Unit notes

An exponential expression changes by a constant factor over equal input intervals. A logarithm answers an exponent question. Multiplying by 1.08 means an 8% increase, while multiplying by 0.92 means an 8% decrease; opposite percentage changes do not exactly undo one another.

For equally spaced inputs, an exponential model has a constant output ratio. In f(x) = ab^x, a is the value at x = 0 and b is the multiplicative factor per one input unit. If b > 1, positive a gives growth; if 0 < b < 1, it gives decay. A factor of 1.08 corresponds to an 8% increase, not a 108% increase.

A constant percentage change produces changing absolute increments because the percentage acts on a changing base. By contrast, a linear model adds the same amount per equal interval. Compare data ratios and differences, and keep the interval size in view.

A monthly factor cannot be used as an annual factor without adjustment. Repeated growth multiplies factors, so an m-period factor is b^m. A doubling or half-life description can be written as a power of 2 or 12 with time divided by the relevant interval.

Continuous exponential models use Ae^kt. Here k is a continuous rate parameter, not always the same number as a stated discrete percentage rate. The equivalent one-unit factor is e^k. Context can limit the time domain even though the expression accepts all real inputs.

log_b(x) is the exponent to which b must be raised to produce x, with b > 0, b ≠ 1, and x > 0 in the real-valued setting. Exponential and logarithmic functions with the same base are inverses. Their graphs reflect across y = x when corresponding domains and ranges are used.

Log rules convert products into sums, quotients into differences, and positive-argument powers into multiples. They do not distribute over addition: log(x + y) is not log x + log y. Check every logarithm’s argument after solving an equation because algebra can produce extraneous candidates.

When bases match, equality of exponential expressions can allow exponents to be equated. Otherwise, take logarithms and use change of base. Keep exact expressions until the final approximation to avoid rounding drift. A calculator’s numerical solution still requires a domain and contextual check.

Combining logarithms can simplify an equation, but each original argument must remain positive. Squaring or exponentiating can change the set of algebraic candidates. An answer should include units and state whether a noninteger time or quantity is meaningful in the context.

A vertical shift changes an exponential model’s horizontal asymptote, while a horizontal shift changes a logarithmic model’s vertical asymptote. The domain of a transformed logarithm comes from requiring its inner expression to be positive. The range and monotonic direction depend on the base and output transformations.

An inverse model changes the question: a population model gives amount from time, while its inverse gives time from amount. Restrict the inverse to outputs actually reachable in the original context. The inverse of a model is not a claim that the process physically runs backward.

Over a limited range, multiple model families can fit similar data. Examine residuals, plausible mechanisms, and parameter meaning. A model that predicts negative counts or unlimited resource use may be inappropriate outside the fitted interval even if its local fit is strong.

Compare relative change and absolute change separately. A rapidly growing percentage on a small base can initially add fewer units than a slow percentage on a large base. Long-run exponential comparisons depend on growth factors, but real constraints can invalidate indefinite extrapolation.

In A(t)=A₀bᵗ, A₀ is the value at t=0 and b is the factor per unit of t. If t changes from years to months, the factor must change: the monthly factor corresponding to annual factor b is b¹¹². Dividing the annual percentage by 12 is only an approximation, not the exact conversion for multiplicative growth.

The form A₀e^kt uses continuous rate k, with b=e^k for a one-unit interval. Doubling time solves 2=e^kt, so t=ln2k for positive k. Half-life solves 12=e^kt for negative k. Keep the time units consistent with k. A model fitted from two observations may describe that interval well but still fail when resources or external conditions change.

For positive arguments, a product inside a logarithm becomes a sum, a quotient becomes a difference, and a power becomes a coefficient. A sum inside a logarithm does not split into a sum of logarithms. Before solving, require every original logarithmic argument to be positive. Later algebra can produce candidates that violate those original restrictions.

To solve an exponential equation with a common base, compare exponents. If a common base is inconvenient, take logarithms and use the power rule. For logarithmic equations, combine logs only when the arguments satisfy their domain conditions, convert to exponential form when helpful, and substitute candidate solutions into the original equation. Squaring or clearing denominators can also introduce extraneous candidates.

An inverse reverses input and output roles. A one-to-one function passes the horizontal-line test, and its graph reflects across y=x to produce the inverse graph. The original domain becomes the inverse range, and the original range becomes the inverse domain. A restricted physical domain can therefore restrict an inverse even when an algebraic expression accepts more inputs.

If a model gives population from elapsed time, its inverse gives elapsed time from population. The inverse’s units reverse accordingly. In exponential decay, a horizontal asymptote at zero becomes a vertical asymptote for the logarithmic inverse. The inverse is not the reciprocal 1f(x); use composition to verify that a proposed inverse reverses the original function on the appropriate domains.

A linear model adds a constant amount over equal input intervals. An exponential model multiplies by a constant factor. In A·bˣ, A is the value at zero and b is the one-unit growth factor. A 6% increase uses b=1.06; a 6% decrease uses b=0.94. The percentage is applied to the changing amount, not repeatedly to the initial amount.

For intervals of length h, use the factor bʰ. If a quantity doubles every three hours, one possible model is A·2^t3, not A·2^3t. Dimensional reasoning helps: the exponent must be dimensionless. For data, compare ratios over equal input spacing before choosing an exponential model.

Sequences describe discrete inputs. An arithmetic sequence has constant differences; a geometric sequence has constant ratios. Decide whether indexing starts at zero or one before writing a formula. A real-world process measured each year may use a discrete model even if a smooth graph is drawn between observations.

The statement log_b(x)=y means bʸ=x, with positive b other than one and positive x. A logarithm asks for an exponent. This meaning explains why log_b(1)=0 and log_b(b)=1. Logarithmic outputs may be negative even though the argument must be positive.

Product, quotient, and power laws follow exponential structure, but there is no law that log(x+y)=log x+log y. Before expanding or combining, state the domain. Algebraic manipulation can hide restrictions or create extraneous candidates, so check final solutions in the original equation.

To solve an exponential equation with an isolated exponential expression, take logarithms and divide by the relevant logarithm. If both sides can be written with the same base, equating exponents may be simpler. A calculator approximation should be delayed until the exact setup is clear.

A vertical shift in A·bˣ+k changes the horizontal asymptote to y=k. It also means the entire output no longer changes by the same multiplicative factor; the distance from the asymptote does. This distinction matters when fitting data that approach a nonzero baseline.

A semilog display can make exponential behavior appear linear because logarithms turn products into sums. Always read the axis scale: equal plotted steps may represent equal ratios rather than equal differences. A good visual fit still requires residual analysis and a plausible context.

Exponential growth cannot continue indefinitely in many physical systems. Use a model within a supported interval and identify what resource or constraint could eventually break it. An inverse prediction far beyond observed data is still extrapolation, even if the algebra is exact.

The forms A·bᵗ and A·e^kt are equivalent when k=ln b. The continuous rate parameter k is not exactly the same number as the one-period percentage change b−1. Keep the interpretation tied to the form used.

For compound growth, identify the period associated with the stated rate and how often the factor is applied. For a half-life model, a factor of 12 is applied once per half-life. Solve for time by dividing the logarithm of the target ratio by the logarithm of the per-time factor.

When comparing models, evaluate a meaningful quantity: time to a threshold, output at a specified input, or rate of relative change over a fixed interval. A larger initial value does not imply a model remains larger forever if growth factors differ.