A table has equally spaced inputs. Which check points to a logarithmic model rather than a linear, quadratic or exponential one?

None of the first differences, the second differences or the ratios of the outputs is constant. Instead the input values change proportionally while the outputs change additively, which is the reverse of the exponential pattern.

True or False? Over a limited range of data, linear, quadratic and exponential models can all give very similar values.

True. Where the rate of change is only slightly varying, all three can produce curves that sit close to the data. It is only over larger intervals that they pull apart, which is why fit alone cannot decide between them.

How does each of a linear, a quadratic and an exponential model behave beyond the range of the data?

A linear model keeps growing at a constant rate without bound, and an exponential model grows or decays without bound. A quadratic model changes direction, so sooner or later it turns and heads the other way.

Why does behavior beyond the data matter when choosing a model?

Because the best model has to make sense both within the data range and for reasonable predictions beyond it. A model that fits the data closely but behaves unrealistically outside it is a poor choice.

What does a residual plot show, and what does it look like when the model is appropriate?

It plots the residuals against the input variable. When the model is appropriate the plot shows no pattern , with the points scattered randomly above and below the horizontal axis.

True or False? A residual plot with roughly as many points above zero as below shows that the model is appropriate.

False. What matters is whether there is a clear pattern , not how the points balance above and below zero. A plot can have balanced counts and still curve systematically, which would mean the model is the wrong one.

A linear regression's residual plot is U-shaped. What does that tell you?

That the linear model is not appropriate , because the residuals follow a systematic curve instead of scattering randomly. The data has a curvature the line cannot capture, so a quadratic or exponential model would fit it better.

When might an underestimate be preferable to an overestimate, and when the reverse?

It depends on the context, and on what an error in each direction would cost. Underestimating the strength of a bridge builds in a safety margin, while overestimating project costs makes sure the budget is sufficient.

Why does an exponential function appear as a straight line on a semi-log plot?

Because on a logarithmically scaled axis, equal multiplicative changes take up equal vertical distances. An exponential function multiplies its output by the same factor at each unit step, so those steps come out equally spaced and the points line up.

True or False? If data lies along a straight line on a semi-log plot, it can be modeled by an exponential function.

True. That is the key rule for reading a semi-log plot, and it is useful in reverse too. If the data does not look linear on a semi-log plot, an exponential model is probably not the best fit.

AP®MathsCollege BoardPrecalculusFlashcardsExponential & Logarithmic FunctionsModeling with Exponential & Logarithmic Functions

Modeling with Exponential & Logarithmic Functions (College Board AP® Precalculus): Flashcards

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FrontSelecting & Constructing Exponential Models
A table has equally spaced inputs. Which check points to a logarithmic model rather than a linear, quadratic or exponential one?

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Cards in this collection (39)

