Logarithms

Basis
Last updated: Tags: Elementary Function

The Exponential Functions article showed you that exp:R(0,)\exp : \mathbb{R} \to (0, \infty) is strictly increasing and surjective onto (0,)(0, \infty). A strictly increasing function is automatically injective, so exp\exp is a bijection — it has a unique inverse. That inverse is one of the most useful functions in all of analysis.

The natural logarithm

Definition. The natural logarithm ln:(0,)R\ln : (0, \infty) \to \mathbb{R} is the inverse of exp\exp. That is:

ln(exp(x))=xfor all xR,(1)\ln(\exp(x)) = x \qquad \text{for all } x \in \mathbb{R}, \tag{1} exp(ln(y))=yfor all y>0.(2)\exp(\ln(y)) = y \qquad \text{for all } y > 0. \tag{2}

Intuitively, ln(y)\ln(y) answers the question: “To what power must ee be raised to obtain yy?” Because exp\exp maps all of R\mathbb{R} bijectively onto (0,)(0, \infty), this question always has a unique answer.

Immediate values

From the definition and what you already know about exp\exp:

  • ln(1)=0\ln(1) = 0, because exp(0)=1\exp(0) = 1.
  • ln(e)=1\ln(e) = 1, because exp(1)=e\exp(1) = e.
  • ln(en)=n\ln(e^n) = n for every nZn \in \mathbb{Z}, because exp(n)=en\exp(n) = e^n.

These three anchor points are worth memorizing; they let you reason about ln\ln quickly without a calculator.

The product rule for logarithms

Theorem (product rule). For all x,y>0x, y > 0:

ln(xy)=lnx+lny.(3)\ln(xy) = \ln x + \ln y. \tag{3}

Proof. Let alnxa \coloneqq \ln x and blnyb \coloneqq \ln y, so exp(a)=x\exp(a) = x and exp(b)=y\exp(b) = y. The functional equation of the exponential — exp(a+b)=exp(a)exp(b)\exp(a + b) = \exp(a)\exp(b) — gives:

xy=exp(a)exp(b)=exp(a+b).xy = \exp(a) \cdot \exp(b) = \exp(a + b).

Applying ln\ln to both sides and using identity (1)(1):

ln(xy)=ln(exp(a+b))=a+b=lnx+lny.\ln(xy) = \ln(\exp(a + b)) = a + b = \ln x + \ln y. \qquad \square

Intuitively, ln\ln converts multiplication into addition — precisely the property that made logarithm tables indispensable for arithmetic long before calculators.

Two immediate corollaries follow from (3)(3). Setting y=1/xy = 1/x gives ln(1/x)=lnx\ln(1/x) = -\ln x. Setting y=xy = x gives ln(x2)=2lnx\ln(x^2) = 2\ln x. In fact, a full power rule holds for all real exponents.

The power rule for logarithms

Theorem (power rule). For all x>0x > 0 and rRr \in \mathbb{R}:

ln(xr)=rlnx.(4)\ln(x^r) = r\ln x. \tag{4}

Proof sketch. For r=nNr = n \in \mathbb{N}, apply the product rule (3)(3) repeatedly:

ln(xn)=ln(xxxn times)=lnx+lnx++lnxn times=nlnx.\ln(x^n) = \ln(\underbrace{x \cdot x \cdots x}_{n\text{ times}}) = \underbrace{\ln x + \ln x + \cdots + \ln x}_{n\text{ times}} = n\ln x.

For rational r=p/qr = p/q with qN+q \in \mathbb{N}^+, note that qln(xp/q)=ln((xp/q)q)=ln(xp)=plnxq \cdot \ln(x^{p/q}) = \ln((x^{p/q})^q) = \ln(x^p) = p\ln x, so ln(xp/q)=pqlnx\ln(x^{p/q}) = \tfrac{p}{q}\ln x. For arbitrary real rr, the equality extends by continuity of ln\ln. \square

The power rule (4)(4) turns exponentiation into multiplication, just as (3)(3) turned multiplication into addition.

The derivative of ln\ln

Theorem. For all x>0x > 0:

ddxlnx=1x.(5)\frac{d}{dx}\ln x = \frac{1}{x}. \tag{5}

Proof. Let y=lnxy = \ln x, so that x=exp(y)x = \exp(y). Differentiate both sides with respect to xx using the inverse function theorem:

1=ddxexp(y)=exp(y)dydx=xdydx.1 = \frac{d}{dx}\exp(y) = \exp(y) \cdot \frac{dy}{dx} = x \cdot \frac{dy}{dx}.

Solving for dydx\frac{dy}{dx}:

dydx=1x.\frac{dy}{dx} = \frac{1}{x}. \qquad \square

This is a striking result: the slope of the graph of ln\ln at the point xx is simply 1/x1/x, with no multiplicative constant or extra function.

Monotonicity and behavior at the boundary

Because (lnx)=1/x>0(\ln x)' = 1/x > 0 for all x>0x > 0, the function ln\ln is strictly increasing on its entire domain (0,)(0, \infty).

The behavior at the two boundary points follows directly from the relationship with exp\exp:

limxlnx=+,limx0+lnx=.\lim_{x \to \infty} \ln x = +\infty, \qquad \lim_{x \to 0^+} \ln x = -\infty.

