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Use decimal logarithms and write $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{{\mathrm{Mg}}^{2+}}/a_{{\mathrm{H}}^{+}}^2)$. The question's concentration notation is interpreted as dimensionless thermodynamic activities, or concentrations divided by a fixed standard concentration with activity coefficients set to one.

For the chlorite–gibbsite reaction, unit solid activities giveHence $y=10.53-0.60x$. It is a descending straight mineral activity diagram boundary, passing through $(0,10.53)$ and $(-5,13.53)$. Its continuation alone does not establish which portions are stable against kaolinite.

Subtract the chlorite–gibbsite reaction from the chlorite–kaolinite reaction. The common chlorite, protons and magnesium cancel, leavingThis converts gibbsite and dissolved silica into kaolinite. Gibbs energies of reaction add with reaction stoichiometry, so decimal logarithms of equilibrium constants also subtract:With unit water and solid thermodynamic activities, $K=a_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}^{-2}$; thus $x=-8.49/2=-4.245$, a vertical boundary. Reversing the reaction would give ${\log }_{10}K=-8.49$ and the same boundary. Its intersection with either sloping line is $(x,y)=(-4.245,13.077)$.

Use base-ten logarithms and dimensionless thermodynamic activities at the stated temperature and pressure. Define $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{\mathrm{N}{\mathrm{a}}^{+}}/a_{{\mathrm{H}}^{+}})$. The concentration notation is an ideal dilute-solution approximation with a consistent concentration reference.

With pure solids and unit water activity, the two reaction quotients are $10^{2y+2x}$ and $10^{2y+6x}$. Equating them to the corresponding equilibrium constants givesThe slopes on the mineral activity diagram are $-1$ and $-3$. The lines meet at $(x_{\ast },y_{\ast })=(-5.515,8.955)$, within the negative silica-logarithm range.

Take base-ten logarithms and use dimensionless thermodynamic activities. The supplied concentration notation is interpreted in the ideal dilute-solution approximation, with a consistent reference concentration. PutWith pure solids and unit water activity, the reaction quotients are $Q_I=10^{2y+2x}$ and $Q_{II}=10^{2y+6x}$. At chemical equilibrium, $Q=K$, so the two lines on the mineral activity diagram areThey have slopes $-1$ and $-3$. Restricting the plotted silica activity to $x<0$ is compatible with their intersection,Their full straight extensions are equilibrium relations for the respective phase pairs; not every extension is a stable boundary when the third mineral is available.

The fugacity coefficient gives $f_{{\mathrm{H}}_2\mathrm{O}}=\varphi P=0.82\times 4500=3690\mathrm{bar}$. Keep the solid pressure correction at the actual $4500\mathrm{bar}$; only the water activity uses the fugacity. ThusThe decrease from part (c) follows because $\varphi <1$ lowers the chemical potential of the water product. All logarithms in these thermodynamic formulae are natural logarithms.

The pH of a chemical solution is the negative base-ten logarithm of its hydrogen ion thermodynamic activity:The thermodynamic activity is a dimensionless quantity, so the logarithm has a dimensionless argument. The definition uses thermodynamic activity, rather than total chemical acid concentration. Single-ion thermodynamic activities require a convention; practical pH measurements are tied to calibrated standards. See IUPAC Gold Book, pH.

In a dilute ideal chemical solution in water, a useful approximation isSquare brackets denote concentration in moles per litre. In water, $[{\mathrm{H}}^{+}]$ is shorthand for the hydronium concentration, often written $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$; it does not describe isolated bare protons. Departures from ideal behavior need activity coefficients. The concentration formula approximates the thermodynamic activity definition.

For an ideal strong acid supplying one proton per initial unit, stipulate a total concentration of $0.001\mathrm{mol}{\mathrm{L}}^{-1}$ and negligible self-ionization of water. Then $[{\mathrm{H}}^{+}]\approx 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}$, soReducing that concentration tenfold by dilution gives $10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ and pH $4$, under the same assumptions. More generally, dividing the hydrogen ion thermodynamic activity by ten raises pH by one. This is why the scale compresses an enormous range: equal steps of pH represent multiplicative changes, rather than equal differences in concentration. These are model calculations, rather than measured samples.

