5. Exponential and logarithmic functions

Shared foundation: positive real powers without logarithmic circularity

Foundation contract. We take the real numbers as a complete ordered field and use induction on nonnegative integers. Finite algebra, absolute value and the definition of a sequence limit are the basic language. No logarithm, real exponential, derivative, integral or complex power law is assumed in this construction. Negative-base irrational powers are not being defined. The usual constant $e>1$ may be chosen as a base after the construction; no differential characterization of $e$ is needed for the algebra here.

1. Integer bounds, rational approximation and positive roots. The positive integers are unbounded: if their supremum were $s$, some integer would exceed $s-1$, and its successor would exceed $s$. Thus the Archimedean property holds. For every real $x$ an integer floor exists, since sufficiently large positive and negative integers bound $x$ and the finite intervening integer list has a largest member not exceeding $x$. Consequently $r_n=\lfloor nx\rfloor/n$ satisfies $r_n\le x<r_n+1/n$. Between distinct reals $u<v$, choose $n$ with $1/n<v-u$; then $(\lfloor nu\rfloor+1)/n$ lies strictly between them. These facts use completeness and integer order, not exponentials.

In the complete ordered real field, for $b>0$ and integer $m\ge1$, let $S=\{t\ge0:t^m\le b\}$. It contains $\min\{1,b\}>0$ and is bounded above by $\max\{1,b\}$, so $r=\sup S>0$ exists. Finite factorization gives, for nonnegative $t,r\le M$,

$$t^m-r^m=(t-r)\sum_{j=0}^{m-1}t^{m-1-j}r^j,\qquad |t^m-r^m|\le mM^{m-1}|t-r|.$$

Take $M=r+1$. If $r^m<b$, choose $0<h<1$ so $mM^{m-1}h<b-r^m$; then $(r+h)^m<b$, contradicting the upper bound $r$. If $r^m>b$, choose $0<h<\min\{1,r\}$ with $mM^{m-1}h<r^m-b$. Then $(r-h)^m>b$, and monotonicity of integer powers on nonnegative numbers makes $r-h$ an upper bound of $S$, contradicting its supremum. Hence $r^m=b$. At $b=0$ the unique nonnegative root is zero. The same factorization shows $t^m>s^m$ whenever $t>s\ge0$, so this positive root is unique.

2. Rational powers and their laws. For $a>0$, define $a^{p/q}$ by the positive $q$th root of $a^p$, where $p\in\mathbb Z$, $q\ge1$ and negative integer powers use reciprocals. If $p/q=p'/q'$, raising the two positive candidate values to $qq'$ gives the same value $a^{pq'}=a^{p'q}$; uniqueness of a positive root proves independence of the fraction. To add rational exponents, put them over a common positive denominator and write all powers as integer powers of its unique positive root. The integer grouping laws then give $a^{r+s}=a^ra^s$ and $a^{-r}=1/a^r$ for rational $r,s$. Positive-root uniqueness applied to integer products likewise gives $(ab)^r=a^rb^r$ for $a,b>0$ and rational $r$: if $r=p/q$, the positive number $a^rb^r$ has $q$th power $a^pb^p=(ab)^p$ and is therefore the defining positive root.

For $a>1$, a positive rational power is greater than one, so $a^r$ is strictly increasing in rational $r$. Induction proves Bernoulli's inequality $(1+d)^n\ge1+nd$ for $d\ge0$: multiplication by $1+d$ adds the nonnegative term $nd^2$ to the next step. With $d=a^{1/n}-1\ge0$ it yields

$$1\le a^{1/n}\le1+\frac{a-1}{n}.$$

Thus $a^{1/n}\to1$, and for rational $|h|\le1/n$, $a^{-1/n}\le a^h\le a^{1/n}$.

