Logarithmic graphs for power and exponential fits
Введено| English | Значение |
|---|---|
| intercept/ˌɪntəˈsept/ | пересечение с осью |
Which horizontal axis distinguishes a power model from an exponential one?
- A curve can become a straight line after taking logs.
- This lesson studies intercept 截距: The transformed vertical value when the chosen horizontal variable is zero.
The method
- 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.
In a log-log fit of y=ax^p, what does the gradient give?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Logging both sides turns the power law into ln y=ln a+p·ln x: gradient p, intercept ln a. For an exponential y=kb^x plot ln y against x instead - the axes decide what the gradient means.
A worked case
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The log-log gradient is ln16/ln4=2.
Where it goes wrong
- 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.
Not true: Plotting ln y against x linearises every power model y=ax^p. A power model is linear in ln y against ln x; ln y against x linearises exponential models.
Find its multiplicative parameter a.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The intercept is ln3, so a=e^(ln3)=3.
Plotting ln y against x linearises every power model y=ax^p.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A power model is linear in ln y against ln x; ln y against x linearises exponential models.
Applying it
- 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.
Find the exponential base b for the given exponential data.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The semilog gradient is ln2, so b=e^(ln2)=2.
Match each model to the plot that linearises it.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The power plot cannot log a nonpositive x, while the exponential plot allows any x; intercepts correspond to ln x=0 or x=0 respectively.
In your course
- AQA A-level Mathematics (7357), specification section F.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Intercept: The transformed vertical value when the chosen horizontal variable is zero.
- For positive observations and parameters, y=ax^p with x>0 gives ln y=ln a+p ln x.
- Watch for: for the power fit the gradient is the exponent, not its logarithm.