Neutrality depends on temperature. In the dilute ideal water model, define the dimensionless ion product $K_w=([{\mathrm{H}}^{+}]/c^{\circ })([{\mathrm{OH}}^{-}]/c^{\circ })$ and $\mathrm{p}{K}_w=-{\log }_{10}{K}_w$. Equal hydronium and hydroxide concentrations implyAt $25{}^{\circ }\mathrm{C}$, the textbook approximation $K_w\approx 10^{-14}$ gives neutral pH $7$. Since $K_w$ varies with temperature, neutrality need not mean pH $7$. Nor is pH restricted to $0$–$14$: the activity definition permits negative pH when $a_{{\mathrm{H}}^{+}}>1$, and pH above $14$ when $a_{{\mathrm{H}}^{+}}<10^{-14}$. High concentrations can invalidate the dilute ideal approximation. See Flowers et al., Chemistry 2e, pH and pOH.

S. P. L. Sørensen introduced pH at the Carlsberg Laboratory in 1909, as recorded in the Carlsberg Foundation historical timeline.

Phase folding of a light curve maps timestamps onto a trial cycle of duration $P>0$. Given a reference time $t_0$, define the phase in cycles bywhere $\mathrm{frac}(x)=x-\lfloor x\rfloor$ is the fractional part, using the floor function also for negative $x$. Plot brightness against ${\varphi }_i$ rather than $t_i$; measurements separated by an integer multiple of $P$ occupy the same phase. For a periodic function $F(t+P)=F(t)$, this reveals the shared shape of many cycles. Folding changes coordinates; it neither averages samples nor chooses the correct $P$ by itself. Welsh et al.'s folded Kepler plot separates the exoplanet transit and orbital brightness modulation. ^[ref].

The second figure uses a completely synthetic trapezoidal dip: $P=3$ days, $t_0=1$ day, and the same $1\%$ depth and four-hour duration as the first figure. Different colours identify four successive cycles. Correct folding superimposes the dips; a wrong trial orbital period spreads them out. If $P^{\prime }=P+ϵ$, the centre of the $n$th repeated event shifts by approximately $-nϵ/P$ cycles when folded on $P^{\prime }$, for $|ϵ|\ll P$. With $P^{\prime }=3.03$ days, the fourth plotted event is displaced by $-3(0.03)/3.03\simeq -0.0297$ cycles. Transit-timing variations can also smear a constant-period fold, so retain the original timestamps and inspect individual events.

One may combine nearby phases with a weighted mean, but the uncertainty reduction requires assumptions. For $N$ independent measurements with common standard deviation $\sigma$ and effectively the same expected brightness within the bin, their arithmetic mean has standard deviation $\sigma /\sqrt{N}$. Thus sixteen independent $0.0006$-precision samples give $0.00015$ precision on that phase-bin mean. Shared calibration errors, changing starspots, and bins wide enough to mix ingress with out-of-transit brightness violate these assumptions. A folded plot can conceal changes between cycles. Gaussian process models can represent correlated variability and interpolate irregular samples, but their predictions and uncertainties depend on their assumed covariance function.

The bottom panels show the same model as relative brightness $f=F/F_0$ and as magnitude difference in one fixed photometric passband with a fixed magnitude zero point:A dip to $f=0.99$ gives $\Delta m\simeq +0.010912$ magnitudes: fainter means a larger apparent magnitude. The figure places larger magnitude differences lower on the page so the two dip shapes are easy to compare; reversing that plotting convention does not change the formula. For $f=1-\delta$ and small $\delta$, expansion of the logarithm gives $\Delta m\simeq (2.5/\ln 10)\delta \simeq 1.085736\delta$. The exact transform is nonlinear: transforming an exposure's mean brightness is generally different from averaging its instantaneous apparent magnitudes. The ordinary logarithm cannot be applied to a nonpositive background-subtracted measurement; keep such measurements and their uncertainties in brightness units for an appropriate statistical model.

