Practice: The Natural Logarithm

(a) The product law. Split the defining integral at $a$:

It remains to show the second integral equals $\ln b$. Substitute $t=au$, so $dt=adu$, and the limits $t=a,ab$ become $u=1,b$:

Hence $\ln (ab)=\ln a+\ln b$. What makes it work. The factor $a$ appears twice, once from $dt$ and once in the denominator $t=au$, and cancels exactly. This happens only for the integrand $1/t$: under $t=au$ the function $t^n$ acquires $a^{n+1}$, which equals 1 only when $n=-1$. Geometrically, stretching an interval by $a$ multiplies its width by $a$ and divides the height of $1/t$ by $a$, leaving the area unchanged. So the product law is a property of the reciprocal specifically, the same exceptional power the power rule cannot integrate. Verified: $\ln 6=1.791759469228=\ln 2+\ln 3$; $\ln 35=3.555348061489=\ln 5+\ln 7$; $\ln 4=1.386294361120=\ln 0.5+\ln 8$, this last with one argument below 1. (b) Domain, sign, and the absolute value. Why . The integrand $1/t$ has a singularity at $t=0$. For the interval from 1 to $x$ lies entirely within $(0,\mathrm{\infty })$ and never meets it, so the integral of a continuous function on a closed bounded interval exists. For $x\le 0$ the path would have to cross $t=0$, where $1/t$ is unbounded, and the integral does not exist. This is the source of the condition imposed on $\ln g(x)$, inherited from the integral rather than stipulated. Why the value is negative below 1. The integrand is positive throughout $(0,\mathrm{\infty })$, so the sign cannot come from it. It comes from orientation: when the upper limit lies below the lower, and ${\int }_1^x=-{\int }_x^1$. The area from $0.5$ to $1$ is $0.693147180560$, so $\ln 0.5=-0.693147180560$, which also matches the quotient law, $\ln (1/2)=\ln 1-\ln 2=-\ln 2$. In general on $(0,1)$, $\ln 1=0$ because the interval is empty, and beyond 1. Why $\ln |u|$. On an interval of negative numbers, $\frac{d}{du}\ln (-u)=\frac{-1}{-u}=\frac{1}{u}$, so $\ln (-u)$ is also an antiderivative of $1/u$ there. Writing $\ln |u|$ covers both intervals in one expression. The restriction: this is valid on any interval not containing 0, and an integral spanning 0 still fails to exist. Verified: ${\int }_{-2}^{-1}\frac{dt}{t}=-0.693147180560$, equal to $\ln |-1|-\ln |-2|=0-\ln 2$. (c) The equation. Domain first. The terms require and , so the domain is . Recording this before the algebra is what makes the final check possible. Combine and solve. By the product law, valid since both arguments are positive on the domain,

Since $\ln$ is strictly increasing, hence injective, the arguments are equal:

Check against the domain. $x=5$ lies in and is valid. $x=-2$ does not; substituting into the original gives $\ln (-2)+\ln (-5)$, neither term defined. Rejected. Why the algebra produced it. The step to $\ln (x(x-3))$ is valid only where both original logarithms exist, namely . The combined expression, however, is defined wherever the product is positive, including , where two negatives multiply to a positive. The manipulation enlarged the domain, and the extra root came from the enlargement. Verify: $\ln 5+\ln 2=1.609437912434+0.693147180560=2.302585092994=\ln 10$. The answer is $x=5$. (d) Differentiation. Simplify first, using all three laws:

The product became a sum, the quotient a difference, the square root a coefficient $\frac{1}{2}$. Differentiate term by term with $\frac{d}{dx}\ln u=\frac{u^{\prime }}{u}$:

At $x=4$: $\frac{2}{4}+\frac{1}{10}-\frac{1}{1}=0.5+0.1-1=-0.4$. Differentiating the original expression numerically at $x=4$ gives $-0.400000$. What simplifying saved. Differentiating the original directly needs the quotient rule, the product rule and the chain rule together on a fraction containing a radical. The laws replaced that with three elementary derivatives. (e) Why $e$ is located, and why slow growth is compatible with unboundedness. $e$ is discovered, not chosen. The derivative $\frac{d}{dx}\ln x=\frac{1}{x}$ is positive for all , so $\ln$ is strictly increasing; it is continuous, being differentiable; and it is unbounded above, since $\ln (2^n)=n\ln 2$ with exceeds any bound for large $n$. By the intermediate value theorem it therefore attains the value 1, and by strict monotonicity at exactly one point. That point is defined to be $e$. Bisecting on the condition ${\int }_1^{x}dt/t=1$ gives $e=2.718281828459$, agreeing with the known value to all twelve places. Nothing about $e$ was assumed. Contrast the usual development, which defines $\ln$ as the inverse of $e^x$ and so must already know what $e$ is and what $e^x$ means for irrational $x$, the latter requiring a limit of rational powers. Here $e^x$ is defined afterwards as the inverse of $\ln$, which supplies $e^{\sqrt{2}}$ directly: the number whose logarithm is $\sqrt{2}$. Verified: $\ln (e^x)=x$ and $e^{\ln x}=x$ at $x=0.5,1,2,5$, exact to twelve places. Unbounded yet slow. These are compatible because unboundedness is a claim about the limit and slowness a claim about the rate. Each multiplication of $x$ by 10 adds only $\ln 10=2.302585$ to the output, a fixed increment for an ever-larger absolute increase in $x$: | $x$ | $\ln x$ |
|---|---|
| $10$ | $2.302585$ |
| ${10}^2$ | $4.605170$ |
| ${10}^6$ | $13.815511$ |
| ${10}^{12}$ | $27.631021$ |
| ${10}^{100}$ | $230.258509$ | The increments never stop, so the function exceeds every bound; but reaching requires . The same pattern appears in the harmonic partial sums, which diverge while tracking $\ln n$, for a related reason: $\sum 1/k$ and $\int dt/t$ are the discrete and continuous versions of the same accumulation.