The exponential function eˣ
The exponential function \(e^x\) is built on the base \(e \approx 2.718\), an irrational constant as fundamental as \(\pi\). What singles out \(e\) from every other base is one property: the curve \(y = e^x\) has, at every point, a slope exactly equal to its own height. No other base does this — \(2^x\) and \(10^x\) grow with slopes proportional to, but not equal to, their heights. This self-replicating behaviour is the reason \(e\) appears in every natural process whose rate of change is proportional to the amount present.
The exponential obeys the index laws you already know, now with a continuous exponent: \(e^a \cdot e^b = e^{a+b}\), \(e^a / e^b = e^{a-b}\), and \((e^a)^b = e^{ab}\). Two reference values anchor the curve: \(e^0 = 1\), so every exponential passes through the point \((0,1)\); and \(e^{-x} = 1/e^x\), so the decaying exponential is simply the growing one reflected across the vertical axis. As \(x \to +\infty\), \(e^x\) climbs without bound; as \(x \to -\infty\), it sinks toward zero but never reaches it.
Natural log vs common log
The logarithm answers the inverse question: to what power must the base be raised to give a number? The natural logarithm \(\ln x\) uses base \(e\), so \(\ln x = y\) means \(e^y = x\). The common logarithm \(\log_{10} x\) uses base 10, so \(\log_{10} x = y\) means \(10^y = x\). The two are exact inverses of their respective exponentials: \(\ln(e^x) = x\) and \(e^{\ln x} = x\); likewise \(\log_{10}(10^x) = x\).
The two logarithms differ only by a constant factor. Since \(x = 10^{\log_{10} x}\) and taking \(\ln\) of both sides gives \(\ln x = (\ln 10)\log_{10} x\), and \(\ln 10 \approx 2.303\), the bridge is
\[ \ln x = 2.303 \, \log_{10} x. \]
Physics decay and circuit problems use \(\ln\) because the natural form \(e^{-\lambda t}\) arises directly from the rate equation. Physical-chemistry quantities such as pH and pK are defined on base 10 and use \(\log_{10}\). Mixing the two is the most expensive routine error in this topic, because every answer is then wrong by the factor 2.303.
The log laws
The log laws convert multiplication into addition and powers into multipliers. They hold for any base; we state them for \(\ln\) because that is the physics workhorse, but the same forms apply to \(\log_{10}\). These four identities are the only algebra you need to push exponentials around.
| Law | Identity | What it does |
|---|---|---|
| Product | \(\ln(ab) = \ln a + \ln b\) | The log of a product is the sum of the logs. |
| Quotient | \(\ln\!\left(\dfrac{a}{b}\right) = \ln a - \ln b\) | The log of a ratio is the difference of the logs. |
| Power | \(\ln(a^n) = n\,\ln a\) | An exponent drops to the front as a multiplier. |
| Change of base | \(\log_b a = \dfrac{\ln a}{\ln b}\) | Re-expresses any base through the natural log. |
Three standard values are worth memorising for fast estimation: \(\ln 1 = 0\) (since \(e^0 = 1\)), \(\ln e = 1\), and \(\ln 2 \approx 0.693\). The last is the constant of the half-life and appears in almost every decay calculation NEET poses. Note the domain restriction — the logarithm of a non-positive number is undefined, because no real power of a positive base yields zero or a negative result.
Derivative and integral
Two calculus facts are all that physics demands of this function, and both are short. The exponential is its own derivative, and the natural log is the integral of the reciprocal.
| Operation | Result | Where it bites |
|---|---|---|
| \(\dfrac{d}{dx}\,e^x\) | \(e^x\) | The function equal to its own slope; underlies all decay. |
| \(\dfrac{d}{dx}\,\ln x\) | \(\dfrac{1}{x}\) | Natural log only; for \(\log_{10}\) multiply by 0.4343. |
| \(\dfrac{d}{dt}\,e^{-\lambda t}\) | \(-\lambda\,e^{-\lambda t}\) | Chain rule: the decay constant comes out front. |
| \(\displaystyle\int \dfrac{1}{x}\,dx\) | \(\ln|x| + C\) | The \(n=-1\) exception to the power rule. |
| \(\displaystyle\int e^{-\lambda t}\,dt\) | \(-\dfrac{1}{\lambda}\,e^{-\lambda t} + C\) | Chain rule in reverse; capacitor and decay totals. |
The chain rule is what makes \(\dfrac{d}{dt}e^{-\lambda t} = -\lambda e^{-\lambda t}\). The inside function is \(-\lambda t\); its derivative \(-\lambda\) is pulled to the front. This single result is the engine behind decay: it says the rate of change of an exponentially decaying quantity is proportional to the quantity itself, with the constant of proportionality being \(-\lambda\). Run that statement backwards — "the rate of decrease is proportional to the amount present" — and you arrive at exactly the differential equation \(dN/dt = -\lambda N\), whose solution is \(N = N_0 e^{-\lambda t}\).
The chain rule and these standard derivatives are developed in full in differentiation — basics for physics, with the reverse process in integration — basics for physics.