Choose the following original phenomenological mathematical model solely to calculate a paired-pulse readout. It is not experimental data, a fit to the teaching experiment, or a simulation of the Hodgkin-Huxley model. The conditioning action potential defines time zero. Test pulses have fixed width $0.2\mathrm{ms}$; $I$ denotes their inward electric current amplitude. Impose an absolute neuronal refractory period $t_{\mathrm{abs}}=2\mathrm{ms}$. No finite test amplitude generates an event for $0<\Delta <t_{\mathrm{abs}}$. After that hold, stipulate a required amplitudewith $I_{\infty }=0.2\mathrm{nA}$, dimensionless $A=2$ and recovery time constant ${\tau }_r=3\mathrm{ms}$. Emit a test event exactly when $I\ge I_{\mathrm{req}}(\Delta )$, including equality. For this ideal event readout, assign events to pulse onset and assume the conditioning event is successfully evoked. There is no calculated membrane potential, spike waveform, response latency or stochastic success probability.

The exponential decay represents a combined recovery deficit; it is neither the Hodgkin-Huxley model's $h$ nor a measured potassium electrical conductance. One such curve cannot separately identify sodium recovery and persistent potassium electrical conductance. A constant pulse-width amplitude criterion also does not establish an equivalent rule for other pulse widths or voltage trajectories.

At the hold boundary, $I_{\mathrm{req}}(2\mathrm{ms})=0.2(1+2)=0.6\mathrm{nA}$. At $\Delta =4\mathrm{ms}$,at $\Delta =6\mathrm{ms}$,Thus a $0.35\mathrm{nA}$ test fails at $1\mathrm{ms}$ because of the imposed hold, fails at $4\mathrm{ms}$ because recovery is insufficient for that amplitude, and succeeds at $6\mathrm{ms}$. At $4\mathrm{ms}$, a $0.5\mathrm{nA}$ pulse would instead succeed: failure of the weaker pulse was relative refractoriness.

For $I>I_{\infty }$, solve $I\ge I_{\mathrm{req}}(\Delta )$ using the logarithm. The earliest allowed delay isFor $I=0.35\mathrm{nA}$ this is $2+3\ln (8/3)\approx 4.942488\mathrm{ms}$; for $I=0.5\mathrm{nA}$ it is $2+3\ln (4/3)\approx 2.863046\mathrm{ms}$. If $I\ge 0.6\mathrm{nA}$, the imposed $2\mathrm{ms}$ hold is the only delay restriction. If $I\le I_{\infty }$, no finite delay passes the rule: the required amplitude stays strictly above its asymptotic baseline. A finite experimental “fully recovered” time therefore also needs a tolerance and a readout definition.

With Total variation distance defined by $d_{\mathrm{TV}}(P,Q)={\sup }_A|P(A)-Q(A)|$ and relative entropy computed using natural logarithms, Pinsker's inequality statesIf $P$ is not absolutely continuous with respect to $Q$, the relative entropy is infinite and the bound is vacuous. A small relative entropy therefore forces small discrepancies in probabilities of all events. For example, $P=(0.6,0.4)$ and $Q=(0.5,0.5)$ have Total variation distance $0.1$ and relative entropy $0.6\log 1.2+0.4\log 0.8\simeq 0.02014$, giving an upper bound about $0.1003$.

A short proof reduces to a binary event. Grouping the probability mass inside and outside $A$, the log-sum inequality gives $D(P\Vert Q)\ge d(p\Vert q)$, with $p=P(A)$ and $q=Q(A)$. For $0<q<1$, the binary relative entropy as a function of $p$ has minimum zero at $p=q$ and second derivative $1/[p(1-p)]\ge 4$. Integrating this derivative bound twice gives $d(p\Vert q)\ge 2(p-q{)}^2$. Endpoint cases follow by limits. Taking the supremum over $A$ proves the inequality. Using base-two logarithms changes the constant by a factor of $\log 2$.

