1.4 Logarithms and the Laws of Logarithms
SL 1.5 SL 1.7
- what a logarithm is, and how to change between index form and logarithmic form
- common logarithms (\(\log\), base 10) and natural logarithms (\(\ln\), base \(\mathrm{e}\))
- the laws of logarithms: product, quotient and power laws
- the change of base formula
- how to simplify and rewrite expressions with logarithms
What is a logarithm?
Look at the equation \(2^x = 150\). In Section 1.5 we will solve equations like this, but first we need a name for the number \(x\).
\(x\) is the exponent we must put on the base \(2\) to get \(150\). We call it the logarithm of 150 to base 2, and we write
\[ x = \log_2 150 . \]
You read \(\log_2 150\) out loud as “log base 2 of 150” or “the logarithm of 150 to the base 2”.
So a logarithm is an exponent. The two statements below say exactly the same thing.
Definition — for \(a > 0\), \(b > 0\), \(a \neq 1\) In the formula booklet
\[ \underbrace{a^x = b}_{\text{index form}} \quad\Leftrightarrow\quad \underbrace{x = \log_a b}_{\text{logarithmic form}} \]
\(\log_a b\) is “the power you put on \(a\) to get \(b\)”.
- \(a\): the base (\(a > 0\), \(a \neq 1\))
- \(x\): the power (the logarithm of \(b\))
- \(b\): the number we want to get (\(b > 0\))
Read \(\log_2 8\) as “the power of 2 that gives 8”. Since \(2^3 = 8\), \(\log_2 8 = 3\).
When does \(\log_a y\) exist? The base follows the same rules as for exponential functions (\(a > 0\), \(a \neq 1\)). And since \(a^x\) is always positive, \(y\) must be positive. So \(\log_2 0\) and \(\log_2 (-4)\) do not exist.
Common and natural logarithms
Two bases are so useful that they have their own buttons on the calculator.
Common and natural logarithms Not in the formula booklet
| Name | Base | Written | Meaning |
|---|---|---|---|
| common logarithm | \(10\) | \(\log x\) (or \(\log_{10} x\)) | \(10^{y} = x \Leftrightarrow y = \log x\) |
| natural logarithm | \(\mathrm{e}\) | \(\ln x\) | \(\mathrm{e}^{y} = x \Leftrightarrow y = \ln x\) |
- \(\mathrm{e}\): Euler’s number, \(\mathrm{e} = 2.718\,28\ldots\)
- \(x\): a positive number
- \(y\): the logarithm of \(x\) (the power)
In IB examinations, \(\log x\) with no base means base \(10\). (Some books write \(\lg x\).)
To evaluate a logarithm numerically, use the GDC: for example \(\ln 20 = 2.995\ldots\) and \(\log 20 = 1.301\ldots\). On the TI-Nspire CX II, the key marked log lets you type any base.
Example 1 (a) Write in logarithmic form: \(\;\) (i) \(3^4 = 81\) \(\quad\) (ii) \(\mathrm{e}^{1.5} \approx 4.48\)
(b) Write in index form: \(\;\) (i) \(\log_5 125 = 3\) \(\quad\) (ii) \(\ln 6 = k\)
The base stays the base; the exponent becomes the answer of the logarithm.
(a) (i) \(\log_3 81 = 4\) \(\qquad\) (ii) \(\ln 4.48 \approx 1.5\)
(b) (i) \(5^3 = 125\) \(\qquad\) (ii) \(\mathrm{e}^{k} = 6\)
Example 2 Without a calculator, find the value of
(a) \(\log_2 32\) \(\qquad\) (b) \(\log_9 3\) \(\qquad\) (c) \(\log_4 8\) (d) \(\log 0.001\)
(e) \(\ln \dfrac{1}{\sqrt{\mathrm{e}}}\)
In each case, ask: “what power of the base gives this number?”
(a) \(2^5 = 32\), so \(\log_2 32 = 5\).
(b) \(9^{\frac{1}{2}} = 3\), so \(\log_9 3 = \dfrac{1}{2}\).
(c) Write both numbers as powers of 2: \(4^{x} = 8 \Rightarrow 2^{2x} = 2^3 \Rightarrow x = \dfrac{3}{2}\).
(d) \(0.001 = 10^{-3}\), so \(\log 0.001 = -3\).
(e) \(\dfrac{1}{\sqrt{\mathrm{e}}} = \mathrm{e}^{-\frac{1}{2}}\), so \(\ln \dfrac{1}{\sqrt{\mathrm{e}}} = -\dfrac{1}{2}\).
Special values
Three results follow straight from the definition.
Special values Partly in the formula booklet
\[ \log_a a^x = x = a^{\log_a x} \quad \text{where } a, x > 0,\ a \neq 1 \]
In the booklet
\[ \log_a 1 = 0 \qquad\quad \log_a a = 1 \qquad (a > 0,\ a \neq 1) \]
- \(a\): the base (\(a > 0\), \(a \neq 1\))
- \(x\): a positive number
So \(\ln \mathrm{e} = 1\), \(\log 10 = 1\), \(\ln 1 = \log 1 = 0\), and \(\mathrm{e}^{\ln 7} = 7\). Taking \(\log_a\) and raising \(a\) to a power undo each other.