3. Supremum extension and the addition law. For $a>1$ and real $x$, define

$$E_a(x)=\sup\{a^r:r\in\mathbb Q,\ r\le x\}.$$

Choose integers $M\le x<N$. Then $a^M$ belongs to the set and $a^N$ is an upper bound. The supremum therefore exists and is at least $a^M>0$; this does not assert a positive lower bound for every member of the defining set. For rational $x$ it equals the rational power $a^x$. With $r_n=\lfloor nx\rfloor/n$ and $v_n=a^{1/n}$, monotonicity on rational exponents gives

$$a^{r_n}\le E_a(x)\le a^{r_n}v_n.$$

An integer $N>x$ bounds $a^{r_n}\le a^N$, so $0\le E_a(x)-a^{r_n}\le a^N(v_n-1)\to0$. Hence $a^{r_n}\to E_a(x)$. For real $x,y$, put $t_n=r_n(x)+r_n(y)$. Then $t_n\le x+y<t_n+2/n$, and

$$a^{t_n}\le E_a(x+y)\le a^{t_n}v_n^2.$$

The rational addition law gives $a^{t_n}=a^{r_n(x)}a^{r_n(y)}\to E_a(x)E_a(y)$, and the factors $a^{t_n}$ are bounded above by one fixed integer power. Since $v_n^2\to1$, the displayed bound proves $E_a(x+y)=E_a(x)E_a(y)$. The elementary product-limit step is justified directly by

$$|u_nv_n-uv|\le |u_n|\,|v_n-v|+|v|\,|u_n-u|,$$

because a convergent $u_n$ satisfies $|u_n|\le|u|+1$ on a tail. Choose $|v_n-v|<\varepsilon/[2(|u|+1)]$ and $|u_n-u|<\varepsilon/[2(|v|+1)]$ on a common tail; the right side is then smaller than $\varepsilon$. The squeeze steps follow by making their explicit nonnegative upper bounds smaller than $\varepsilon$.

The defining sets show that $E_a$ is nondecreasing, $E_a(0)=1$, and the addition law gives $E_a(-x)=1/E_a(x)$. If $h>0$, choose a rational $0<r<h$; then $E_a(h)\ge a^r>1$. Therefore $x<y$ implies $E_a(y)=E_a(x)E_a(y-x)>E_a(x)$: the extension is strictly increasing.

4. Continuity and every positive output. If $|h|\le1/n$, then

$$a^{-1/n}\le E_a(h)\le a^{1/n}.$$

Both bounds approach one; for $v_n=a^{1/n}\ge1$, $0\le1-v_n^{-1}\le v_n-1\to0$. Given any tolerance, choose $n$ so the bounds differ from one by less than that tolerance, then take $|h|<1/n$. This proves continuity at zero. The addition law $E_a(x+h)=E_a(x)E_a(h)$ proves continuity at every $x$, with the tolerance divided by the fixed positive factor $E_a(x)$.

Bernoulli gives $E_a(n)=a^n\ge1+n(a-1)$, which is unbounded; $E_a(-n)=1/a^n\to0$. Fix $u>0$. The real set $T=\{t:E_a(t)\le u\}$ is therefore nonempty and bounded above. Let $x=\sup T$. If $E_a(x)<u$, continuity gives a small $h>0$ with $E_a(x+h)<u$, contradicting the upper bound $x$. If $E_a(x)>u$, continuity gives a small $h>0$ with $E_a(x-h)>u$; monotonicity then makes $x-h$ an upper bound of $T$, contradicting the least upper bound. Thus $E_a(x)=u$. Strict monotonicity makes this preimage unique. We have constructed a continuous bijection $\mathbb R\to(0,\infty)$ without logarithms.

5. All positive bases and the remaining real-exponent laws. Set $1^x=1$. For $0<a<1$ define $a^x=E_{1/a}(-x)$; for $a>1$ write $a^x=E_a(x)$. These definitions agree with rational powers, preserve the addition law and continuity, and give strict decrease and the same positive range when $0<a<1$.

For $a,b>0$, choose rationals $r_n\to x$. Continuity in the exponent and the rational law $(ab)^{r_n}=a^{r_n}b^{r_n}$ give $(ab)^x=a^xb^x$. For the power-of-a-power law, first let $r=p/q$ be rational with $q>0$. The addition law proves $(a^{x/q})^q=a^x$, so positive-root uniqueness gives $(a^x)^{1/q}=a^{x/q}$. Integer powers and reciprocals now give $(a^x)^{p/q}=a^{xp/q}$. Finally take rationals $r_n\to y$ and use continuity of both exponent functions in $(a^x)^{r_n}=a^{xr_n}$ to obtain $(a^x)^y=a^{xy}$. Every base used in these root and continuity steps is positive.

