Part IV · Series, Exponentials, and Integral Calculus

Chapter 11 — The Exponential and Logarithmic Functions

Why This Matters

The equation $$y'=y$$ describes a quantity whose rate of change is proportional to its current value. It appears in ideal growth and decay models and, after scaling, in first-order circuit responses. The exponential function is the function that solves it. The logarithm reverses exponential relationships.

$\exp(x)$ is defined by its derivative, not by repeated multiplication. The exponential function generalises exponentiation to all real inputs. Its defining property is $\frac{d}{dx}e^x = e^x$ — the function equals its own rate of change.

Why e^x equals its own derivative

Starting point. Define $f(x) = e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ .

\[f'(x) = \frac{d}{dx}\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right)\]
Differentiate the power series term-by-term (valid for all real $$x$$ ).reversible
\[f'(x) = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \frac{4x^3}{4!} + \cdots\]
Apply the power rule $\frac{d}{dx} x^n = n x^{n-1}$ to each individual term.reversible
\[f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = f(x)\]
Simplify coefficients $\frac{n}{n!} = \frac{1}{(n-1)!}$ : the original series reappears identically.reversible

Conclusion. The function $$f(x) = e^x$$ is its own derivative: $$f'(x) = f(x)$$ with initial condition $$f(0) = 1$$ . It is the unique function satisfying this property.

The base e as a natural limit

From

Continuous compounding: compound interest with rate $$r = 1$$ split into $$n$$ intervals yields $\left(1 + \frac{1}{n}\right)^n$ .
The derivative definition requires $\lim_{h\to0} \frac{e^h - 1}{h} = 1$ .

Letting $$n = 1/h$$ , the limit becomes $\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = e \approx 2.71828$ .
The number $$e$$ is not an arbitrary constant: it is the exact base that makes the derivative of $$b^x$$ have scale factor $$1$$ (since $\frac{d}{dx}b^x = b^x \ln b$ ).

Result. $e = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828$ . The natural base simplifies all calculus derivatives by eliminating extra $\ln b$ scale factors.

11.1 Defining $\exp(x)$

Exponential models in engineering. Whenever a quantity changes at a rate proportional to its current size, the response is exponential: growth ( $$k > 0$$ ) or decay ( $$k < 0$$ ).

Deriving the time constant

Starting point. For a first-order system with decay rate $k = 1/\tau$ : $y(t) = A e^{-t/\tau}$ .

\[y(0) = A\]
At $$t = 0$$ , the initial value is $$A$$ .reversible
\[y(\tau) = A e^{-1} \approx 0.368 A\]
After one time constant $t = \tau$ , the quantity falls to $1/e \approx 36.8\%$ of its initial value (a $63.2\%$ drop).reversible
\[y(3\tau) = A e^{-3} \approx 0.0498 A\]
After three time constants $t = 3\tau$ , under $5\%$ of the initial value remains.reversible
\[y(5\tau) = A e^{-5} \approx 0.0067 A\]
After five time constants $t = 5\tau$ , less than $1\%$ remains — practically settled in engineering circuits.reversible

Conclusion. The time constant $\tau = 1/k$ sets the natural timescale: $1\tau \to 36.8\%$ , $3\tau \to 5\%$ , $5\tau \to <1\%$ .

Identify the initial value $$A = y(0)$$ .
Determine the rate constant $$k$$ or time constant $\tau = 1/k$ from the physical circuit parameters (e.g. $\tau = RC$ ).
Write the model: $y(t) = A e^{-t/\tau}$ (decay) or $y(t) = A(1 - e^{-t/\tau})$ (step response).
Check units: the exponent $-t/\tau$ must be dimensionless.

The exponential function can be defined by its power series

\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.

The ratio test shows convergence for every real $$x$$ . Differentiating term by term returns the same series:

\frac d{dx}\exp(x)=\exp(x).

