Read semilog scales and test exponential evidence

Try

A graph can show a tenfold increase with the same vertical distance as another tenfold increase. That is useful when amounts span several orders of magnitude, but only if you read the scale correctly. On a logarithmic vertical axis, moving from 1 to 10 takes the same distance as moving from 10 to 100. On a linear vertical axis, those changes are 9 and 90 units and occupy different distances.

Begin with positive dimensionless response ratios: divide a response by a fixed positive reference amount expressed in the same units. The common logarithm log₁₀(ratio) tells which power of 10 produces the ratio. Thus 1,10,100 have coordinates 0,1,2. A logarithmic axis can label these positions with the original values 1,10,100 rather than their logarithms. The graph’s horizontal input axis remains linear: equal horizontal distances represent equal differences in x, not equal input ratios. This is a semilog graph, not a graph with both axes logarithmic.

Predict

Which value lies halfway between 1 and 10 on the log axis: 5.5 or √10? Write the exponent at that midpoint before calculating the original value. The exponent is halfway between 0 and 1, so the value is 10^(1/2)=√10≈3.16228. In contrast, the linear midpoint is the arithmetic mean 5.5. A logarithmic midpoint is the geometric mean of its positive endpoint values.

Original ratio	Linear coordinate	Log₁₀ coordinate
1	1	0
2	2	Approximately 0.30103
5	5	Approximately 0.69897
10	10	1
20	20	Approximately 1.30103
50	50	Approximately 1.69897
100	100	2

Draw a vertical log axis with equal distances between 1,10,100. Place 2 and 5 between 1 and 10 using the table, then place 20 and 50 one decade higher. Keep a separate horizontal axis with uniformly spaced inputs 0,1,2,3,4. Now predict whether a baseline-shifted exponential must form a straight line on the raw log-y graph.

Experiment

The workbench starts with engineered synthetic pairs (0,7),(1,9),(2,13),(3,21),(4,37). Input x is dimensionless; response y is measured in sensor units. Use one sensor unit as the reference, so each numerical excess represents a dimensionless ratio to that reference. First keep the assumed baseline at 0. Reveal the fit, inspect the linear-response and semilog-excess panels, and record the transformed table, line coefficients and residual evidence.

Compare ordinary and semilog evidence

Predict which assumed baseline will make the adjusted observations lie on a straight semilog trace. The default data are synthetic. The baseline k is supplied by you, never estimated here. State the units and reference quantity in your written model record.

Observed pairs, one x,y pair per line0, 7 1, 9 2, 13 3, 21 4, 37 Assumed baseline k (−100 to 100)

Use 3–40 rows with x and y between −100 and 100. Every adjusted response y−k must be at least 0.000001. Nonpositive adjusted values cannot appear on a real logarithmic axis; invalid rows are never silently removed. Negative raw responses can be valid when the supplied baseline is lower.

Reveal semilog fit and comparisonReset semilog investigation

Record your assumed baseline, raw data, both axis meanings, coefficient equation and a fresh prediction check. A straight pattern at observed points is evidence compatible with exponential behavior of y−k, not proof of a global law. Floating-point calculations are rounded for display. Editing observations or baseline clears the old fit and query; editing the query clears only its result. This temporary workbench saves no responses and awards no mastery.

Then set the assumed baseline to 5 and repeat. The adjusted responses y−5 are 2,4,8,16,32. Their log₁₀ values increase by log₁₀2 at every unit step. The adjusted semilog line is therefore log₁₀(y−5)=log₁₀2+x log₁₀2 when y is expressed numerically in sensor units. Converting back gives y=5+2·2^x. Both the original-response and adjusted-log fitting SSE are zero in exact arithmetic for this engineered dataset.

For baseline 0, the log-line fit instead has intercept about 0.7979041534 and slope 0.1814184153. Its equation is y≈6.279197648·10^(0.1814184153x), with original-response SSE approximately 17.02055308 sensor-units squared and log SSE approximately 0.007435192. These coefficients minimize the selected log-space criterion; they are not claimed to minimize original-response SSE. Retain enough digits to avoid inventing discrepancies from rounded coefficients.

