Conditions on the Base and the Argument

Idea

$\log_a x$ is not defined for every pair of numbers. Three conditions have to hold: the base must be positive, the base cannot be $1$ , and the argument must be positive.

None of these is a rule someone invented. Each one is inherited from the exponential $a^y = x$ that the logarithm reverses. If the exponential cannot do the job, there is nothing for the logarithm to name.

Rule

$a > 0 \quad\text{and}\quad a \ne 1 \quad\text{and}\quad x > 0$

the value $y$ has no restriction at all

How it is used

1. Why the base must be positive

A negative base breaks down between the whole-number powers. $(-4)^{1/2}$ is not a real number, so $(-4)^y$ has no unbroken curve to reverse.

(-4)^1 = -4,

(-4)^{1/2} = ?,

(-4)^2 = 16

There is no smooth path from $-4$ to $16$ , so there is no inverse to define.

2. Why the base cannot be 1

Every power of $1$ is $1$ .

1^0 = 1,

1^5 = 1,

1^{-3} = 1

So $1^y = x$ has no solution when $x \ne 1$ , and infinitely many when $x = 1$ . Neither case gives a single answer, which is what $\log_1 x$ would have to return.

3. Why the argument must be positive

A positive base raised to any real power is positive. It gets close to zero but never reaches it, and never crosses below.

2^{10} = 1024,

2^{-10} = \frac{1}{1024},

2^y \ne 0

So $\log_2 0$ and $\log_2(-8)$ ask for a power that does not exist.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 2 · Blue: y = 2ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -1.

4. Reading the conditions off an expression

Set up one inequality per condition and take what they all allow.

Expression	Conditions	Allowed
$\log_3(x - 2)$	$x - 2 > 0$	$x > 2$
$\log_5(7 - x)$	$7 - x > 0$	$x < 7$
$\log_x 9$	$x > 0$ , $x \ne 1$	$x > 0$ , $x \ne 1$
$\log_{x-4} 5$	$x - 4 > 0$ , $x - 4 \ne 1$	$x > 4$ , $x \ne 5$

Worked example

For which values of $x$ is $\log_{x-1}(6 - 2x)$ defined?

Remember

the base must satisfy $a > 0$ and $a \ne 1$ the argument must satisfy $x > 0$ the answer is what all the conditions allow at once

The base is $x - 1$ , so it must be positive.

x - 1 > 0 \;\Rightarrow\; x > 1

The base also cannot equal $1$ .

x - 1 \ne 1 \;\Rightarrow\; x \ne 2

The argument is $6 - 2x$ , so it must be positive.

6 - 2x > 0 \;\Rightarrow\; x < 3

Keep only what every line allows.

1 < x < 3, \quad x \ne 2

Explore

Drag the base slider and watch the green logarithm curve.

Two things never change however far you drag. The curve stays to the right of the $y$ -axis, because the argument must be positive, and it always passes through $(1, 0)$ . What does change is the direction: above a base of $1$ the curve rises, below it the curve falls. Slide the base towards $1$ itself and watch the curve flatten out until it has nothing left to say. That is the case the definition rules out.