Let $e=\exp(1)$ . The addition law

\exp(x+y)=\exp(x)\exp(y)

implies $\exp(x)=e^x$ for real $$x$$ . It also gives $e^{-x}=1/e^x$ , so $$e^x>0$$ for all real $$x$$ .

Why the exponential is unusually stable. Its power-series coefficients are chosen so that differentiating shifts every term into the same series. That self-reproduction is the analytic reason the exponential models processes whose rate is proportional to their current amount.

e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots

11.2 Exponential growth and decay

The function $Ae^{kt}$ has initial value $$A$$ . If $$k>0$$ , it grows; if $$k<0$$ , it decays. Its derivative is

\frac d{dt}(Ae^{kt})=kAe^{kt}.

The time constant $\tau>0$ is often used in decay as $e^{-t/\tau}$ . After one time constant the remaining fraction is $e^{-1}\approx0.368$ ; after five it is about $$0.0067$$ . These are model-based landmarks, not abrupt physical cutoffs.

Engineering consequence. Exponential change appears when the instantaneous rate is proportional to the amount currently present. Equal time intervals then multiply the amount by the same factor rather than adding the same amount.

\frac{dy}{dt}=ky\;\Longrightarrow\;y(t)=Ae^{kt}

Exponential growth and decay curves sharing an initial value.

11.3 Logarithms as inverses

The natural logarithm $\ln(x)$ undoes the exponential. Because $$e^x > 0$$ for all real $$x$$ , its inverse $\ln(x)$ is defined strictly for $$x > 0$$ .

Derivative of ln(x) via inverse function rule

Starting point. Let $y = \ln x$ , so $$x = e^y$$ .

\[\frac{d}{dx}(x) = \frac{d}{dx}(e^y)\]
Differentiate both sides with respect to $$x$$ using the chain rule.reversible
\[1 = e^y \frac{dy}{dx}\]
The derivative of $$e^y$$ with respect to $$x$$ is $e^y \frac{dy}{dx}$ .reversible
\[\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\]
Solve for $\frac{dy}{dx}$ and substitute back $$e^y = x$$ .reversible

Conclusion. $\frac{d}{dx}\ln x = \frac{1}{x}$ for all $$x > 0$$ . This fills the missing power rule gap: $\int x^{-1} dx = \ln|x| + C$ .

Because $$e^x$$ is strictly increasing and positive, it has an inverse $\ln x$ on $$x>0$$ :

\ln(e^x)=x,\qquad e^{\ln x}=x\quad(x>0).

The addition law for exponentials becomes logarithm laws:

\ln(ab)=\ln a+\ln b,\qquad \ln(a^r)=r\ln a,

when the real expressions are defined. Differentiating $e^{\ln x}=x$ gives

e^{\ln x}\frac d{dx}\ln x=1,

so $\frac d{dx}\ln x=1/x$ for $$x>0$$ .

For $u(x)\ne0$ on an interval,

\frac d{dx}\ln|u(x)|=\frac{u'(x)}{u(x)}, \qquad \int\frac{u'(x)}{u(x)}\,dx=\ln|u(x)|+C.

Undoing a scale factor in the exponent. The logarithm answers the exponent question: which power of the base gives this positive number? Because an exponential with positive base is one-to-one, its logarithm reverses it. This also explains why logarithm laws come from exponent laws rather than from memorised symbol patterns.

\ln(e^x)=x,\qquad e^{\ln y}=y\;(y>0)

11.4 General bases

For $$a>0$$ , $a\ne1$ ,

\log_a x=\frac{\ln x}{\ln a},\qquad x>0.

This gives

\frac d{dx}a^x=a^x\ln a.

Example — solving an exponential equation

Solve $5e^{-t/2}=0.2$ . Divide by $$5$$ :

e^{-t/2}=0.04.

Take natural logs, which is legal because both sides are positive:

-\frac t2=\ln(0.04),\qquad t=-2\ln(0.04).

