Modeling with Logarithmic Functions
Introduced| English | Meaning |
|---|---|
| logarithmic model/ˌlɒɡəˈrɪθmɪk ˈmɒdl/ | logarithmic model |
| parameters/pəˈræmɪtəz/ | Values that specify a model or function, or named inputs that a subroutine receives. |
| domain/dəˈmeɪn/ | The set of input values for which a function is defined. |
| interpolation/ɪnˌtɜːpəˈleɪʃn/ | interpolation |
| extrapolation/ekˈstræpəleɪʃn/ | Estimating a value outside the range of observed data by extending a pattern or model. |
| regression/rɪˈɡreʃn/ | A statistical method for modelling how a response variable is related to one or more explanatory variables. |
| natural log/ˈnætʃərəl lɒɡ/ | natural log |
| residuals/rɪˈsɪdʒuːəlz/ | An observed value minus the value predicted by a model. |
Recover elapsed time from an amount
- Suppose an amount doubles each hour, starting at 100. Amounts $100,200,400,800$ correspond to elapsed times $0,1,2,3$ hours: multiplying the input amount by 2 adds 1 to the output time.
- A logarithmic model 对数模型 expresses this inverse question. The amount grows exponentially with time, while time is a logarithmic function of the amount. Which variable is the input matters.
Build a model from a zero and a proportion
- The time model has a zero at amount $N=100$, where elapsed time is 0. With each doubling adding one hour, the model is $T(N)=\log_2(N/100)$. It gives $T(800)=3$.
- Its mathematical input condition is $N>0$, but a context starting at time zero uses $N\ge100$. Fractional time predictions also assume the exponential rule describes continuous growth between observations.
When is an increasing, nonconstant logarithmic model a reasonable candidate to check?
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A logarithm turns multiplicative input changes into additive output changes. Check the variables, residuals and context; a concave-down appearance alone is insufficient.
An amount starts at 100 and doubles each hour. Which model returns elapsed hours from amount N?
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The model must return 0 at N=100 and add 1 each time N doubles. The ratio N/100 normalises the start.
Determine a model from two observations
- Fix base 10 and use the unshifted form $y=a\log_{10}x+c$. The points $(1,1)$ and $(10,3)$ give $c=1$ and $a+1=3$, so the model is $y=2\log_{10}x+1$.
- For a fixed base $b>0$, $b\ne1$, two distinct positive inputs determine the coefficients of this unshifted form. Subtract the observation equations, then recover the constant. An unknown input shift adds another parameter and requires additional information.
For $y=a\log_{10}x+c$ through (1,1) and (10,3), what is a?
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The first point gives c=1; the second gives a+1=3, so a=2.
Interpret parameters with their base and units
- The parameters 参数 have precise meanings: in $y=a\log_b x+c$, the value at $x=1$ is $c$, and multiplying a positive input by $b$ changes the output by $a$. For $2\log_{10}x+1$, each tenfold input increase adds 2.
- The constant is not the output at zero: the logarithm is undefined there. With base greater than 1, a negative $a$ gives decreasing values. A zero fitted coefficient suggests a constant model instead of a nonconstant logarithm. Interpret coefficients only with the stated base and input units.
The parameters of a fitted model can be interpreted in terms of the real situation.
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Interpret coefficients with their base and units. In y=a log_b(x)+c, c is the model output at x=1; multiplying the positive input by b changes the output by a.
For $y=2\log_{10}x+1$, what is the model output at x=1?
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log₁₀1=0, so the constant 1 is the output at one input unit, not at zero.
For $y=2\log_{10}x+1$, what does the coefficient 2 mean?
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A base-10 logarithm increases by 1 when its input is multiplied by 10. The coefficient scales that change to 2.
Evaluate $y=-2\log_{10}x+5$ at x=100.
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The logarithm is 2, so y=−4+5=1. A negative coefficient with base greater than 1 gives a decreasing model.
Separate domain from the observed input range
- The domain 定义域 of an unshifted logarithmic expression is $x>0$. If only inputs 1 and 10 were observed, predicting at $x=5$ is interpolation 内插, while predicting at $x=100$ is extrapolation 外推.
- The formula $2\log_{10}x+1$ predicts about 2.398 at 5. An extrapolated value can be mathematically defined without being supported by the observations. State the data range, assumptions and any physical restrictions with the prediction.
A model is fitted only to inputs 1 and 10. A prediction at input 100 is…
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100 lies outside the observed input range [1,10]. The formula can be defined there without the data validating it.
Justify any shift before fitting it
- A shifted example $P(t)=10\ln(t+1)+5$ is mathematically defined for $t>-1$. In a context restricted to $t\ge0$, it starts at $P(0)=5$; the shift makes time zero usable because the log argument is 1.
- Here $P(e-1)=15$. Equal multiplicative changes belong to the adjusted input $t+1$, not to raw time $t$. A shift should come from the context or be estimated with sufficient data; do not insert one merely to hide an invalid input.