Growth, decay and the time constant
Whenever the rate of change of a quantity is proportional to the quantity itself, the solution is exponential. A positive proportionality gives growth, \(N = N_0 e^{+kt}\); a negative one gives decay, \(N = N_0 e^{-\lambda t}\). The two share a shape and a method — take the natural log of both sides to linearise, turning the curve into a straight line whose slope is the rate constant.
| Decay: \(N = N_0 e^{-\lambda t}\) | Growth: \(N = N_0 e^{+kt}\) | |
|---|---|---|
| Behaviour | Falls toward 0 as \(t \to \infty\) | Rises without bound |
| Characteristic time | Half-life \(T_{1/2} = \dfrac{\ln 2}{\lambda}\) | Doubling time \(T = \dfrac{\ln 2}{k}\) |
| Linear form | \(\ln N = \ln N_0 - \lambda t\) | \(\ln N = \ln N_0 + kt\) |
| Slope of \(\ln N\) vs \(t\) | \(-\lambda\) | \(+k\) |
| NEET examples | Radioactivity, RC discharge, cooling | RC charging (as \(1-e^{-t/RC}\)), avalanche |
The time constant \(\tau\) is the natural ruler of an exponential. Writing decay as \(N = N_0 e^{-t/\tau}\), the constant \(\tau\) is the time for \(N\) to fall to \(1/e \approx 37\%\) of its initial value. For radioactivity \(\tau = 1/\lambda\); for an RC circuit \(\tau = RC\); for an LR circuit \(\tau = L/R\). The half-life is a different reference level on the same curve — the time to reach 50% — and the two are linked by \(T_{1/2} = \tau \ln 2 \approx 0.693\,\tau\). NEET routinely asks students to slide between \(\tau\), \(\lambda\) and \(T_{1/2}\), and to read \(\lambda\) off the slope of a \(\ln N\) versus \(t\) plot.
Half-life from the decay constant
The half-life is defined as the time for a decaying quantity to fall to half its starting value. Setting \(N = N_0/2\) in \(N = N_0 e^{-\lambda t}\) gives \(\tfrac{1}{2} = e^{-\lambda T_{1/2}}\). Taking the natural log of both sides and using \(\ln(1/2) = -\ln 2\):
\[ -\ln 2 = -\lambda T_{1/2} \quad\Longrightarrow\quad T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}. \]
The power of the half-life is that it repeats. After each interval \(T_{1/2}\) the amount halves again, independent of where you started, because the exponential has no memory. After \(n\) half-lives the fraction remaining is \((1/2)^n\): one half, then a quarter, then an eighth, and so on. This is captured equivalently by \(N = N_0 (1/2)^{t/T_{1/2}}\), which is just \(N_0 e^{-\lambda t}\) rewritten.
Worked examples
Simplify \(\ln\!\left(\dfrac{a^3 b}{c^2}\right)\) into a sum of single logarithms, and hence express \(t\) from \(N = N_0 e^{-\lambda t}\).
Apply quotient then product then power. The quotient law splits numerator from denominator: \(\ln(a^3 b) - \ln(c^2)\). The product law splits the numerator: \(\ln(a^3) + \ln b - \ln(c^2)\). The power law lowers the exponents: \(\boxed{3\ln a + \ln b - 2\ln c}\).
Now invert the decay equation. Divide by \(N_0\): \(N/N_0 = e^{-\lambda t}\). Take \(\ln\): \(\ln(N/N_0) = -\lambda t\). By the quotient law, \(\ln N - \ln N_0 = -\lambda t\), so \(t = \dfrac{1}{\lambda}\ln\!\left(\dfrac{N_0}{N}\right)\). The same log laws that simplify an algebraic expression are what extract time from a decay curve.
A radioactive sample has a decay constant \(\lambda = 0.0231~\text{s}^{-1}\). (a) Find its half-life. (b) What fraction of the sample remains after 90 s?
(a) Half-life. \(T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.0231} = 30~\text{s}\).
(b) Fraction after 90 s. Ninety seconds is exactly three half-lives, since \(90/30 = 3\). The fraction remaining is \((1/2)^3 = 1/8\). As a check via the exponential: \(e^{-\lambda t} = e^{-0.0231 \times 90} = e^{-2.079} = e^{-3\ln 2} = (1/2)^3 = 1/8\). Only \(12.5\%\) of the sample survives.
A capacitor discharges through a resistor with time constant \(\tau = RC = 2~\text{ms}\). How long does the charge take to fall to \(20\%\) of its initial value? Take \(\ln 5 \approx 1.609\).
Set up the ratio. The charge follows \(q = q_0 e^{-t/\tau}\). We want \(q/q_0 = 0.20 = 1/5\), so \(e^{-t/\tau} = 1/5\).
Take the natural log. \(-t/\tau = \ln(1/5) = -\ln 5\), hence \(t = \tau \ln 5 = 2~\text{ms} \times 1.609 = 3.22~\text{ms}\). Note the natural log is mandatory here — using \(\log_{10} 5 \approx 0.699\) would give \(1.4~\text{ms}\), a wrong answer off by the factor 2.303.
The exponential toolkit in one breath
- \(e \approx 2.718\); \(y = e^x\) has slope equal to its height, so \(\dfrac{d}{dx}e^x = e^x\) and \(\dfrac{d}{dx}\ln x = \dfrac{1}{x}\).
- \(\ln\) is base \(e\), \(\log_{10}\) is base 10, and \(\ln x = 2.303\log_{10}x\). Physics decay uses \(\ln\).
- Log laws: \(\ln(ab) = \ln a + \ln b\), \(\ln(a/b) = \ln a - \ln b\), \(\ln(a^n) = n\ln a\). There is no rule for \(\ln(a+b)\).
- Decay: \(N = N_0 e^{-\lambda t}\). Linearise as \(\ln N = \ln N_0 - \lambda t\), slope \(-\lambda\).
- Half-life \(T_{1/2} = (\ln 2)/\lambda = 0.693/\lambda\); time constant \(\tau = 1/\lambda\); \(T_{1/2} = 0.693\,\tau\).
- After \(n\) half-lives the fraction left is \((1/2)^n\).