The first limit holds because exp(t)+\exp(t) \to +\infty as t+t \to +\infty, so its inverse must also grow without bound. The second holds because exp(t)0\exp(t) \to 0 as tt \to -\infty. Together these confirm that ln\ln is a bijection from (0,)(0, \infty) onto all of R\mathbb{R}, the mirror image of the fact that exp\exp maps R\mathbb{R} bijectively onto (0,)(0, \infty).

Natural logarithm as an integral

There is a second way to arrive at ln\ln that makes the derivative formula (5)(5) transparent from the outset. Define:

L(x)1x1tdt(x>0).(6)L(x) \coloneqq \int_1^x \frac{1}{t}\,dt \qquad (x > 0). \tag{6}

By the Fundamental Theorem of Calculus, L(x)=1/xL'(x) = 1/x and L(1)=0L(1) = 0. One can verify that LL satisfies the same product rule as ln\ln (changing variables in the integral), and since both are continuous functions that agree at x=1x = 1 and have the same derivative, they are identical:

lnx=1x1tdt.\ln x = \int_1^x \frac{1}{t}\,dt.

This integral representation is sometimes taken as the definition of ln\ln, with Euler’s number then recovered as the unique e>0e > 0 satisfying 1e1tdt=1\int_1^e \frac{1}{t}\,dt = 1.

General exponentials via the logarithm

Now that ln\ln is available, you can define bxb^x for any base b>0b > 0 and any real exponent xRx \in \mathbb{R} — including irrationals.

Definition. For b>0b > 0:

bx    exp(xlnb).(7)b^x \;\coloneqq\; \exp(x \ln b). \tag{7}

When xx is a rational number p/qp/q, this agrees with the usual arithmetic meaning of bp/qb^{p/q} (you can verify this using the power rule (4)(4)). Definition (7)(7) extends it seamlessly to all real exponents.

Derivative. Differentiating (7)(7) by the chain rule:

ddxbx=exp(xlnb)lnb=(lnb)bx.\frac{d}{dx}\,b^x = \exp(x \ln b) \cdot \ln b = (\ln b)\,b^x.

The extra factor lnb\ln b is what distinguishes bxb^x from exe^x: when b=eb = e, lne=1\ln e = 1 and the factor disappears, which is precisely why ee is the “natural” base.

Logarithms to an arbitrary base

Definition. For b>0b > 0 with b1b \neq 1, the logarithm to base bb is:

logb(x)    lnxlnb(x>0).(8)\log_b(x) \;\coloneqq\; \frac{\ln x}{\ln b} \qquad (x > 0). \tag{8}

By this definition, logb(x)=y\log_b(x) = y if and only if by=xb^y = x (substitute (7)(7) to verify), which matches the familiar meaning from school algebra.

Change-of-base formula

From definition (8)(8), for any two valid bases aa and bb:

loga(x)=lnxlna=lnxlnblnblna=logb(x)1logb(a).\log_a(x) = \frac{\ln x}{\ln a} = \frac{\ln x}{\ln b} \cdot \frac{\ln b}{\ln a} = \log_b(x) \cdot \frac{1}{\log_b(a)}.

Equivalently:

loga(x)=logb(x)logb(a).\log_a(x) = \frac{\log_b(x)}{\log_b(a)}.

In practice, scientific calculators provide log10\log_{10} (common logarithm) and ln\ln; the change-of-base formula lets you reach any other base through either one.

Summary

  • The natural logarithm ln:(0,)R\ln : (0, \infty) \to \mathbb{R} is the inverse of exp\exp: ln(exp(x))=x\ln(\exp(x)) = x and exp(lny)=y\exp(\ln y) = y.
  • Key values: ln1=0\ln 1 = 0, lne=1\ln e = 1, ln(en)=n\ln(e^n) = n for all nZn \in \mathbb{Z}.
  • Product rule: ln(xy)=lnx+lny\ln(xy) = \ln x + \ln y, derived from the functional equation exp(a+b)=exp(a)exp(b)\exp(a+b) = \exp(a)\exp(b).
  • Power rule: ln(xr)=rlnx\ln(x^r) = r\ln x for all rRr \in \mathbb{R} and x>0x > 0.
  • Derivative: (lnx)=1/x(\ln x)' = 1/x, proved via the inverse function theorem.
  • ln\ln is strictly increasing on (0,)(0, \infty); lnx+\ln x \to +\infty as xx \to \infty and lnx\ln x \to -\infty as x0+x \to 0^+.
  • Integral representation: lnx=1x1tdt\ln x = \int_1^x \frac{1}{t}\,dt, an alternative definition that makes the derivative formula immediate.
  • General exponential: bxexp(xlnb)b^x \coloneqq \exp(x \ln b), with derivative (lnb)bx(\ln b)\,b^x.
  • Logarithm base bb: logb(x)lnxlnb\log_b(x) \coloneqq \frac{\ln x}{\ln b}; the change-of-base formula is loga(x)=logb(x)logb(a)\log_a(x) = \frac{\log_b(x)}{\log_b(a)}.