# IB Class Notes

## Logarithms

Aims: By the end of this note, you will be able to

1. define a logarithm,
2. manipulate logarithm laws, and
3. apply change of base.

### What is logarithm?

Well, ¡° a logarithm is an index.¡± Let us try to make sense of this statement.

Let us consider a very simple exponential function, say $y = 2^x$. Let us plot the function.

Notice that for each positive value of y there is ONLY one value of $x$ that satisfies $y = 2^x$. What is the corresponding value of $x$ when $y=2$ ? Answer $x=1$. Thus, given a function such as $y = 2^x$ and any positive value of y say $y_1$, we can ask for the corresponding value of $x$.

The solution could be rephrased into ¡°$x$ is the logarithm of $y_1$ to the base of 2 and we write $x = log_2 y_1$.¡±

Therefore, $y = 2^x$ is equivalent to $x = log_2 y$ [because we are dealing with the same graph.]

You are right to guess that logarithm is just the ¡°inverse¡± of the index or exponential function (see the table below); the power to which a base must be raised to yield a given number.
That is in general, $log_b a$ stands for "How much does the base '$b$' has to be raised to obtain the value '$a$' ?"
Thus, $log_b a = x \iff b^x a$ where '$x$' is the solution.

#### Questions

1. Evaluate the following:
1. $log_2 64$
2. $log_3 (\frac{1}{243})$
3. $log_{10} 1000$
4. $log_5 (125)^{-1}$
5. $log_4 (2\times 8)$
2. Express the following in logarithmic notation:
1. $27 = 3^3$
2. $1728 = (12)^3$
3. $27 = (81)^{\frac{3}{4}}$
4. $a^{10} = b$
5. $p^q = r$
The commonly used base is 10 and the irrational number e. Logarithm in base 10 is usually denoted as ¡°log¡± and sometimes ¡°lg.¡± Logarithm in base $e$ is simply denoted by ¡°ln¡± and also known as ¡°natural logarithm.¡±

When solving a problem, you are free to choose logarithms in any base but usually for convenience sake we will restrict ourselves to one consistent base, say ¡°ln.¡±

 Note: $ln 5 = log_e 5$ And $ln e = log_e e = 1$ $e$ is also known as the Euler number. $e$ is the limit to $\left[ 1 + \frac{1}{n} \right]^n$ has $n$ tends to infinity. When $n$ = 1 then the number = 2 When $n$ = 10 then the number = 2.59374... When $n$ = 100 then the number = 2.7048... When $n$ = 10,000 then the number = 2.7181... $e$ = 2.7182818284590... [$e$ is an irrational number.]

(1a) $2^6 = 64$ so $log_2 {64} = 6$
(1b) $3^{-5}$ so $log_3 {\frac{1}{243}} = -5$
(1c) $3$; (1d)$-3$; (1e) $log_4 (16) = 2$ because $2^4 = 16$
(2a) $log_3 {27} = 3$ ; (2b) $log_{12} {1728} = 3$ ; (2c) $log_{81} {27} = \frac{3}{4}$
(2d) $log_a b = 10$ ; (2e) $log_p r = q$

 Index Laws Logarithm Laws 1 $a^m \times a^n = a^{m+n}$ $log_a x + log_a y = log_a (xy)$ 2 ${a^m} \div {a^n} = a^{m-n}$ $log_a x - log_a y = log_a \frac{x}{y}$ 3 $\left( a^m \right) ^n = a^{mn}$ $log_a x^y = y log_a x$ 4 $a^0 = 1$ $log_a 1 = 0$ 5 $a^{-x} = \frac{1}{a^x}$ $log_a {\frac{1}{x}} = log_a x^{-1} = -log_a x$ 6 $a^1 = a$ $log_a a = 1$

Let us deduce Logarithm law 4 from Index law 4.

$a^0 = 1 \iff log_a 1 = 0$ by definition.

Use the definition of logarithm and deduce Logarithm law 6 from Index law 1.

Let us deduce the Logarithm law 1.
We start by letting $x = a^m$ and $y = a^n$ where $a$ is a positive real number and $m$ and $n$ are some real numbers.
Thus, we also have the following by applying the definition of logarithm
$log_a x = m$ and $log_a y = n$.

Then,
$xy = a^m \times a^n$
$xy = a^{m+n}$ by Index Law 1.

By applying the definition of logarithm, we obtain
$log_a {xy} = m + n$
$log_a {xy} = log_a x + log_a y$ which is the desired result.

Using the above method, deduce Logarithm law 2 from Index law 2.

Let us deduce Logarithm law 3 from Index law 3.

Let us start by letting $x = a^m$ and by the definition of logarithm we also have $log_a x = m$ where again $a$ is a positive real number and m is some real number.