6. Inverse logarithms, with no cycle. For $a>0$, $a\ne1$, define $\log_a$ to be the inverse of the bijection just proved. A strictly increasing function has a strictly increasing inverse; a strictly decreasing one has a strictly decreasing inverse, since the two possible input orders give the corresponding strict output orders. This proves the monotonicity used in logarithmic inequalities. The product, quotient, power and change-of-base logarithm proofs in this chapter now rest on the independently constructed exponent laws and inverse identities. If $\exp(x)=e^x$ and $\ln=\log_e$, then $a=e^{\ln a}$ and the now proved power law gives $a^x=\exp(x\ln a)$ as a consequence, not a circular definition used to justify those same laws.

Learning goals and conventions

Use exponent laws, convert between logarithmic bases, solve exponential equations and inequalities, and model constant proportional growth. Here $e^x$ is the real exponential; its calculus construction is studied later. We use its standard properties: continuity, strict increase, $e^{x+y}=e^xe^y$, $e^0=1$, and range $(0,\infty)$.

For $a>0$, define $a^x=e^{x\ln a}$, where $\ln:(0,\infty)\to\mathbb R$ is the inverse of $e^x$. For $a\ne1$, define $\log_a x=\ln x/\ln a$. The base restriction matters: $a=1$ gives a constant exponential with no inverse, and negative bases do not give a real exponential on all real inputs.

Theorem: logarithm laws

For positive $x,y$ and real $r$,

$$\ln(xy)=\ln x+\ln y,\qquad \ln(x/y)=\ln x-\ln y,\qquad \ln(x^r)=r\ln x.$$

Proof. Applying the exponential addition law gives $e^{\ln x+\ln y}=xy$. Uniqueness of the inverse value proves the first formula. The identity $e^{-\ln y}=1/y$ then proves the quotient law. The definition $x^r=e^{r\ln x}$ gives the power law. None of these identities asserts $\ln(x+y)=\ln x+\ln y$, which is generally false.

The map $a^x$ is strictly increasing for $a>1$ and strictly decreasing for $0<a<1$, since the sign of $\ln a$ reverses the order in $x\ln a$. Its inverse has the same monotonicity. Taking a logarithm to solve an inequality therefore requires checking both positivity and whether the chosen base reverses inequalities.

Models and exact solutions

A model $Q(t)=Q_0e^{kt}$ with $Q_0>0$ has constant proportional rate $k$ when differentiated. If $k>0$, the doubling time is $\ln2/k$. If $k<0$, the half-life is $-\ln2/k$. These expressions have units of time only when $k$ has inverse-time units. The model is an assumption about a process, not a consequence of any two measured values.

Worked example: logarithms can introduce false candidates. Solve $\ln(x-1)+\ln(x+2)=\ln4$. The domain is $x>1$. Combining logs gives $(x-1)(x+2)=4$, so $x^2+x-6=0$ and $x=2$ or $-3$. Only $2$ lies in the original domain. Checking the product alone would retain the invalid candidate $-3$.

Worked example: solve a growth calibration. Suppose $Q(0)=80$ and $Q(3)=120$ under the exponential model. Then $e^{3k}=3/2$, so $k=\ln(3/2)/3$ and $Q(t)=80(3/2)^{t/3}$. The doubling time is $3\ln2/\ln(3/2)$. Keeping this exact expression avoids rounding the rate prematurely.

Worked example: a decreasing base. The inequality $(1/3)^{2x-1}\ge9$ is $3^{1-2x}\ge3^2$, hence $1-2x\ge2$ and $x\le-1/2$. Equivalently, comparing exponents in base $1/3$ reverses the inequality.

Graded practice

1. Foundation. Simplify $\log_2 8+\log_2(1/4)$ and $\ln(e^{3x})$.
2. Core. Solve $e^{2x}-5e^x+6=0$.
3. Core. Solve $\ln(x+3)-\ln(x-1)\le\ln2$ on its real domain.
4. Proof. Prove the change-of-base formula from the definition of an inverse logarithm, without assuming that formula in advance.

Full solutions

1. The first expression is $3-2=1$. The second is $3x$ for every real $x$, because $e^{3x}>0$.
2. Put $u=e^x>0$. Then $(u-2)(u-3)=0$, and both positive roots are allowed. Thus $x=\ln2$ or $x=\ln3$.
3. The domain is $x>1$. Since $\ln$ is increasing, the inequality is $(x+3)/(x-1)\le2$. The denominator is positive there, so $x+3\le2x-2$, hence $x\ge5$. This already respects the original domain.
4. Let $y=\log_a x$ mean $a^y=x$, with $a>0$, $a\ne1$ and $x>0$. By $a^y=e^{y\ln a}$ and inverse uniqueness, $y\ln a=\ln x$. Since $\ln a\ne0$, division gives $y=\ln x/\ln a$. More generally $\log_a x=\log_b x/\log_b a$ for any admissible base $b$, by substituting the natural-log expressions and cancelling.