The pKa is the negative base-ten logarithm of the dimensionless acid dissociation constant. A lower pKa means a larger acid dissociation constant and a stronger tendency to donate a proton under the same standard state, solvent, and temperature conventions. The Henderson-Hasselbalch equation places equal weak acid and conjugate base thermodynamic activities at pH equal to pKa. See Flowers et al., Chemistry 2e, the Henderson-Hasselbalch equation.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

Consider independent event times with hazard function $h_i(t)=\lambda (t)e^{{\beta }^T{x}_i}$ and baseline cumulative hazard function $\Lambda (t)={\int }_0^t\lambda (s)ds$. For optimization over $\beta$, first hold $\lambda$ and $\Lambda$ fixed and independent of $\beta$. Under ignorable independent censoring, an observation $(y_i,{\delta }_i)$ has log-likelihood contributionSet ${\mu }_i=\Lambda (y_i)e^{{\beta }^T{x}_i}$. A Poisson regression with response ${\delta }_i$ and log link has log-likelihood contribution ${\delta }_i\log {\mu }_i-{\mu }_i-\log ({\delta }_i!)$, which differs from ${\ell }_i$ only by terms independent of $\beta$. Thus the same $\beta$ maximizes both criteria. The linear predictor is ${\beta }^T{x}_i+\log \Lambda (y_i)$, with $\log \Lambda (y_i)$ a known offset whenever $0<\Lambda (y_i)<\infty$.

For uncensored observations every response is one. This is an equivalence of objective functions: the event indicators need not actually be Poisson counts. A related split-interval construction handles piecewise-constant hazard functions.^[ref]

If the baseline is unspecified, its integrated values are not known offsets; fitting an ordinary generalized linear model as if they were known does not estimate an arbitrary Cox proportional-hazards model. One must estimate the baseline jointly, use a suitable baseline representation, or eliminate it through Cox partial likelihood. Zero exposure also needs separate handling rather than taking the logarithm of zero.

A positive real number is a real number $x$ with $x>0$. Zero is excluded; allowing zero gives a nonnegative real number. Both $1/2$ and $\sqrt{2}$ are positive.

The sum, product, and reciprocal of positive real numbers are positive. Consequently the positive real numbers form a group under multiplication, and the natural logarithm converts that multiplication into addition: $\log (xy)=\log x+\log y$. Positivity is also the domain restriction that makes the real logarithm well-defined.

A power law is a relation $f(x)=C{x}^p$, with fixed constants $C$ and $p$. For positive $x$, any real number $p$ is allowed. Its useful feature is a simple scaling rule:The same proportional change of input produces the same proportional change of output, independently of the starting value. The exponent controls that response; the coefficient fixes the magnitude.

For example, the area of a circle is $A(r)=\pi r^2$. Doubling the radius multiplies its area by four, and tripling it multiplies the area by nine. The volume of a sphere is $V(r)=4\pi r^3/3$: the same changes multiply its volume by eight and twenty-seven. The distinction explains why similarly shaped objects do not preserve their ratio of surface area to volume when their size changes.

When the output is positive, a logarithm turns the power law into a straight-line relation. Using positive reference values $x_0$ and $f_0=f(x_0)$,These ratios are dimensionless quantities, so the formula also applies to physical measurements with units. Plotting both ratios on a logarithmic scale gives a line of slope $p$. Changing a unit changes the numerical coefficient, while the exponent stays the same. Positive $p$ describes increasing values; negative $p$ describes decreasing values; $p=0$ gives a constant. The zero coefficient is a degenerate case with no useful logarithmic slope.

Exact scaling is stronger than regular variation. The function $x^2\log x$ has a doubling ratio tending to four, but its logarithmic correction keeps growing. The diagram contrasts it with the exact power law $x^2$.

A power law for a probability density and a power law for a tail probability have different exponents. For a Pareto distribution with $\alpha >0$ and lower endpoint one,The tail probability has exponent $-\alpha$; differentiating it and changing sign gives the probability density exponent $-\alpha -1$. Confusing them changes the interpretation of a fitted exponent. The normalizing coefficient and lower endpoint also matter: a decreasing power law is not automatically a valid probability density over the entire positive real line.

The appearance of a line over a finite logarithmic range is evidence to investigate, rather than a proof of an exact power law or its mechanism. In their study of power-law distributions, Clauset, Shalizi and Newman combine maximum likelihood estimation, goodness-of-fit tests and comparisons with alternative models; their abstract explains why ordinary least-squares fitting can give misleading results for tail data. See Aaron Clauset, Cosma Rohilla Shalizi and M. E. J. Newman, Power-law distributions in empirical data, SIAM Review 51 (2009), 661–703. The scaling rule is mathematically exact when it defines a model; whether a dataset supports that model is a question of statistical inference.