Use the separate prediction control at x=2.5 without adding a new row. The unshifted model predicts approximately 17.84227348; the baseline-5 model predicts 5+2·2^2.5≈16.31370850. Only after recording both, compare with a supplied fresh measurement of 16.4±0.2. The interval [16.2,16.6] contains the shifted prediction and excludes the unshifted one. This supports the shifted candidate on this check; it does not prove that the baseline is physically 5.

Understand

A straight line in transformed coordinates has the form log₁₀(y/yref)=α+βx, where yref is a fixed positive reference in the same units as y. Raising 10 to both sides recovers y=yref·10^α·10^(βx). The intercept determines the initial amount A=yref·10^α; the slope determines the factor per input unit, 10^β. Across an input change h, the response factor is 10^(βh). If x measures hours, β has units per hour so the exponent βx is dimensionless. Do not interpret a two-hour graph interval as a one-hour factor.

With an assumed baseline k, apply the same reasoning to (y−k)/yref. Then y=k+A·10^(βx). Every adjusted response must be strictly positive. A zero or negative adjusted value prevents this real logarithmic fit; the workbench refuses the whole dataset rather than silently dropping inconvenient observations. Its numerical domain also requires adjusted responses at least 10^−6 and finite bounded inputs. That computational floor is stricter than mathematical positivity and is a tool limit, not a theorem about logarithms. A different baseline might make all values positive, but that alone does not justify the baseline or establish exponential behavior.

The workbench fits the baseline you supply; it does not estimate an unknown baseline. Selecting 5 because it straightens these training points is a model-selection action. A straight-looking finite plot is not proof of a global exponential law, and changing a baseline after seeing validation results uses those results for development. Preserve the original prediction and obtain new evidence for a subsequent independent check.

Changing logarithm base changes the line’s coordinates, not the response family. Since ln u=ln(10)·log₁₀u, both intercept and slope are multiplied by ln(10). Exponentiating consistently produces the same original response. Also distinguish this workflow from fitting a logarithmic response y=a+b ln x: here the response or its excess is transformed, while x stays linear. In the previous logarithmic-response lesson, the input was transformed and the response stayed on its original scale.

Apply

Compare candidate errors on the same original observations. Original-response residuals are observed y minus predicted y; their squares have the same units for both baseline choices, so their original SSEs can be compared on these same rows. A log residual compares the logarithms of adjusted responses. Changing the baseline changes that adjustment and the implied error criterion, so the two log SSEs do not supply a common original-response error measure. Even a smaller original training SSE does not by itself establish better prediction elsewhere.

Try a new exact case: enter (0,4),(2,10),(4,28) and set the assumed baseline to 1. Recover the transformed line and predict at x=3 before opening the feedback. Draw original-response and adjusted semilog sketches with labeled axes and exact landmarks.

Check the new model and scales

The excesses are 3,9,27. Across a two-unit step the factor is 3, so the per-unit factor is √3. The line has intercept log₁₀3 and slope (log₁₀3)/2. The response is y=1+3·3^(x/2), predicting 1+9√3≈16.58845727 at x=3. The adjusted semilog points are equally spaced vertically and horizontally; the original-response points are not collinear. Baseline 1 is still supplied rather than estimated. Exact fitting does not certify an unmeasured physical process.

If an observation were y=1 at one input in this changed case, the adjusted value would be zero. Report that the chosen transformation cannot include the full dataset. Do not delete the row, replace zero with a small arbitrary positive number, or label a different dataset as the same analysis. Investigate whether the baseline, measurement convention or model family needs revision.

Continue

Keep a reproducible record: raw pairs, reference units, assumed baseline, transformed table and tick convention, line coefficients, original-response equation, original-scale residuals, and a prediction made before inspecting new evidence. Mark which inputs are inside the calibration range and which would be extrapolation. The tool’s numerical bounds do not themselves establish a physically valid range.

Return to shifted exponential identification when the baseline is unknown rather than stipulated. Compare with logarithmic response models to explain why transforming x and transforming y answer different modeling questions. A successful calculation demonstrates execution of this workflow; independent physical validation needs evidence beyond a fitted graph.