The base stays the base. In both forms $a$ is the base; the logarithm $c$ is the index. Moving between the two forms is the single most useful skill on this page.
| Index form | Log form | Reading it |
|---|---|---|
| $2^5 = 32$ | $\log_2 32 = 5$ | $2$ to the power $5$ is $32$ |
| $10^{-2} = 0.01$ | $\log_{10} 0.01 = -2$ | Logs can be negative |
| $9^{1/2} = 3$ | $\log_9 3 = \tfrac12$ | Logs can be fractions |
| $a^0 = 1$ | $\log_a 1 = 0$ | True for every base |
| $a^1 = a$ | $\log_a a = 1$ | True for every base |
Find (a) $\log_4 64$ (b) $\log_8 2$ (c) $\log_3 \tfrac{1}{27}$ (d) $\log_{1/2} 4$.
Because logs are indices, each law of indices has a matching law of logs. All the logs in one law must have the same base.
| Law of indices | Law of logarithms |
|---|---|
| $a^m \times a^n = a^{m+n}$ | $\log_a x + \log_a y = \log_a(xy)$ |
| $a^m \div a^n = a^{m-n}$ | $\log_a x - \log_a y = \log_a\left(\dfrac{x}{y}\right)$ |
| $\left(a^m\right)^k = a^{mk}$ | $k\log_a x = \log_a\left(x^k\right)$ |
Write $2\log 3 + \log 4 - \log 6$ as a single logarithm, and simplify.
Write $\log_a\left(\dfrac{x^3 y}{z^2}\right)$ in terms of $\log_a x$, $\log_a y$ and $\log_a z$.
Given $\log_a 2 = p$ and $\log_a 3 = q$, express (a) $\log_a 12$ (b) $\log_a 1.5$ (c) $\log_a 0.75$ in terms of $p$ and $q$.
Two bases have their own calculator buttons.
| Name | Written | Base | Inverse of |
|---|---|---|---|
| Common logarithm | $\log_{10} x$ or $\lg x$ | $10$ | $10^x$ |
| Natural logarithm | $\ln x$ | $e \approx 2.718$ | $e^x$ |
The laws are identical for $\ln$. Two facts are used constantly: $\ln\left(e^k\right) = k$ and $e^{\ln x} = x$.
Simplify (a) $\ln\left(\dfrac{e^3}{\sqrt e}\right)$ (b) $e^{2\ln 3}$ (c) $e^{\ln 4 + 1}$.
A single log equal to a number is solved by rewriting it in index form. More involved equations — with several logs, or with the unknown in an index — are on the Exponential and Log Equations page.
Solve (a) $\log_3 x = 4$ (b) $\log_x 49 = 2$ (c) $\log_2(x + 1) = 5$ (d) $\ln(2x) = 3$.
Solve $\log_2 x + \log_2 8 = 7$.
Definition
$\log_a b = c \iff a^c = b$
Special values
$\log_a 1 = 0$, $\log_a a = 1$
Multiply
$\log x + \log y = \log xy$
Divide
$\log x - \log y = \log \dfrac{x}{y}$
Power
$k\log x = \log x^k$ — do this first.
Reciprocal
$\log\dfrac1x = -\log x$
No law
for $\log(x+y)$ or $\dfrac{\log x}{\log y}$.
Domain
Only positive numbers have logs.
Natural log
$\ln e^k = k$, $e^{\ln x} = x$
Other bases
$\log_a b = \dfrac{\ln b}{\ln a}$
Write in logarithm form: (a) $2^5 = 32$ (b) $10^{-3} = 0.001$ (c) $7^0 = 1$. Write in index form: (d) $\log_5 125 = 3$.
▶ Show solution
(a) $\log_2 32 = 5$ (b) $\log_{10} 0.001 = -3$ (c) $\log_7 1 = 0$
(d) $5^3 = 125$
Without a calculator, find (a) $\log_3 81$ (b) $\log_2 \tfrac18$ (c) $\log_{25} 5$ (d) $\log_4 8$.