A table has equally spaced inputs. Which check points to a logarithmic model rather than a linear, quadratic or exponential one?
None of the first differences, the second differences or the ratios of the outputs is constant.
Instead the input values change proportionally while the outputs change additively, which is the reverse of the exponential pattern.
Fill in the two gaps.
When data is not perfectly exponential, a graphing calculator can fit an exponential $______$ model, choosing $a$ and $b$ by the method of $______$ squares.
The completed sentence is:
When data is not perfectly exponential, a graphing calculator can fit an exponential regression model, choosing $a$ and $b$ by the method of least squares.
The resulting model can then predict outputs at inputs not in the original data, including non-integer ones.
Why does an exponential regression give a better model than using just the first two data points?
A regression uses all the data, whereas two points ignore everything else.
For the values $10, 17, 28, 48$ the first ratio alone suggests a base of $1.7$ , while the regression gives about $1.683$ , and the two predict noticeably different values.
Data grows exponentially by a factor of $3$ each unit and equals $10$ at $t = 2$ . Write a model.
$f (t) = 10 \cdot 3^{(t - 2)}$ .
That is the horizontal shift of $y = 10 \cdot 3^{t}$ which moves the known value from $t = 0$ across to $t = 2$ .
True or False?
The ratio test for an exponential model only works when the inputs increase by $1$ .
False.
The input intervals need only be equal in length, not equal to $1$ .
With inputs $0, 2, 4, 6, 8$ and outputs $96, 48, 24, 12, 6$ the ratio is $0.5$ across each interval of length $2$ , so an exponential model fits.
In $f (x) = a b^{x}$ , what does the base $b$ tell you about percent change?
It is the growth factor, the number the output is multiplied by each time the input rises by one unit.
If $b > 1$ the percent increase per unit is $(b - 1) \times 100$ , and if $0 < b < 1$ the percent decrease is $(1 - b) \times 100$ .
True or False?
An increase of 6.1 percent per quarter gives a growth factor of $0.061$ .
False.
The growth factor is $1.061$ , because it is $1 + \frac{r}{100}$ rather than $\frac{r}{100}$ .
For a decrease of $r$ percent the decay factor is $1 - \frac{r}{100}$ in the same way.
A population grows by 5 percent per month, and $t$ is measured in years. Write the model.
$P (t) = P_{0} {(1.05)}^{12 t}$ .
The monthly growth factor is $1.05$ , and there are $12 t$ months in $t$ years, so the exponent counts the number of growth periods.
A growth rate is given per quarter but $t$ is in years. How do you check the exponent is the right way round?
Substitute a simple value and count the growth periods: at $t = 1$ year there should be $4$ quarters.
So the exponent has to come out as $4$ when $t = 1$ , which means $4 t$ rather than $\frac{t}{4}$ .
A substance decays by 10 percent per decade, and $t$ is measured in years. Write the model.
$A (t) = A_{0} {(0.90)}^{\frac{t}{10}}$ .
The decay factor per decade is $0.90$ , and there are $\frac{t}{10}$ decades in $t$ years, so this time the exponent divides rather than multiplies.
The model is $f (d) = 2^{d}$ with $d$ in days. Rewrite it to show the growth factor per week.
$f (d) = {(2^{7})}^{\frac{d}{7}} = 128^{\frac{d}{7}}$ .
So the quantity grows by a factor of $128$ every $7$ days.
The general move is $b^{x} = {(b^{k})}^{\frac{x}{k}}$ for any nonzero $k$ .
A radioactive substance has a half-life of $8$ days. Write a model for the amount left after $d$ days.
$h (d) = A_{0} {(0.5)}^{\frac{d}{8}}$ .
A half-life of $8$ days means the amount is multiplied by $0.5$ every $8$ days, so the exponent counts how many $8$ -day periods have passed.
Fill in the two gaps.
In the model $P (t) = 500 {(1.03)}^{t}$ with $t$ in years, the quantity grows by $______$ percent for every one $______$ that passes.
The completed sentence is:
In the model $P (t) = 500 {(1.03)}^{t}$ with $t$ in years, the quantity grows by 3 percent for every one year that passes.
The growth factor always applies per unit of whatever the input variable measures.
True or False?
Over a limited range of data, linear, quadratic and exponential models can all give very similar values.
True.
Where the rate of change is only slightly varying, all three can produce curves that sit close to the data.
It is only over larger intervals that they pull apart, which is why fit alone cannot decide between them.
Fill in the two gaps.
A context with a fixed amount added each period suggests a $______$ model, while a context with a constant percent change, such as compound interest, suggests an $______$ model.
The completed sentence is:
A context with a fixed amount added each period suggests a linear model, while a context with a constant percent change, such as compound interest, suggests an exponential model.
A quantity that rises and then falls, such as the height of a projectile, points instead to a quadratic model.
How does each of a linear, a quadratic and an exponential model behave beyond the range of the data?
A linear model keeps growing at a constant rate without bound, and an exponential model grows or decays without bound.
A quadratic model changes direction, so sooner or later it turns and heads the other way.
Why does behavior beyond the data matter when choosing a model?
Because the best model has to make sense both within the data range and for reasonable predictions beyond it.
A model that fits the data closely but behaves unrealistically outside it is a poor choice.
Revenue is $10, 16, 25, 40, 65$ in months $1$ to $5$ . Is a quadratic or an exponential model more appropriate?
An exponential model.
The second differences are $3, 6, 10$ , which are not constant, so a quadratic does not fit well.
The ratios are about $1.6, 1.563, 1.6$ and $1.625$ , roughly constant, so the outputs are approximately proportional over equal intervals.
Define residual.
A residual is the difference between the actual value of the dependent variable and the value the model predicts.
In symbols, residual $=$ actual value $-$ predicted value.
What does a residual plot show, and what does it look like when the model is appropriate?
It plots the residuals against the input variable.
