Algebra, functions, graphs, and logarithms

Name the quantities and the domain

A symbol can stand for a physical quantity, including its unit. In a function, each allowed input has exactly one output. The set of allowed inputs is its domain. Naming inputs, outputs and fixed parameters makes a mathematical rule easier to use.

Our original hypothetical model follows one chosen point on a cart along a straight track. A fixed track origin defines position x, with right positive. Time t is measured from the start. We stipulate constant velocity v = +0.40 m/s and initial position x0 = −0.50 m throughout 0 ≤ t ≤ 4 s, including both endpoints. The rule is x(t) = x0 + v t. These are supplied exact model values, not measurements.

Here t is the input, x(t) is the output, and x0 and v are fixed parameters. The notation x(t) means position at time t; it is not x multiplied by t. At t = 2 s, substitution gives x(2 s) = −0.50 m + (+0.40 m/s)(2 s) = +0.30 m. The seconds cancel in the product, leaving a length that can be added to x0.

Rearrange with valid operations

An equation stays equivalent when the same valid operation is applied to both sides. To find a time from x = x0 + v t, subtract x0 from both sides: x − x0 = v t. If v ≠ 0, divide both sides by v: t = (x − x0) / v. Division requires a nonzero denominator.

Carry the quantities and units through the rearrangement. The numerator x − x0 is a displacement, measured here in metres. Its division by velocity has units m / (m/s) = s. After finding a candidate time, substitute it into the original equation and check that it belongs to the declared domain.

If v = 0 in a separate constant-position case, the rule becomes x(t) = x0. A requested position equal to x0 occurs at every allowed time; a different requested position occurs at none. Dividing by zero would lose this distinction. A function can give one output for each input without giving a unique input for every requested output.

Connect a table with a labeled graph

The table gives five exact evaluations of our cart rule. On the accompanying graph, the horizontal axis is time t in seconds and the vertical axis is position x in metres. Both axes use linear scales: equal numerical intervals have equal spacing along that axis. A plotted point at t = 1 s and x = −0.10 m gives the cart point's position at that time.

The displayed straight segment follows the stipulated constant-velocity rule throughout 0 ≤ t ≤ 4 s. It is a model curve, not a set of measurements. Five listed values alone would not justify a unique curve between them; the stated rule supplies that information.

A position–time graph has one time axis and one position axis. Its line is not a drawing of a track in two spatial directions. Read its height as position and its horizontal coordinate as time. Position and changes in position are defined relative to the chosen frame and positive direction.

Read changes and intercepts

For two distinct times t1 and t2 with t2 > t1, define Δx = x2 − x1 and Δt = t2 − t1. The Greek letter Δ means change. The graph's change in position divided by its change in time, Δx / Δt, is the average velocity over that interval. For our constant-velocity model this ratio is +0.40 m/s for every such interval inside the domain.

The vertical-axis intercept is x0 = −0.50 m at t = 0. The horizontal-axis crossing occurs when x = 0: 0 − (−0.50 m) = (+0.40 m/s)t, so t = 1.25 s. These are different quantities: an initial position and a crossing time. A negative graph height means left of the origin; it does not by itself mean motion to the left.

Use the printed ticks and units to calculate a gradient. Stretching a drawing horizontally or vertically changes the line's apparent angle without changing the labeled Δx / Δt. Graph height, numerical slope and visual angle are therefore different pieces of information.

Respect the model's stated range

The cart rule supplies positions between its listed times because constant velocity was stipulated throughout the full interval. For example, evaluating it at 2.5 s is permitted. This does not establish an interpolation rule for arbitrary measured points or for every physical process.

The same algebraic expression can be evaluated at 5 s, giving +1.50 m. However, 5 s lies outside our declared model domain. That calculation supplies no prediction about the actual cart after 4 s. Additional physical information would be needed to extend the allowed interval.

Keep the distinction between exact calculation within a supplied model and the accuracy of a real description. A graph or a successful rearrangement does not establish that an actual cart obeys the rule. Actual applications also need suitable observations, assumptions and accuracy checks. Our stipulated decimal values do not establish real measurement precision.

Read a logarithm as an exponent

Powers of ten describe multiplicative steps: 10² = 100, 10¹ = 10, 10⁰ = 1 and 10⁻¹ = 1/10 = 0.1. A base-10 logarithm asks which exponent produces a positive input r. Write log10(r) = y when 10^y = r. Thus log10(100) = 2 and log10(0.1) = −1. Base 10 is the common-logarithm base.

More generally, log_b(r) = y means b^y = r, for a positive base b ≠ 1 and a positive input r. The base must be named because changing it changes the output. For example, the same input 16 gives log_2(16) = 4 because 2⁴ = 16, while log_4(16) = 2 because 4² = 16.

