Logarithmic graphs for power and exponential fits
| English | 中文 | Pinyin |
|---|---|---|
| intercept/ˌɪntəˈsept/ | 截距 | jié jù |
A curve can become a straight line after taking logs. Which horizontal axis distinguishes a power model from an exponential one?
- A curve can become a straight line after taking logs. Which horizontal axis distinguishes a power model from an exponential one?
- This lesson studies intercept 截距: The transformed vertical value when the chosen horizontal variable is zero.
Choose the mathematical structure
- For positive observations and parameters, y=ax^p with x>0 gives ln y=ln a+p ln x. Plot ln y against ln x: gradient p, intercept ln a. For y=kb^x with k,b>0, ln y=ln k+x ln b. Plot ln y against x: gradient ln b, intercept ln k. Use the same log base for both logarithmic axes and recover parameters with that base’s inverse.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines intercept?
The transformed vertical value when the chosen horizontal variable is zero.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
Power data (x,y)=(1,3),(2,12),(4,48) give gradient [ln48−ln3]/[ln4−ln1]=2, intercept ln3, hence y=3x². Exponential data (0,5),(1,10),(2,20),(3,40) give a semilog gradient ln2 and intercept ln5, hence y=5×2^x. With base-10 logs the same exponential gradient is log₁₀2≈0.30103, so b=10^0.30103≈2 rather than e^0.30103. A new power-data observation (3,30) exceeds the fitted prediction 27 by residual 3; a perfectly straight calibration set does not prove the model beyond its observed range.
Logarithmic graphs for power and exponential fits
For positive observations and parameters, y=ax^p with x>0 gives ln y=ln a+p ln x
State the valid input, plotted variable or time unit before using the logarithmic/exponential relationship.
Find the power exponent for the given power data.
The log-log gradient is ln16/ln4=2.
Test a tempting shortcut
- For the power fit the gradient is the exponent, not its logarithm. For the exponential fit the gradient is the logarithm of the base, not the base itself. The power graph’s intercept corresponds to x=1 because ln x=0 there; the exponential intercept corresponds to x=0. Nonpositive observations cannot be logged directly. An exponential log-y plot may include x=0 or negative x; the log-log power plot cannot.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Plotting ln y against x linearises every power model y=ax^p. This claim is false. Explain which definition or assumption it violates.
Find its multiplicative parameter a.
The intercept is ln3, so a=e^(ln3)=3.
Plotting ln y against x linearises every power model y=ax^p.
For the power fit the gradient is the exponent, not its logarithm. For the exponential fit the gradient is the logarithm of the base, not the base itself. The power graph’s intercept corresponds to x=1 because ln x=0 there; the exponential intercept corresponds to x=0. Nonpositive observations cannot be logged directly. An exponential log-y plot may include x=0 or negative x; the log-log power plot cannot.
Interpret a new situation
- Label transformed axes and estimate gradient from well-separated points on the fitted line, not from a single data point. Recover each intercept parameter before returning to original units. Substitute predictions into the original model and examine residuals there. State the fitted range, measurement uncertainty and whether extrapolation is justified; systematic departures can suggest a different model.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the exponential base b for the given exponential data.
The semilog gradient is ln2, so b=e^(ln2)=2.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · F. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The transformed vertical value when the chosen horizontal variable is zero. Choose the relationship, show the method, check its assumptions and interpret the result.