For $x\ge 2$, the harmonic number $H_{\lfloor x\rfloor }$ exceeds $\log x$. The finite Euler product over the primes at most $x$ contains every summand of this harmonic number. Taking logarithms and applying the geometric-series bound for the logarithmic remainder bounds the total remainder by $\frac{1}{2}{\sum }_{n=2}^{\infty }[n(n-1){]}^{-1}=1/2$, proving the displayed bound.

Suppose a consistent linear multistep method has a simple principal polynomial root and its analytic logarithm satisfies $\log w(z)=z+C{z}^{2r}+O(z^{2r+1})$ near zero, with real coefficients. If $C(-1{)}^r>0$, then at $z=i\epsilon$ the real part of $\log w$ is positive to leading order. The same remains true after giving $z$ a small negative real part of smaller magnitude than that leading term. Thus a growing amplification factor occurs inside the left half-plane, ruling out A-stability. For an order-three method with $C>0$, choose $z=-C{\epsilon }^4/2+i\epsilon$.

For a fixed finite number of nonnegative terms and $a_L\to \infty$, use ${\max }_j{p}_{j,L}\le {\sum }_j{p}_{j,L}\le r{\max }_j{p}_{j,L}$. The logarithm of $r$, divided by the large-deviation speed, tends to zero, and a finite maximum commutes with the limit superior. If all individual logarithmic limits exist, the limit of the sum is their maximum.

Under the finite-alphabet, positive-reference, closed-convex set assumptions above, every feasible law $r\in \mathcal{C}$ satisfiesThus a feasible law's excess relative entropy above the optimum is at least its relative entropy from the optimizer. The information projection supplies a quantitative comparison with every other feasible law, not merely the location of a minimum.^[ref]

For affine expected value constraints, the result becomes a Pythagorean identity for relative entropy. In relative entropy minimization under a mean constraint, the logarithm of $q_i^{\ast }/p_i$ is an affine function of the outcome $a_i$. Any feasible $r$ has the same normalization and mean as $q^{\ast }$, making the residual in the three-point identity for relative entropy zero.^[ref]

In bits, the quantum Pinsker inequality isThe trace-distance-preserving binary measurement converts the trace distance into classical total variation distance without loss. Apply the data-processing inequality for quantum relative entropy to that measurement, then classical Pinsker's inequality. The factor $\ln 2$ is omitted when relative entropy uses natural logarithms.

Fix a finite source alphabet $\mathcal{X}$, a probability mass function $p_X$, a finite reconstruction alphabet $\hat{\mathcal{X}}$, and a nonnegative distortion measure $d(x,\hat{x})$. The information rate-distortion function isThe source probability distribution stays fixed; the conditional distribution of the reconstruction is what we optimize. Each feasible test channel must keep expected distortion at most $D$. The objective is mutual information, measured in bits when the logarithms have base two. If no test channel satisfies the constraint, the infimum is $+\infty$.^[ref]

Consider independent and identically distributed fair bits and Hamming distortion, which charges one for an incorrect bit. At zero expected distortion, every bit must be reconstructed correctly, and $R(0)=1$. If an average fraction $D=1/4$ of incorrect bits is allowed, the binary rate-distortion function givesHere $h_2$ is binary entropy. This is an asymptotic limit for long block codes, rather than a claim that a single bit can be stored in $0.18872$ physical bits. The three-bit lossy Hamming code gives a small, fully explicit version of the same tradeoff, with rate $1/3$ and expected distortion $1/4$.

Rate-distortion theory separates the question of what is possible from the design of a practical encoder and decoder. It also forces the fidelity criterion to be explicit. Hamming distortion treats all symbol mismatches alike; mean squared error instead penalizes larger numerical errors more strongly. Those objectives can favor different reconstructions of the same data. An average error criterion does not by itself control the error on every individual input.

Claude Shannon's Coding Theorems for a Discrete Source With a Fidelity Criterion by Claude Shannon was published in 1959. Its formulation fixes a cost for replacing one source letter by another and studies the communication rate compatible with the allowed average cost. The important shift is to quantify which distinctions may be discarded, rather than requiring exact recovery of every letter.^[ref]