The logarithm with respect to a particular base is the inverse of exponentiation of that base. The ''base-$a$ logarithm of $x$ is denoted \[\log_a x\]
If $x = a^n$, for $n$ any number, $a \gt 0$, and $x \gt 0$, then equivalently $\log_ax = n$.
The most common bases are $10$ and $ e$. Conventionally, $\log$ written without a specific base means $\log_{10}$, and $\ln$ means $\log_ e$. A logarithm to the base $ e$ is called the natural logarithm, because of its mathematical properties.
Some useful results to remember are \begin{align} \log(10) &= 1, & \ln ( e) &= 1,\\ \log_a (a) &= 1, & \log_a (1) &= 0,\\ \log_a (a ^x) &= x, & a ^{\log_a(x)} &= x. \end{align}
\begin{align} \log_a(xy) &= \log_a(x) + \log_a(y) & \textbf{(1)} \\ \log_a\left(\dfrac{x}{y}\right) &= \log_a(x) - \log_a(y) & \textbf{(2)} \\ \log_a(x^b) &= b\log_a(x) & \textbf{(3)} \\ \log_b(a) &= \dfrac{1}{\log_a(b)} & \textbf{(4)} \\ \log_a(x) &= \dfrac{\log_b(x)}{\log_b(a)} & \textbf{(5)} \end{align}
Simplify $3\log (x) - 4\log (x+3) + \log (y)$.
First, use law 3 to bring the constant coefficients inside the logarithms. \[3\log (x) - 4\log (x+3) + \log (y) = \log (x^3) - \log ((x+3)^4) + \log (y)\] Then combine the three terms using laws 1 and 2. \begin{align} \log (x^3) - \log ((x+3)^4) + \log (y) &=\log (x^3y) - \log( (x+3)^4)\\ &=\log \left(\dfrac{x^3y}{(x+3)^4}\right) \end{align}
Simplify $2\log (\sqrt{27}) + 2\log (3x) - \dfrac{1}{3} \log (x)$.
Use law 3 to obtain: \begin{align} 2\log (\sqrt{27}) + 2\log (3x) - \dfrac{1}{3} \log (x)&=\log (\sqrt{27}^2) + \log ((3x)^2)- \log ((x)^\frac{1}{3})\\ &= \log (27) + \log (9x^2) - \log (x^\frac{1}{3}) \end{align} Then combine the terms using laws 1 and 2. \begin{align} \log (27) + \log (9x^2) - \log (x^\frac{1}{3}) &= \log(243x^2) - \log(x^\frac{1}{3}) \\ &= \log \left(\dfrac{243x^2}{x^\frac{1}{3} }\right) \\ &= \log \left(243x^{\frac{5}{3} }\right) \end{align} Finally, notice that $243 = 3^{5} = 27^{5/3}$ and use law 3 to bring out a factor of $\frac{5}{3}$. \[\log \left(243x^{\frac{5}{3} }\right) = \frac{5}{3} \log(27x)\]
So we have obtained: \[2\log (\sqrt{27}) + 2\log (3x) - \dfrac{1}{3} \log (x)=\dfrac{5}{3} \log(27x)\]
When asked to solve an equation where the unknown variable appears in the power, the log law $\log(a^b) = b\log a$ can be used to bring the power down and make it the subject of the equation.
You must take the logarithm of both sides the equation, not just the term with the variable in the power. Then, the other log laws and rearranging can be used to solve the equation.
Solve $3^x=5^{x-2}$.
As the unknown appears in the power, take the logarithm of both sides and then use log laws and rearranging to separate the $x$ out.
\[3^x=5^{x-2}\] Take the logarithm of both sides. \[\log (3^x) = \log (5^{x-2})\] Use law 3. \[x\log (3) = (x-2)\log( 5)\] Expand the brackets. \[x\log( 3) = x\log( 5) - 2\log( 5)\] Collect the $x$ terms on one side. \begin{align} x\log( 3) - x\log (5) &= - 2\log( 5)\\ x\log (5) - x\log (3) &= 2\log (5)\\ x(\log (5) - \log (3)) &= 2\log (5) \end{align} Use law 2. \[x\log \left(\frac{5}{3}\right) = 2\log (5)\] Divide through by $\log \left(\frac{5}{3}\right)$ to get $x$ on its own. \[x = \frac{2\log (5)}{\log\left(\frac{5}{3}\right)}\]
Note: It is usually better to leave a logarithmic solution in exact algebraic form unless you are asked for a numeric answer.
Prof. Robin Johnson simplifies the expression $\log_{10}(5) + \log_{10}(\sqrt{5})-\log_{10}(25)$.
Prof. Robin Johnson solves the equation $5^x = 7 \times 3^{1-x}$.
Prof. Robin Johnson shows how $y^2=3x^3$ can be expressed as a straight line by taking logs.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Test yourself: Numbas test on solving equations using logarithms