Summary and term definitions

An exponential function has a fixed positive base and variable exponent; a logarithm is its inverse. Proportional growth scales current size by a constant rate. A doubling time or half-life is a model-specific elapsed time, not the exponent itself. Logarithm laws require positive arguments. Solve algebraically only after recording the domain, then reject candidates that violate it.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: bounded monotone functions are integrable

For nondecreasing $f$, take a uniform partition with $n$ intervals of width $(b-a)/n$. On interval $i$, the infimum and supremum are the endpoint values. Hence

$$U-L=\frac{b-a}{n}\sum_{i=1}^n[f(x_i)-f(x_{i-1})] =\frac{(b-a)(f(b)-f(a))}{n}\longrightarrow0.$$

The criterion proves integrability even if $f$ has jump discontinuities. For a nonincreasing function apply the result to $-f$.

Bounded proper Riemann integrals: mesh equivalence and elementary properties

Let $a<b$ be finite, and let $f:[a,b]\to\mathbb R$ be bounded. For a partition $P$, write $L(f,P),U(f,P)$ for the lower and upper Darboux sums. We use real completeness and the proved Darboux criterion: equality of lower and upper integrals is equivalent to arbitrarily small $U-L$. This packet does not assume continuity of $f$.

Darboux implies every fine tagged sum converges. Let the Darboux integral be $I$. Given $\varepsilon>0$, choose a fixed partition $P_0$ with gap below $\varepsilon/2$. Choose $M>0$ with $|f|\le M$, and let $k$ be the number of interior partition points of $P_0$. For any other partition $P$, mark as bad every cell whose closure contains one of those $k$ points. Their total length is at most $2k\|P\|$, because each point belongs to at most two cells. Every remaining cell lies within a single $P_0$ cell. The contribution of all good cells to the oscillation sum $U(f,P)-L(f,P)$ is at most the gap for $P_0$; the bad cells contribute at most $2M(2k\|P\|)$. Thus

$$U(f,P)-L(f,P)<\varepsilon/2+4Mk\|P\|.$$

Take $\|P\|<\varepsilon/[8M\max(1,k)]$. Every tagged sum $S(P)$ and $I$ lie between $L(f,P)$ and $U(f,P)$, so $|S(P)-I|<\varepsilon$, independently of tags.

The converse. Suppose all sufficiently fine tagged sums are within $\eta$ of a common finite $I$. Fix one such fine partition. Each cell supremum and infimum can be approximated by actual function values with error below $\eta/(b-a)$. The resulting upper-approximating tagged sum exceeds $U(f,P)-\eta$, and the lower-approximating sum is below $L(f,P)+\eta$. Therefore $U(f,P)<I+2\eta$ and $L(f,P)>I-2\eta$. Since $\eta$ can be arbitrarily small, the Darboux criterion gives integrability. Lower and upper bounds on tagged sums then identify its integral as $I$. Suprema need not be attained for this argument.

Linearity, constants and order. For integrable $f,g$ and real $\alpha,\beta$, on any common tagged partition the exact finite identity is $S(\alpha f+\beta g)=\alpha S(f)+\beta S(g)$. Its limit exists, so the equivalence just proved gives integrability and the corresponding linear integral identity. The constant function $c$ has every sum equal to $c(b-a)$. If $f\le g$, their tagged sums with the same tags obey the same order; their limits do too. Consequently $|\int f|\le M(b-a)$ whenever $|f|\le M$.

Absolute values. The reverse triangle inequality gives $\operatorname{osc}_J|f|\le\operatorname{osc}_J f$ on every cell $J$: apply $||f(x)|-|f(y)||\le|f(x)-f(y)|$ and approximate the suprema/infima. Thus every Darboux gap for $|f|$ is at most the gap for $f$, and $|f|$ is integrable. Since $-|f|\le f\le|f|$, order and linearity give $|\int_a^b f|\le\int_a^b|f|$.

