MHF4U Introduction to Logarithms: Notes and Practice

You already know how to answer questions like "what is $2^{5}$ ?" — you multiply five copies of 2 and get 32. Logarithms ask the same question backwards: "32 is 2 to the power of what?" The answer is 5, and the notation for it is $lo g_{2} 32 = 5$ . That is the whole idea. A logarithm is not a new operation to be afraid of; it is a name for the exponent you were already hunting for whenever you solved an equation by matching bases. Until now you could only solve $2^{x} = 32$ because 32 happened to be a nice power of 2 — but what about $2^{x} = 30$ ? There is definitely an exponent that works, sitting somewhere between 4 and 5; you just had no way to name it. Logarithms give it a name, $x = lo g_{2} 30$ , and once a number has a name you can work with it, graph it, and estimate it.

The definition

lo g_{b} x = y ⟺ b^{y} = x (b > 0, b \neq = 1, x > 0)

The logarithmic equation on the left and the exponential equation on the right say exactly the same thing.

Read the left side out loud as "the logarithm of $x$ , base $b$ , equals $y$ ", but think of it as " $y$ is the exponent you put on $b$ to get $x$ ". The base $b$ is written small and low, the same place it sits in $b^{y}$ . The number $x$ inside the logarithm is called the argument.

Fig 1The exponential and logarithmic forms are two ways of writing one relationship between a base, an exponent, and a result.

VideoJensenMath, an Ontario teacher, introduces the logarithm as a way to solve for an unknown exponent and shows how to evaluate logarithms on a calculator.Watch on YouTube

Converting between the two forms

Exponential form	Logarithmic form	Said out loud
$2^{5} = 32$	$lo g_{2} 32 = 5$	5 is the exponent on 2 that gives 32.
$1 0^{3} = 1000$	$lo g_{10} 1000 = 3$	3 is the exponent on 10 that gives 1000.
$3^{- 2} = \frac{1}{9}$	$lo g_{3} \frac{1}{9} = - 2$	−2 is the exponent on 3 that gives one ninth.
$5^{0} = 1$	$lo g_{5} 1 = 0$	0 is the exponent on 5 that gives 1.
$7^{1/2} = 7$	$lo g_{7} 7 = \frac{1}{2}$	One half is the exponent on 7 that gives root 7.
$b^{y} = x$	$lo g_{b} x = y$	y is the exponent on b that gives x.

Almost every logarithm problem starts by converting to whichever form is easier to work with. The safest way to convert is to say the definition to yourself: base to the power of the answer equals the argument.

Evaluating logarithms in your head

To evaluate $lo g_{b} x$ without a calculator, rewrite the argument as a power of the base. If $x = b^{n}$ , then the answer is simply $n$ . When the argument is not obviously a power of the base, write both as powers of some smaller common base — usually 2, 3 or 5 — and compare exponents, exactly as you did when solving exponential equations by matching bases.

$lo g_{5} 125 = 3$ , because $125 = 5^{3}$ .
$lo g_{2} \frac{1}{16} = - 4$ , because $\frac{1}{16} = 2^{- 4}$ — a fraction argument gives a negative logarithm.
$lo g_{9} 3 = \frac{1}{2}$ , because $9^{1/2} = 9 = 3$ — an argument between 1 and the base (for a base greater than 1) gives a logarithm between 0 and 1.
$lo g_{b} 1 = 0$ for every valid base, because $b^{0} = 1$ always.
$lo g_{b} b = 1$ for every valid base, because $b^{1} = b$ .
$lo g_{1/2} 8 = - 3$ , because $(\frac{1}{2})^{- 3} = 2^{3} = 8$ .

Common logarithms (base 10)

Base 10 is used so often — because our number system is base 10 — that it gets to drop its subscript. When you see $lo g x$ with no base written, it means $lo g_{10} x$ . This is the "log" button on a calculator. So $lo g 100 = 2$ , $lo g 1 = 0$ , and $lo g 0.001 = - 3$ , since $0.001 = 1 0^{- 3}$ .

The three restrictions, and where they come from

The definition only makes sense when

b > 0

b \neq = 1

, and

x > 0

. Each restriction is inherited directly from the exponential function

y = b^{x}

that the logarithm undoes — they are not arbitrary rules bolted on afterwards.

Why the restrictions exist

The base must be positive: if $b$ were negative, something like $(- 4)^{1/2} = - 4$ would not be a real number, so the exponential function would have gaps and could not be reversed.
The base cannot be 1: $1^{y} = 1$ for every exponent $y$ , so $lo g_{1} 7$ would be asking "1 to what power gives 7?" — no answer at all — and $lo g_{1} 1$ would have infinitely many answers.
The argument must be positive: $b^{y}$ is always positive for a positive base, so no exponent can ever produce 0 or a negative result. That makes $lo g_{2} 0$ and $lo g_{2} (- 8)$ undefined.

Undefined, not zero

A logarithm of zero or of a negative number is undefined — it does not equal 0, and it is not "negative infinity" at this level. If an answer to an equation forces a logarithm to take a non-positive argument, that answer must be thrown out. You will use this constantly in the last topic of this unit.

Introduction to Logarithms

Key things to remember

The definition

Converting between the two forms

Evaluating logarithms in your head

Common logarithms (base 10)

Why the restrictions exist

Worked examples

Where marks get lost

Flashcards 8

Practice