\left(\frac1x\right)$
• $k=-\frac12$: $-\frac12\log_ax=\log_a\left(\frac{1}{\sqrt x}\right)$ (Also $k=\frac12$ gives $\frac12\log_ax=\log_a\sqrt x$.) The same laws hold for $\ln$.

Use the laws to combine many logs into one, or to split one into simple parts. Example: $\log_{10}12=\log_{10}(2^2\times3)=2a+b$ when $a=\log_{10}2$, $b=\log_{10}3$. Example: $\log_{10}\frac16=-\log_{10}6=-(a+b)$. Example: $\ln\frac{x^3}{\sqrt y}=3\ln x-\frac12\ln y$. Example: $2\ln5+\ln4=\ln25+\ln4=\ln100$. Exact values without a calculator: $\log_48$: let $4^y=8$, so $2^{2y}=2^3$ and $y=\frac32$.

For an equation containing logarithms of the same base: (1) state the domain (every argument must be positive), (2) combine into a single logarithm using the laws, (3) convert to a power, (4) solve, and (5) check each answer against the domain. Example: $\log_2(x+6)-\log_2(x-1)=3$ needs $x>1$. Then $\frac{x+6}{x-1}=2^3=8$, so $x+6=8x-8$ and $x=2$, which is valid. Example: $2\log_3x-\log_3(x+4)=1$ gives $\frac{x^2}{x+4}=3$, so $x^2-3x-12=0$ and $x=\frac{3\pm\sqrt{57}}{2}$. The negative root is rejected because $\log_3x$ needs $x>0$.