Restriction and interval addition. For $a<c<b$, refine any small-gap partition by $c$. The nonnegative gaps on $[a,c]$ and $[c,b]$ sum to the refined whole gap; each can therefore be arbitrarily small. Both restrictions are integrable. Join partitions of the two pieces with meshes tending to zero. The exact sum on the joined partition is the sum of the two tagged sums, so passage to their three limits gives $\int_a^b f=\int_a^c f+\int_c^b f$. Conversely, if both restrictions are bounded integrable, join two partitions with gaps below $\varepsilon/2$; the resulting whole gap is below $\varepsilon$, proving whole integrability and the same identity. At a shared endpoint both restrictions use the same value of $f$; no endpoint term has positive interval length. Set equal-bound integrals to zero and reversed-bound integrals to the negative. Splitting a finite interval and cancellation now extend addition to every ordering of its three endpoints.

These proofs justify the estimates used in accumulation, continuous FTC, integration by parts and proper substitution. They assert bounded proper Riemann statements only, and do not replace the separate convergence definitions at singularities or infinite endpoints.

From a proper area logarithm to the natural exponential

Use proper integrals of continuous functions on finite closed intervals, their proved additivity/order/linearity, the completeness and Archimedean properties, and the intermediate value theorem. Define

$$L(x)=\int_1^x\frac{dt}{t}\qquad(x>0),$$

with oriented bounds. Every integral here is proper: its interval stays a positive distance from zero.

Continuity and strict increase. For $0<u<v$, additivity and order give $L(v)-L(u)=\int_u^vdt/t\ge(v-u)/v>0$. Near any fixed $a>0$, restrict $x>a/2$. Then $|L(x)-L(a)|\le(2/a)|x-a|$, proving continuity without using a logarithmic derivative.

The product law from actual sums. Fix $u>0$. Scaling a tagged partition of the interval between $1$ and $v>0$ by $u$ scales its mesh by $u$ and its widths by $u$. The corresponding sums satisfy

$$\sum_i\frac{u\Delta s_i}{u\xi_i}=\sum_i\frac{\Delta s_i}{\xi_i}.$$

Both are sums of continuous proper integrands, so their all-mesh limits are equal. For $v<1$ reverse orientation; $v=1$ is zero directly. Hence $\int_u^{uv}dt/t=L(v)$ and

$$L(uv)=L(u)+L(v).$$

In particular $L(1)=0$, $L(1/u)=-L(u)$ and $L(2^n)=nL(2)$. As $L(2)>0$, the latter values are unbounded above by the Archimedean property; the reciprocal values are unbounded below. For any real $y$, choose positive arguments with $L$ below and above $y$. Continuity and the intermediate value theorem give an $x>0$ with $L(x)=y$. Strict increase makes that $x$ unique.

Construct and control the inverse. Define $E:\mathbb R\to(0,\infty)$ as this inverse. It is strictly increasing and continuous. To prove continuity at $y_0=L(a)$, choose $0<d<\min(a,\varepsilon)$. The numbers $L(a-d)<y_0<L(a+d)$ have positive gaps. If $|y-y_0|$ is smaller than both gaps, monotonicity forces $a-d<E(y)<a+d$, so $|E(y)-a|<\varepsilon$. The product law and uniqueness yield $E(s+t)=E(s)E(t)$, $E(0)=1$ and $E(-t)=1/E(t)$. Define $e=E(1)>1$ and $a^t=E(tL(a))$ for $a>0$. It follows that $a^{s+t}=a^sa^t$ and $(a^s)^t=a^{st}$, since $L(E(z))=z$. For integer exponents these equal repeated products; for rational $m/n$, $n>0$, the definition is the unique positive $n$th root of $a^m$. Thus it agrees with the elementary positive rational powers and gives their continuous real extension. In particular $E(t)=e^t$, and the area logarithm is its inverse $\ln=L$.