The laws of logarithms
Laws of logarithms — for \(a, x, y > 0\), \(a \neq 1\) In the formula booklet
| Law | |
|---|---|
| Product law | \(\log_a xy = \log_a x + \log_a y\) |
| Quotient law | \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\) |
| Power law | \(\log_a x^{m} = m \log_a x\) |
- \(a\): the base (\(a > 0\), \(a \neq 1\))
- \(x\): a positive number
- \(y\): a positive number
- \(m\): a real number (the power)
Why the product law works. Let \(\log_a x = p\) and \(\log_a y = q\), so \(x = a^{p}\) and \(y = a^{q}\). Then
\[ xy = a^{p} \times a^{q} = a^{p+q} \quad\Longrightarrow\quad \log_a (xy) = p + q = \log_a x + \log_a y . \]
The laws of logarithms are the laws of exponents, written in logarithmic form.
Example 3 Without a calculator, simplify
(a) \(\log_3 \left(9\sqrt{3}\right)\) \(\qquad\) (b) \(\log_6 4 + \log_6 9\) \(\qquad\) (c) \(\log_5 50 - \log_5 2\) \(\qquad\) (d) \(2\log 5 + \log 4\)
(a) \(9\sqrt{3} = 3^2 \times 3^{\frac{1}{2}} = 3^{\frac{5}{2}}\), so \(\log_3 \left(9\sqrt{3}\right) = \dfrac{5}{2}\).
(b) Product law: \(\log_6 (4 \times 9) = \log_6 36 = 2\).
(c) Quotient law: \(\log_5 \dfrac{50}{2} = \log_5 25 = 2\).
(d) Power law first, then product law: \(\log 5^2 + \log 4 = \log (25 \times 4) = \log 100 = 2\).
Example 4 Express \(3\log x - \dfrac{1}{2}\log y + 2\) as a single logarithm.
First write the number \(2\) as a logarithm with base 10: \(2 = \log 10^{2} = \log 100\).
Use the power law on the other terms: \(3\log x = \log x^{3}\) and \(\dfrac{1}{2}\log y = \log y^{\frac{1}{2}} = \log \sqrt{y}\).
\[ \log x^{3} - \log \sqrt{y} + \log 100 = \log \left(\frac{100x^{3}}{\sqrt{y}}\right) \]
Change of base
Your GDC has buttons for \(\log\) and \(\ln\). To work with any other base, use this formula.
Change of base — for \(a, b, x > 0\) and \(a, b \neq 1\) In the formula booklet
\[ \log_a x = \frac{\log_b x}{\log_b a} \]
In particular, \(\log_a x = \dfrac{\ln x}{\ln a} = \dfrac{\log x}{\log a}\).
- \(a\): the old base
- \(x\): a positive number
- \(b\): the new base (any base you like, often \(\mathrm{e}\) or \(10\))
Example 5 (a) Use your GDC to find \(\log_7 20\), correct to 3 significant figures.
(b) Without a calculator, show that \(\log_8 32 = \dfrac{5}{3}\).
(a) \(\log_7 20 = \dfrac{\ln 20}{\ln 7} = 1.539\ldots \approx 1.54\).
(b) Change to base 2, because \(8\) and \(32\) are both powers of 2:
\[ \log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{5}{3}. \]
Writing logarithms in terms of other logarithms
A common examination question gives you \(\log_a\) of some numbers as letters, and asks you to write another logarithm in terms of those letters. Break the number into a product or quotient of the given numbers.
Example 6 Given that \(\log_2 3 = a\) and \(\log_2 5 = b\), express each of the following in terms of \(a\) and \(b\).
(a) \(\log_2 45\) \(\qquad\) (b) \(\log_2 0.6\) \(\qquad\) (c) \(\log_3 5\)
(a) \(45 = 3^2 \times 5\), so \(\log_2 45 = 2\log_2 3 + \log_2 5 = 2a + b\).
(b) \(0.6 = \dfrac{3}{5}\), so \(\log_2 0.6 = \log_2 3 - \log_2 5 = a - b\).
(c) Change to base 2: \(\log_3 5 = \dfrac{\log_2 5}{\log_2 3} = \dfrac{b}{a}\).
Example 7 Given that \(\log_3 p = x\), \(\log_9 q = y\) and \(pq = 3^{k}\), express \(k\) in terms of \(x\) and \(y\).