We use real logarithms here. Their input must be greater than zero. A negative logarithm output is allowed: it can describe a positive ratio smaller than one. It does not mean that the logarithm's input was negative. Zero and negative inputs have no real logarithm under these conditions.

Form a physical ratio before taking its logarithm

For physical quantities, a logarithm's input must be dimensionless. Compare quantities of the same kind by dividing a positive quantity by a positive reference expressed in compatible units. Specify both the quantity and its reference; official metrology guidance also makes that requirement explicit for its named logarithmic ratio quantities.

For our separate original comparison, Q is a stipulated positive energy-transfer magnitude and Qref = 0.020 J is its fixed positive reference. Define our comparison index as L = log10(Q / Qref). This is a dimensionless index defined for this lesson. The table shows how factors of ten in Q correspond to steps of one in L. No time evolution, physical response or measured energy dataset is being supplied.

Convert both energies together if you change units. Since 1 mJ = 0.001 J, Q = 2 J and Qref = 0.020 J can also be written as 2000 mJ and 20 mJ. Both give the same dimensionless ratio 100 and index L = 2. The index states a comparison with the reference; it is not an energy of 2 J. Taking log10(2 J) would leave a dimensional input and would not define this comparison.

Worked examples

Try it yourself

1. Use the stated cart rule to find x at t = 1.5 s. Check the units and domain.

   Show the answer and reasoning

   x(1.5 s) = +0.10 m, at an allowed time.

   The product (+0.40 m/s)(1.5 s) = +0.60 m has length units. Add x0 = −0.50 m to obtain +0.10 m. The time 1.5 s belongs to the inclusive interval from 0 to 4 s.

2. In a separate constant-position model, x0 = −0.50 m and v = 0 throughout the same allowed interval 0 ≤ t ≤ 4 s. Which times give x = −0.50 m? Which give x = +0.50 m? Can dividing by v answer these questions?

   Show the answer and reasoning

   Every allowed time gives −0.50 m; none gives +0.50 m. Division by v is invalid because v = 0.

   The rule reduces to x(t) = −0.50 m for every allowed input. One output can correspond to many inputs. Inspecting this rule answers the questions without dividing by zero.

3. For the original cart rule with v = +0.40 m/s, rearrangement for x = +1.50 m gives t = 5 s. Does this establish that the cart reaches that position at 5 s?

   Show the answer and reasoning

   No. The candidate time is outside the declared 0–4 s model domain.

   The arithmetic is (1.50 m − (−0.50 m)) / (0.40 m/s) = 5 s. It evaluates an extension of the algebraic expression. The stipulated physical model supplies no behavior after 4 s, so the calculation does not establish the cart's later position.

4. On the cart graph at t = 1 s, x = −0.10 m. Is the cart moving left? Use the points at 1 s and 2 s to justify the direction.

   Show the answer and reasoning

   It is moving right under the stated constant-velocity model. The negative position only places it left of the origin.

   Position changes from −0.10 m to +0.30 m in 1 s: Δx / Δt = (+0.30 m − (−0.10 m)) / (2 s − 1 s) = +0.40 m/s. This positive velocity indicates motion in the declared positive direction.

5. For Qref = 0.020 J, find the lesson's index for the exact positive magnitude Q = 0.002 J. Would the same real-logarithm definition permit Q = 0 with this reference?

   Show the answer and reasoning

   For 0.002 J, L = −1. For Q = 0, this real logarithm is undefined.

   The positive dimensionless quotient is 0.002 J / 0.020 J = 0.1. Since 10⁻¹ = 0.1, log10(0.1) = −1. With zero Q, the quotient would be zero, outside the real logarithm's positive-input domain. A negative output does not require a negative input.

6. A positive dimensionless ratio is 16. Compare log_2(16) with log_4(16). Are both bases permitted for real logarithms?

   Show the answer and reasoning

   log_2(16) = 4 and log_4(16) = 2. Both bases are positive and different from one, so both are permitted.

   2⁴ = 16 while 4² = 16. The same ratio can therefore have different logarithm outputs under different explicitly chosen bases. The input is positive and dimensionless.

What to carry forward

• State the meaning, unit and allowed domain of a model's quantities and variables.
• Rearrange with valid operations, then check the candidate result in the original equation and domain.
• Read graph labels and ticks before interpreting heights, changes or intercepts.
• A model curve requires its stated rule and conditions; listed points and exact algebra do not establish real behavior outside the model's range.
• A logarithm gives an exponent for an explicitly chosen positive base different from one, with a positive real input.
• For physical logarithmic comparisons, divide compatible quantities to form a dimensionless ratio and name the reference.