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$L (1 / x) = - L (x)$ , $L (x^{n}) = n L (x)$ , and in particular $L (2^{n}) = n L (2)$

Statement

For $x > 0$ ,

$L (1 / x) = - L (x) .$

For every integer $m$ ,

$L (x^{m}) = m L (x),$

and in particular $L (2^{m}) = m L (2)$ . Moreover, $L (2) > 0$ .

Facts & Assumptions

Given: $x > 0$ and an integer exponent $m$ .

[L1]

$L (x y) = L (x) + L (y)$ for positive $x, y$ (The integral logarithm satisfies $L (x y) = L (x) + L (y)$ for all positive $x$ and $y$ ).

[L2]

$L$ is strictly increasing on $(0, \infty)$ (The integral logarithm is continuous and strictly increasing on $(0, \infty)$ ).

[F1]

Natural powers are defined recursively by $x^{0} = 1$ and $x^{n + 1} = x^{n} x$ ; negative integer powers are reciprocal positive powers (Integer powers $a^{m}$ ).

[L3]

A property holding at $0$ and inherited from $n$ to $n + 1$ holds for every natural number (The principle of mathematical induction).

Proof

technique · direct

1.1

Setting both inputs equal to $1$ in [L1] gives $L (1) = 2 L (1)$ , hence $L (1) = 0$ . Applying [L1] to $x (1 / x) = 1$ then gives $L (1 / x) = - L (x)$ .

L1algebra

2.1

For natural $n$ , the identity $L (x^{n}) = n L (x)$ holds at $n = 0$ because $x^{0} = 1$ and $L (1) = 0$ . If it holds at $n$ , then $L (x^{n + 1}) = L (x^{n} x) = L (x^{n}) + L (x) = (n + 1) L (x) .$

step 1.1F1L1algebra

2.2

Since $2 > 1$ and $L (1) = 0$ , strict increase gives $L (2) > 0$ .

step 1.1L2algebra

3.1

Induction [L3] proves the power identity for every natural exponent.

step 2.1L3

4.1

If $m < 0$ , write $m = - n$ with $n > 0$ . Then $x^{m} = 1 / x^{n}$ , so steps 1.1 and 3.1 give $L (x^{m}) = - L (x^{n}) = - n L (x) = m L (x)$ . Thus the formula holds for every integer.

F1step 1.1step 3.1algebra

5.1

Substitute $x = 2$ in step 4.1, together with the natural and zero cases, to obtain $L (2^{m}) = m L (2)$ for every integer $m$ .

step 3.1step 4.1step 2.2∎

Depends on

Used by

The integral logarithm is unbounded above and below Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 61 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

OpenStax, Calculus Volume 1, Section 6.7 (standard reference, not scraped)

L(1/x)=−L(x)L(1/x)=−L(x), L(xn)=nL(x)L(xn)=nL(x), and in particular L(2n)=nL(2)L(2n)=nL(2)