When the model is appropriate the plot shows no pattern, with the points scattered randomly above and below the horizontal axis.
True or False?
A residual plot with roughly as many points above zero as below shows that the model is appropriate.
False.
What matters is whether there is a clear pattern, not how the points balance above and below zero.
A plot can have balanced counts and still curve systematically, which would mean the model is the wrong one.
A linear regression's residual plot is U-shaped. What does that tell you?
That the linear model is not appropriate, because the residuals follow a systematic curve instead of scattering randomly.
The data has a curvature the line cannot capture, so a quadratic or exponential model would fit it better.
Fill in the two gaps.
When a model's predicted value is greater than the actual value it is an $______$ , and that corresponds to a $______$ residual.
The completed sentence is:
When a model's predicted value is greater than the actual value it is an overestimate, and that corresponds to a negative residual.
It follows from residual $=$ actual $-$ predicted, so an underestimate gives a positive residual instead.
When might an underestimate be preferable to an overestimate, and when the reverse?
It depends on the context, and on what an error in each direction would cost.
Underestimating the strength of a bridge builds in a safety margin, while overestimating project costs makes sure the budget is sufficient.
A logarithmic model fits sales well up to day $91$ , but sales then start to fall. Why does the error increase after that?
Because the model keeps increasing while the actual quantity is decreasing.
The two agree at day $91$ , but from then on the gap between predicted and actual values grows wider every day.
A logarithmic model of that form always increases when its coefficient is positive, so it can never follow a falling quantity.
Fill in the two gaps.
A logarithmic model is appropriate when the $______$ values change proportionally over equal-length intervals of the $______$ values.
The completed sentence is:
A logarithmic model is appropriate when the input values change proportionally over equal-length intervals of the output values.
That is the reverse of an exponential model, where it is the outputs that change proportionally.
How do you find $a$ and $b$ in a model $f (x) = a + b \ln (x + c)$ from two data points?
Take $c$ as given, which it usually is, then substitute each data point to get two equations in $a$ and $b$ .
Solving that pair of simultaneous equations gives both constants.
True or False?
In $f (x) = a + b \ln (x + c)$ , the first data point always gives $a$ directly.
False.
It does so only when that point makes the argument of the logarithm equal to $1$ , because $\ln (1) = 0$ then removes the $b$ term.
Questions are often arranged so that this happens, but it is not automatic.
For $S (t) = a + b \ln (t + 1)$ , $S (0) = 15$ and $S (5) = 24$ . Find $a$ and $b$ .
$a = 15$ and $b \approx 5.023$ .
At $t = 0$ the argument is $1$ and $\ln (1) = 0$ , so the first equation gives $a = 15$ at once.
Then $15 + b \ln 6 = 24$ , so $b = \frac{9}{\ln 6} = 5.022995 \dots$
Inputs double for each increase of $1$ in the output, and the output is $0$ when $x = 4$ . Write a model $f (x) = \log_{b} (k x)$ .
$f (x) = \log_{2} (\frac{x}{4})$ .
The inputs doubling for each unit rise in the output makes the base $b = 2$ .
The zero at $x = 4$ needs $k x = 1$ there, since $\log_{b} (1) = 0$ , which gives $k = \frac{1}{4}$ .
How do you fit a logarithmic model when the data does not fit one exactly?
Use a logarithmic regression on a graphing calculator, usually the LnReg function.
It finds the best-fit values of $a$ and $b$ in a model of the form $a + b \ln x$ .
Define semi-log plot.
A semi-log plot is a graph in which one axis uses a logarithmic scale instead of a standard linear one.
In this course it is the $y$ -axis that is logarithmically scaled, with the $x$ -axis kept standard.
Why does an exponential function appear as a straight line on a semi-log plot?
Because on a logarithmically scaled axis, equal multiplicative changes take up equal vertical distances.
An exponential function multiplies its output by the same factor at each unit step, so those steps come out equally spaced and the points line up.
How can you recognise a semi-log plot from the graph itself?
The horizontal grid lines are unevenly spaced, bunching together towards the top of each decade.
The labelled values climb in powers of the base, such as $1, 10, 100, 1000$ , rather than in equal steps.
What advantage does a semi-log plot have over testing the ratios of the output values?
It reveals exponential behaviour even when the data carries a vertical shift, as in $y = a b^{x} + k$ .
The ratio test needs the constant $k$ removed first, whereas for large $x$ the exponential term dominates and the logged values still trend linear.
That makes the plot a more robust visual test.
True or False?
If data lies along a straight line on a semi-log plot, it can be modeled by an exponential function.
True.
That is the key rule for reading a semi-log plot, and it is useful in reverse too.
If the data does not look linear on a semi-log plot, an exponential model is probably not the best fit.
Taking $\log_{n}$ of both sides of $y = a b^{x}$ produces a straight line. Fill in the two gaps.
$\log_{n} y = \log_{n}______+ x \log_{n}______$
The completed linearization is:
$\log_{n} y = \log_{n} a + x \log_{n} b$
This has the form $Y = c + m x$ , with slope $\log_{n} b$ and $y$ -intercept $\log_{n} a$ .
A semi-log plot using base $10$ is linear with slope $0.3$ and $y$ -intercept $1.5$ . Find the exponential model.
$y \approx 31.623 \cdot 2^{x}$ .
The base is $b = 10^{0.3} = 1.995 \dots$ , which is about $2$ , since the slope is $\log_{10} b$ .
The initial value is $a = 10^{1.5} = 31.622 \dots$ , since the intercept is $\log_{10} a$ .
$P$ takes the values $5, 15, 45, 135, 405$ at $t = 0$ to $4$ . What do the base- $10$ logarithms show, and what is the model?
The logarithms are about $0.699, 1.176, 1.653, 2.130, 2.607$ , rising by $0.477$ each time, so the linearized data lies on a straight line.
The slope $0.477$ gives $b = 10^{0.477} = 3$ , and the intercept $0.699$ gives $a = 10^{0.699} = 5$ .
So the model is $P (t) = 5 \cdot 3^{t}$ .

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