What is the mathematical domain of $P(t)=10\ln(t+1)+5$?
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The argument needs t+1>0. A particular context may impose the narrower restriction t≥0, but that is different from the mathematical domain.
Fit scattered observations with regression
- A logarithmic regression 回归 fits $y$ against a logarithm of the positive input $x$. For a specified base and shift, least squares chooses the coefficients to minimise the sum of squared vertical errors. It need not pass through every observation, but it can pass through some or all of them.
- For illustrative observations $(1,1.2)$, $(10,2.6)$ and $(100,5.2)$, the least-squares base-10 fit is $\hat y=2\log_{10}x+1$. Technology may report an equivalent form using the natural log 自然对数 instead. Do not confuse regressing $y$ on $\ln x$ with taking $\ln y$.
Technology can find a logarithmic ____ that best fits a data set.
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For the chosen base and fixed shift, logarithmic regression fits y against log(input). Least squares minimises squared vertical residuals within that family, without proving that the family is appropriate.
Inspect residuals before trusting a prediction
- Residuals 残差 are observed values minus fitted values. At input 10 in the illustrative data, the model predicts 3 while the observation is 2.6, giving residual $-0.4$. The point lies below the fitted curve.
- The three residuals are $0.2,-0.4,0.2$, so their squared sum is 0.24. This is a fit within the chosen family, not proof that the model is appropriate. Three observations are limited evidence: examine patterns, compare plausible families and check further data when available.
Construct a model and inspect its predictions
Fit (1,1) and (10,3), then compare scattered, decreasing and constant observations. Distinguish interpolation from extrapolation.
An observation is 2.6 and the fitted value is 3. What is the residual?
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Residual=observed −fitted=2.6−3=−0.4, so the observation is below the fitted curve.
Use natural logarithms with the same meaning
- The natural log gives the time model as $T(N)=\dfrac{\ln(N/100)}{\ln2}$. At $N=300$, it predicts about 1.585 hours. If the observed amounts span 100 to 800, this is interpolation under the continuous-growth assumption.
- Changing the logarithm's base rescales its coefficient and leaves the fitted curve unchanged. In $y=a\ln x+c$, multiplying the input by $e$ adds $a$ to the output. Retain full coefficients for calculations, then round the final prediction with its units.
For $T(N)=\ln(N/100)/\ln 2$, estimate T(300) to three decimal places.
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ln 3/ln 2≈1.585 hours, assuming continuous doubling between observations.
Choose a family from the input-output pattern
- A growing exponential has proportional outputs over equal input increments. Its logarithmic inverse has equal output increments over proportional inputs. The same doubling process can therefore be written as an exponential amount-against-time model or a logarithmic time-against-amount model.
- Shape alone is insufficient. A decaying exponential also changes rapidly and then flattens, but towards a horizontal asymptote. A bounded growth process can appear concave down over a limited window. Check the variables, ratios, residuals and context before choosing a family.
For a growing exponential and its logarithmic inverse, which input-output contrast is correct?
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Reversing input and output swaps the roles of additive and multiplicative change. An exponential is not always increasing, and shape alone does not choose a model.
Respect a bounded outcome
- In an illustrative score example, $S(h)=20\log_{10}h+40$ matches scores 40 at 1 study hour and 60 at 10 hours. The coefficient means a tenfold increase adds 20 points; 40 is the model output at one hour, not at zero hours.
- A score is capped at 100, but the formula predicts 120 at 10,000 hours. It may be a local approximation, not a global description of a ceiling. Predicting at 100 hours also extrapolates beyond the two observations, even though the numerical prediction remains below 100.
Construct the model from its input-output pattern, a zero and ratio, observations or regression. State the base, units, input restrictions and observed range. A useful fit must be checked against residuals and context, especially before extrapolation.
Select all true statements about fitting and using a nonconstant logarithmic model.
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The unrestricted nonconstant formula is unbounded. A fit may still be useful locally for a bounded outcome, but extrapolation must respect physical limits and evidence.
The score model $S(h)=20\log_{10}h+40$ must be valid for every h>0 because its mathematical formula is defined there.
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A real score is capped at 100, but this formula predicts 120 at h=10000. Mathematical definition does not establish validity outside the observed range or physical limits.
Explain a prediction and its limits
- For $y=2\log_{10}x+1$, evaluate $y$ at $x=100$ and explain the calculation.
- If the model used only $(1,1)$ and $(10,3)$, classify this prediction and explain what the two coefficients mean. Distinguish mathematical validity from evidence that a prediction is reliable.
For $y=2\log_{10}x+1$, find y at x=100. If the model used only (1,1) and (10,3), classify the prediction and interpret both coefficients.
4 plus 1 is 5.
For $y=2\log_{10}x+1$, what is y at x=100?
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2 times 2 plus 1 is 5. Other options — 2 (the logarithm alone); 100 (the input).