Exponentials and logarithms

Review: exponents, roots and percentages

For a positive integer n, aⁿ is a product of n equal factors. For a ≠ 0, we define a⁰ = 1 and a⁻ⁿ = 1/aⁿ to preserve the laws of powers. If a > 0, rational exponents can also be interpreted through roots: a^1/2 = √a and a^2/3 = (∛a)². A positive base allows the power to be extended to all real exponents.

The rules a^x+y = aˣaʸ and (aˣ)ʸ = a^xy express the multiplicative structure. They do not allow aˣ + aʸ = a^x+y. For example, 3² + 3³ = 36, whereas 3⁵ = 243. Always identify whether the operation between powers is addition, multiplication or division.

An increase rate p, written as a decimal, corresponds to the factor 1 + p. Increasing by 8% uses 1,08; decreasing by 8% uses 0,92. If the initial value is Q₀ and the factor per period is q, after n periods Qₙ = Q₀qⁿ. The exponent counts complete periods from the initial instant n = 0, not the position in a list starting at 1.

A variable in the exponent

The basic exponential function is f(x) = aˣ, with a > 0 and a ≠ 1, defined for every real x. If a > 1, it is strictly increasing; if 0 < a < 1, it is strictly decreasing. Its values are always positive, its range is ]0, +∞[ and f(0) = 1. The horizontal axis is an asymptote: the function can approach zero without reaching it.

The expression x³ is a power function, not an exponential function: its variable is in the base. The expression (−2)ˣ does not define a real function for every x, since, for example, x = 1/2 would require the square root of a negative number. Base 1 would give the constant function 1, without a logarithmic inverse that distinguishes exponents.

A model Q(t) = Q₀aᵗ, with Q₀ > 0, multiplies the value by a whenever t increases by one unit. If time is measured in hours, the exponent must be understood as the number of one-hour periods. A more dimensionally explicit form is Q(t) = Q₀a^t/τ, where τ is the period's duration in the same unit as t.

The inverse question: what is the exponent?

For a > 0, a ≠ 1 and y > 0, log_a y is defined as the unique number x such that aˣ = y. Uniqueness follows from the monotonicity of the exponential: the same positive output cannot correspond to two distinct exponents. The logarithm is not division by a; it is the exponent being sought.

For example, log₂ 32 = 5 because 2⁵ = 32; log₄(1/16) = −2 because 4⁻² = 1/16. We have log_a 1 = 0 and log_a a = 1. The inverse identities are log_a(aˣ) = x for every real x and a^{logₐ y} = y for y > 0.

The logarithmic function has domain ]0, +∞[ and range ℝ. Swapping input and output reflects the graphs of the exponential and its logarithm across the line y = x. For a base greater than 1, the logarithm increases; for a base between 0 and 1, it decreases. The argument cannot be zero or negative within the real numbers, because no power of a positive base produces those values.

Why products become sums

If u and v are positive, write u = aʳ and v = aˢ. Then uv = a^r+s, so log_a(uv) = r + s = log_a u + log_a v. The same reasoning with division gives log_a(u/v) = log_a u − log_a v. For u > 0 and real k, log_a(uᵏ) = k log_a u.

These properties come from operations with exponents. There is no general rule turning log(u + v) into log u + log v. One counterexample is enough: in base 10, log(10 + 10) = log 20 ≈ 1,3010, whereas log 10 + log 10 = 2. Applying a property to the wrong type of operation changes the value.

The change-of-base formula allows us to calculate log_a y = ln y/ln a, where ln is the logarithm with base e, approximately 2,71828. You can also use base-10 logarithms in both numerator and denominator. The formula follows from y = aˣ by applying the same logarithm to both sides: log y = x log a. Since a ≠ 1, the denominator is not zero.

Solve equations and preserve the domain

If two powers have the same positive base different from 1, aˣ = aʸ implies x = y. For example, 5^2x−1 = 125 = 5³ gives 2x − 1 = 3, so x = 2. If there is no convenient common base, we apply logarithms to both positive sides and isolate the unknown in the exponent.

For equations containing logarithms, record the argument conditions before combining expressions. In log₂(x − 1) + log₂(x − 3) = 3, we need x > 3. The product rule gives log₂[(x − 1)(x − 3)] = 3, so (x − 1)(x − 3) = 8. The quadratic equation may produce candidates outside the original domain; they must be rejected.

In inequalities, monotonicity determines whether the order is preserved. Applying a logarithm with base greater than 1 preserves the order between positive arguments; a base between 0 and 1 reverses it. The same care is needed when dividing by ln a, which is negative when 0 < a < 1. Checking test values helps verify the direction of the resulting interval.

Doubling time, decay and limits

If Q(t) = Q₀aᵗ and a > 1, the doubling time T satisfies Q(t + T)/Q(t) = aᵀ = 2. Therefore T = ln 2/ln a. It does not depend on the starting instant t within this model, because the factor per period is constant. If 0 < a < 1, the halving time satisfies aᵀ = 1/2 and is T = ln(1/2)/ln a, positive because two negative numbers are divided.

For Q(t) = Q₀e^kt, the product kt must be dimensionless: if t is in days, k has unit day⁻¹. k > 0 gives growth and k < 0 gives decay. This continuous form is useful, but does not demonstrate that a real process grows without limits. Resources, environmental conditions or changes of regime can invalidate the model.

A time answer may require rounding appropriate to the context. If a count is observed only at the end of whole periods and the threshold is exceeded at t ≈ 4,2, the first whole-period record above the threshold will be at period 5. Do not simply round to 4 if the question requires the first period satisfying the condition.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

A logarithm is a fraction between the argument and the base.

It is the exponent the base needs to produce the argument.

log(u + v) = log u + log v.

A sum of logarithms corresponds to the logarithm of a product of positive arguments.

All algebraic candidates work in a logarithmic equation.

Each original argument must remain strictly positive.

Exponential bases between zero and one preserve the order of the exponents.

They are decreasing and reverse that comparison.