Change each statement to index form, using base \(3\) for both.
\[ \log_3 p = x \Rightarrow p = 3^{x}, \qquad \log_9 q = y \Rightarrow q = 9^{y} = 3^{2y}. \]
Then \(pq = 3^{x} \times 3^{2y} = 3^{x + 2y}\). Comparing with \(pq = 3^{k}\): \(\;k = x + 2y\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\log (x + y) = \log x + \log y\) | no simple rule for \(\log(x + y)\) | The product law is for \(\log(xy)\). |
| \(\dfrac{\log x}{\log y} = \log x - \log y\) | \(\log \dfrac{x}{y} = \log x - \log y\) | The quotient law is for the log of a quotient. |
| \((\log x)^2 = 2\log x\) | \(\log x^{2} = 2\log x\) | The power law needs the power inside the log. |
| \(\log_2 8 = 4\) (“8 ÷ 2”) | \(\log_2 8 = 3\) | A log is an exponent: \(2^3 = 8\). |
| \(\log_3 0 = 0\) | \(\log_3 0\) does not exist | \(3^x\) is never \(0\). |
| \(\ln 5 \times \ln 2 = \ln 10\) | \(\ln 5 + \ln 2 = \ln 10\) | Adding logs multiplies the numbers inside. |
Exercises
1 Write in logarithmic form.
(a) \(2^{6} = 64\)
(b) \(10^{-2} = 0.01\)
(c) \(\mathrm{e}^{0} = 1\)
(a) \(\log_2 64 = 6\)
(b) \(\log 0.01 = -2\)
(c) \(\ln 1 = 0\)
2 Write in index form.
(a) \(\log_7 49 = 2\)
(b) \(\ln y = 4\)
(c) \(\log_a b = c\)
(a) \(7^{2} = 49\)
(b) \(y = \mathrm{e}^{4}\)
(c) \(a^{c} = b\)
3 Without a calculator, find
(a) \(\log_5 625\)
(b) \(\log_2 \dfrac{1}{16}\)
(c) \(\log_{27} 3\)
(d) \(\log_8 4\)
(a) \(5^4 = 625\), so \(4\). \(\qquad\) (b) \(2^{-4} = \dfrac{1}{16}\), so \(-4\).
(c) \(27^{\frac{1}{3}} = 3\), so \(\dfrac{1}{3}\).
(d) \(8^{x} = 4 \Rightarrow 2^{3x} = 2^{2} \Rightarrow x = \dfrac{2}{3}\).
4 Without a calculator, find
(a) \(\log_3 54 - \log_3 2\)
(b) \(\log_4 2 + \log_4 32\)
(c) \(\dfrac{\log 27}{\log 3}\)
(a) \(\log_3 \dfrac{54}{2} = \log_3 27 = 3\) \(\qquad\) (b) \(\log_4 64 = 3\)
(c) \(\dfrac{\log 3^{3}}{\log 3} = \dfrac{3\log 3}{\log 3} = 3\) (or: change of base gives \(\log_3 27 = 3\)).
5 Express \(2\ln x + \ln (x + 1) - \ln 3\) as a single logarithm.
\[ \ln x^{2} + \ln (x + 1) - \ln 3 = \ln \left(\frac{x^{2}(x + 1)}{3}\right) \]
6 Write \(\log \left(\dfrac{x^{3}\sqrt{y}}{z^{2}}\right)\) in terms of \(\log x\), \(\log y\) and \(\log z\).
\[ \log x^{3} + \log y^{\frac{1}{2}} - \log z^{2} = 3\log x + \frac{1}{2}\log y - 2\log z \]
7 Given that \(\log_a 2 = p\) and \(\log_a 3 = q\), express in terms of \(p\) and \(q\):
(a) \(\log_a 12\)
(b) \(\log_a \dfrac{9}{8}\)
(c) \(\log_a \dfrac{a^{2}}{3}\)
(a) \(12 = 2^2 \times 3\): \(\;2p + q\)
(b) \(\dfrac{9}{8} = \dfrac{3^2}{2^3}\): \(\;2q - 3p\)
(c) \(\log_a a^{2} - \log_a 3 = 2 - q\)
8 Show that \(\log_4 x = \dfrac{1}{2}\log_2 x\) for \(x > 0\).
Change of base to base 2:
\[ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} = \frac{1}{2}\log_2 x . \]
9 Simplify
(a) \(5^{\log_5 7}\)
(b) \(\mathrm{e}^{2\ln 3}\)
(c) \(\log_2 \left(8^{x}\right)\)
(a) \(7\) \(\qquad\) (b) \(\mathrm{e}^{\ln 3^{2}} = \mathrm{e}^{\ln 9} = 9\) \(\qquad\) (c) \(\log_2 2^{3x} = 3x\)
10 Given that \(\log_2 p = x\), \(\log_8 q = y\) and \(pq = 4^{k}\), express \(k\) in terms of \(x\) and \(y\).
\(p = 2^{x}\) and \(q = 8^{y} = 2^{3y}\), so \(pq = 2^{x + 3y}\).
Also \(4^{k} = 2^{2k}\). Equating exponents: \(2k = x + 3y\), so \(k = \dfrac{x + 3y}{2}\).