The positive answer is expected: a positive decaying exponential reaches a smaller positive threshold at a later time.

Converting between general bases and base $$e$$ . Every exponential $$b^x$$ and logarithm $\log_b x$ can be rewritten in base $$e$$ using natural logarithms.

Base conversion formulas

Starting point. Let $b > 0, b \neq 1$ . Use the identity $b = e^{\ln b}$ .

\[b^x = (e^{\ln b})^x = e^{x \ln b}\]
Rewrite general base $$b^x$$ in natural exponential form.reversible
\[\frac{d}{dx}(b^x) = \frac{d}{dx}(e^{x \ln b}) = (\ln b) e^{x \ln b} = (\ln b) b^x\]
Differentiate using the chain rule: an extra factor of $\ln b$ appears.reversible
\[y = \log_b x \iff b^y = x \iff y \ln b = \ln x \iff y = \frac{\ln x}{\ln b}\]
Apply natural log to both sides of $$b^y = x$$ to derive the change-of-base formula.reversible

Conclusion. $b^x = e^{x \ln b}$ and $\log_b x = \frac{\ln x}{\ln b}$ . In calculus, always convert general bases to base $$e$$ before differentiating or integrating.

To differentiate $$b^x$$ : write $b^x = e^{x \ln b}$ , then differentiate to get $(\ln b) b^x$ .
To differentiate $\log_b x$ : write $\log_b x = \frac{\ln x}{\ln b}$ , then differentiate to get $\frac{1}{x \ln b}$ .
Notice that when $$b = e$$ , $\ln e = 1$ , recovering the simpler natural formulas.

11.5 Exponentials, logarithms, and model checks

The equation $e^{u}=e^{v}$ permits $$u=v$$ because the exponential is one-to-one. Taking $\ln$ of an equation is valid only when both sides are positive real numbers. A logarithm can turn a product into a sum and an exponential unknown into a linear unknown, but it cannot repair a negative or zero argument.

Example — solve and interpret a decay threshold

If $Q(t)=12e^{-t/5}$ mC, find when $$Q=3$$ mC:

3=12e^{-t/5},\qquad \frac14=e^{-t/5}.

Taking natural logarithms gives

\ln\frac14=-\frac{t}{5},\qquad t=5\ln4.

The positive time agrees with decay. A negative answer would signal an algebraic sign error or a threshold above the initial quantity.

Example — logarithmic differentiation of a variable power

For $$y=x^x$$ on $$x>0$$ , take logs before differentiating:

\ln y=x\ln x.

Then $y'/y=\ln x+1$ , hence

\frac{d}{dx}x^x=x^x(\ln x+1).

Neither the ordinary power rule nor the $$a^x$$ rule alone applies because both base and exponent vary.

Engineering Example — time constant from two measurements

If a decaying voltage follows $v(t)=V_0e^{-t/\tau}$ , then

\ln v=\ln V_0-\frac{t}{\tau}.

A plot of $\ln v$ against $$t$$ is a straight line of slope $-1/\tau$ , provided the measurements are positive and the single-exponential model is appropriate. This is a mathematical linearisation of an exponential model.

Sanity checks for exponential models. Always check asymptotic limits, dimensions, and signs when setting up exponential equations.

Initial value: at $$t = 0$$ , does $$y(0)$$ match the known initial condition? (For $y = A e^{-kt}$ , $$y(0) = A$$ . For $y = A(1 - e^{-kt})$ , $$y(0) = 0$$ .)
Long-term limit: as $t \to \infty$ , does $$y(t)$$ approach the correct physical equilibrium? (Decay $\to 0$ , charging $\to V_s$ .)
Exponent units: the argument of an exponential must be dimensionless ( $$kt$$ or $t/\tau$ has unit $$1$$ ).
Monotonicity: check whether the physical quantity should be strictly increasing or decreasing.

Charging capacitor model check