$\left( x \right) ^y = \left( a^m \right)^y$
$\left( x \right) ^y = a^{my}$ ---by Index Law 3
$log_a x^y = my$ --- by definition
$log_a x^y = ym$ --- multiplication is commutative
$log_a x^y = y log_a x$ --- replacing $m$ with $log_a x$.

Express $log \frac{a^2b^3}{{10}^2c^{\frac{1}{4}}}$ in terms of $log a, log b$ and $log c$.
Apply Logarithm law 1 and 2.
\begin{align} log \frac{a^2b^3}{{10}^2c^{\frac{1}{4}}} &= log {a^2b^3} - log {10}^2c^{\frac{1}{4}} \\ &= log a^2 + log b^3 - log {10}^2 - log c^{\frac{1}{4}} \\ &= 2log a + 3log b - 2 log {10} -\frac{1}{4}log c \\ &= 2log a + 3log b - 2 - \frac{1}{4}log c \end{align}

Solve for $x$ in $log_7 x + log_7 3 = log_7 \left( x-1 \right)$
\begin{align} log_7 \left( x3 \right) &= log_7 \left(x-1 \right) \\ 3x &= x -1 \\ 2x &= - 1 \\ x &= \frac {-1}{2} \end{align}
Really, this question DOES NOT have any real solution because logb a is undefined when a is less than or equal to zero for any base b not equal to zero. Answer: No solution in real.

Solve for $x$ in $e^{x+1} = 0.64$

\begin{align} x + 1 &= ln \left( 0.64 \right) \\ x &= ln \left( 0.64 \right) -1 \\ x &\approx -1.45 \text{(accurate to 3 significant figures).} \end{align}

Solve for $x$ in the following :

1. $3^x = 45$
2. $2^{2x-1} = 125$
3. $e^{2-x} = \frac{1}{4}$

Solutions:

1. Usually we will choose base $e$ or 10 for these types of problems. Here I have chosen base 10. We use logarithm on both sides so that the index can be "brought down" using Logarithm Law 3.
\begin{align} log 3^x &= log {45} \\ x log 3 &= log {45} \\ x &= \frac{log {45}}{log 3} \\ x &\approx 3.46 \text{(3 significant figures).} \end{align}
The last line is obtained by pressing the calculator. The second last line is an exact answer. When no calculator is available then we should report the second last line ast the final answer.

2. \begin{align} log 2^{2x-1} &= log {125} \\ \left( 2x -1 \right) log 2 &= log {125} \\ 2x -1 &= \frac {log {125}}{log 2} \\ x &= \frac{\frac {log {125}}{log 2}+1}{2} \\ x &\approx 3.98 \text{(3 significant figures).} \end{align}
The last line is obtained by pressing the calculator. The second last line is an exact answer. When no calculator is available then we should report the second last line ast the final answer.

3. Since the question involves $e$ then we will use natural logarithm.
\begin{align} ln e^{2-x} &= ln {\frac{1}{4}} \\ \left(2-x \right) ln e &= ln {\frac{1}{4}} \\ 2-x &= ln {\frac{1}{4}} \\ x &= 2 - ln {\frac{1}{4}} \\ x &\approx 3.39 \text{(3 significant figures).} \end{align}
The last line is obtained by pressing the calculator. The second last line is an exact answer. When no calculator is available then we should report the second last line ast the final answer.

### Change of base

any given exponential function $a^x = b$ , we can ¡°take the logarithm to base $c$ of both sides of the equation.¡± All a , b and c are positie real numbers.

\begin{align} log_c a^x &= log_c b \\ x log_c a &= log_c b \\ x &= \frac{log_c b}{log_c a} \end{align}
Notice that the base value b is written in the denominator (at the base) after changing of base is completed.

With change of base, we can now express $log_7 {25}$ as $\frac{log 25}{log 7}$ or $\frac{ln 25}{ln 7}$ and the latter two expressions allow us to use calculator to evaluate $log_7 {25}$ . Whether you use "ln" that is base e or "log" which is base 10 will give you the same answer for the ratio obtained from applying change of base. But never use "log" in the numerator and "ln" in the denominator or vice versa. Keep to "ln" throughout or "log" throughout. In the following examples, I will use some "log" and some "ln".
Other examples:

1. $log_{14} 3 = \frac{log 3}{log {14}}$
But $log_{14} 3 \neq \frac{ log 3}{ln 14} \neq \frac{ln 3}{log 14}$
2. $log_9 {145} = \frac{log 145}{log 9}$
3. $log_{\frac{1}{6}} \frac{3}{2} = \frac{ln \frac{3}{2}}{ln \frac{1}{6}}$
4. $log_{0.3} 1.7 = \frac{ln 1.7}{ln 0.3}$
5. $log_{\frac{1}{9}} 1.99 = \frac{log 1.99}{log \frac{1}{9}}$

Other Useful formula are the following:

1. $a^x = e^{x ln a}$
2. $log_a a^x = x = a^{log_a x}$

You may want to show that these statements are true.

Exercises.

Email KokMing Lee