▶ Show solution
(a) $3^4 = 81$, so $4$.
(b) $\tfrac18 = 2^{-3}$, so $-3$.
(c) $25^{1/2} = 5$, so $\tfrac12$.
(d) $4^{3/2} = \left(\sqrt4\right)^3 = 8$, so $\tfrac32$.
Show that $\log_{10} 5 + \log_{10} 4 - \log_{10} 2 = 1$.
▶ Show solution
$\log_{10}\left(\dfrac{5 \times 4}{2}\right) = \log_{10} 10 = 1$.
Write $3\ln x - 2\ln y + \ln 5$ as a single logarithm.
▶ Show solution
Power law first: $\ln x^3 - \ln y^2 + \ln 5$.
Combine: $\ln\left(\dfrac{5x^3}{y^2}\right)$.
Given $\log_a 2 = p$ and $\log_a 3 = q$, express in terms of $p$ and $q$: (a) $\log_a 18$ (b) $\log_a \tfrac89$ (c) $\log_a \sqrt6$.
▶ Show solution
(a) $18 = 2 \times 3^2$: $\;p + 2q$.
(b) $\tfrac89 = \dfrac{2^3}{3^2}$: $\;3p - 2q$.
(c) $\sqrt6 = (2 \times 3)^{1/2}$: $\;\tfrac12(p + q)$.
Find the exact value of $2\log_6 3 + \log_6 4$.
▶ Show solution
$2\log_6 3 = \log_6 9$, so the sum is $\log_6 36$.
$6^2 = 36$, so the value is $2$.
Solve (a) $\log_x 64 = 3$ (b) $\log_5(2x - 1) = 2$ (c) $\ln(x - 3) = 1$.
▶ Show solution
(a) $x^3 = 64$, so $x = 4$.
(b) $2x - 1 = 5^2 = 25$, so $x = 13$.
(c) $x - 3 = e$, so $x = 3 + e \approx 5.718$.
Write in the form $A + B\ln x$: (a) $\ln\left(e^2x^3\right)$ (b) $\ln\sqrt{ex}$ (c) $\ln\left(\dfrac{x}{e^4}\right)$.
▶ Show solution
(a) $\ln e^2 + \ln x^3 = 2 + 3\ln x$
(b) $\tfrac12\ln(ex) = \tfrac12(\ln e + \ln x) = \tfrac12 + \tfrac12\ln x$
(c) $\ln x - \ln e^4 = -4 + \ln x$
Prove that $\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$.
▶ Show solution
Let $m = \log_a x$ and $n = \log_a y$, so $x = a^m$ and $y = a^n$.
Then $\dfrac{x}{y} = \dfrac{a^m}{a^n} = a^{m-n}$.
Converting to log form: $\log_a\left(\dfrac{x}{y}\right) = m - n = \log_a x - \log_a y$. $\blacksquare$
The pH of a solution is defined by $\text{pH} = -\log_{10} H$, where $H$ is the concentration of hydrogen ions in moles per litre.
(a) Find the pH of a solution with $H = 3.2 \times 10^{-4}$, to 2 decimal places. (b) Blood has a pH of $7.4$. Find $H$. (c) Show that multiplying $H$ by $10$ lowers the pH by exactly $1$. (d) Solution A has pH $3$ and solution B has pH $5$. How many times more concentrated in hydrogen ions is A than B?
▶ Show solution
(a) $\text{pH} = -\log_{10}\left(3.2 \times 10^{-4}\right) \approx 3.49$.
(b) $-\log_{10} H = 7.4 \Rightarrow \log_{10} H = -7.4 \Rightarrow H = 10^{-7.4} \approx 3.98 \times 10^{-8}$ moles per litre.
(c) New pH $= -\log_{10}(10H) = -\left(\log_{10}10 + \log_{10}H\right) = -1 - \log_{10}H = (\text{old pH}) - 1$.
(d) $H_A = 10^{-3}$ and $H_B = 10^{-5}$, so $\dfrac{H_A}{H_B} = 10^{2} = 100$. A is $100$ times as concentrated.