Derivatives without a cycle. For $a>0$, the average of $1/t$ on the interval between $a$ and $a+h$, with the sign of $h$ and orientation both retained, lies between its smallest and largest values on that interval. Both bounds tend to $1/a$. Thus

$$\frac{L(a+h)-L(a)}h\longrightarrow\frac1a.$$

For $y_0=L(a)$ and $h\ne0$, put $v=E(y_0+h)\ne a$. Inverse continuity gives $v\to a$, and

$$\frac{E(y_0+h)-E(y_0)}h =\left(\frac{L(v)-L(a)}{v-a}\right)^{-1}\longrightarrow a=E(y_0).$$

The denominator limit is nonzero, so the reciprocal is justified. This proves $L'=1/x$ and $E'=E$ from proper area and inverse continuity, with neither l'Hopital nor a previously assumed exponential derivative. The chain rule then gives $(x^r)'=rx^{r-1}$ on $x>0$ by $x^r=E(rL(x))$, and $(b^x)'=(\ln b)b^x$ for $b>0$. The case $b=1$ is the constant one. This supplies the normalization and identity bridge between the area logarithm, the inverse exponential and the real-power notation.

The discussion concerns real one-variable functions.

1. Chain and quotient rules

Let $g$ be differentiable at an interior point $a$, and $f$ differentiable at $g(a)$ on a neighborhood of that point. Then $f\circ g$ is differentiable at $a$ and

$$(f\circ g)'(a)=f'(g(a))g'(a).$$

Proof. Differentiability of $f$ gives $f(g(a)+u)-f(g(a))=f'(g(a))u+u\epsilon(u)$, where $\epsilon(u)\to0$ as $u\to0$; define $\epsilon(0)=0$. Put $u=g(a+h)-g(a)$. Differentiability of $g$ implies $u\to0$ and $u/h\to g'(a)$. Thus the difference quotient is $(f'(g(a))+\epsilon(u))(u/h)$ and converges to the claimed product. This argument still works when $u=0$ for some nonzero $h$; dividing by $u$ would not.

If $f,g$ are differentiable at $a$ and $g(a)\ne0$, then

$$\left(\frac fg\right)'(a)=\frac{f'(a)g(a)-f(a)g'(a)}{g(a)^2}.$$

Indeed continuity of $g$ makes it nonzero near $a$. The reciprocal difference quotient equals $-[g(a+h)-g(a)]/[h g(a+h)g(a)]$, whose limit is $-g'(a)/g(a)^2$; apply the product rule to $f(1/g)$.

Example. For every real $x$, $F(x)=\log(1+x^2)$ is differentiable because $1+x^2>0$, and the chain rule gives $F'(x)=2x/(1+x^2)$.

Identifying the natural exponential and the model's rate

The positive-real-power construction above applies to every fixed base $a>1$; by itself it does not identify which number is the natural base. Here we choose that base explicitly using the separately established area construction: let $L(x)=\int_1^x dt/t$ for $x>0$, let $E=L^{-1}$, and set $e=E(1)>1$. The area result proves $E(0)=1$, $E(s+t)=E(s)E(t)$, continuity, $E'(t)=E(t)$ and $L'(x)=1/x$. This is the normalization used for $e$ here; no separate series definition of $e$ is being identified.

The supremum construction and $E$ agree for base $e$. Indeed, the addition law gives $E(n)=e^n$ for integers $n$, including negative integers by reciprocals. For any positive integer $q$, $E(1/q)>0$ and $E(1/q)^q=E(1)=e$; uniqueness of the positive $q$th root gives $E(1/q)=e^{1/q}$ in the supremum construction. Hence $E(p/q)=e^{p/q}$ for every integer $p$ and positive integer $q$. For any real $x$, choose rationals $r_j\to x$. Both $E$ and the supremum-constructed function $x\mapsto e^x$ are continuous, so taking limits proves $E(x)=e^x$. Since both are bijections from $\mathbb R$ onto $(0,\infty)$, their inverses agree: $L=\ln$. Therefore

$$\frac{d}{dx}e^x=e^x,\qquad \frac{d}{dx}\ln x=\frac1x\quad(x>0).$$

For a constant $C$, pulling it outside the difference quotient proves $(Cg)'=Cg'$ whenever $g$ is differentiable. For real constants $C,k$, this rule and the chain rule give $f(t)=Ce^{kt}$ and $f'(t)=kCe^{kt}=kf(t)$ for every real $t$. This remains true if $C=0$ or $k=0$. Since $e^{kt}>0$, the ratio $f'(t)/f(t)=k$ is defined precisely when $C\ne0$. Thus the model with $Q_0>0$ has the stated proportional rate. This calculation verifies the displayed model; uniqueness of solutions to an initial-value equation $f'=kf$ requires a further argument, for example a separately established real mean value theorem, and is not asserted here.