Exponential models, transformations, and Log Mode

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Read interval factors and shifts, compare equivalent exponential forms, solve for a target value by graphing both sides, and use automatic Log Mode with suitable data.

Difficulty: Advanced
Estimated time: 75 minutes
Domains: Advanced Math Problem Solving and Data Analysis

What you’ll learn

Read a shifted, stretched exponential model like $c+ab^{(x-h)/k}$ one piece at a time.
Find its reference input and its baseline, and how much it multiplies by over each interval and over each single unit.
Turn a growth or decay multiplier into a percent change.
Spot the equivalent answer choice by full-curve overlap, and rewrite a model by hand when a question needs the exact form.
Find when a model reaches a target by graphing both sides.
Tell automatic Log Mode apart from ordinary nonlinear regression.

Prerequisites

Why this matters on the SAT

Read the multiplier the exponent actually describes

In SAT exponential models, the most useful facts often hide in the exponent. A base might describe the change every $5$ minutes, not every minute. And a number added on the end can mean the base multiplies only the amount above that number, not the whole output.

SAT example

After a ventilation system is turned on, the estimated concentration of a gas in a parking garage is modeled by

y=48(0.84)^{\frac{x-10}{5}}+7,

where $y$ is the concentration, in parts per million, and $x$ is the number of minutes after the system is turned on, with $x\ge10$ .

For each increase of $1$ minute in $x$ , the amount by which the concentration exceeds $7$ parts per million is multiplied by a constant factor. Which choice is closest to that factor?

Fast Desmos solution

Start with the $+7$ . It's a baseline, so the multiplier works on the amount above $7$ , which is $y-7$ . That's exactly the amount the question asks about.

Next, look at the exponent. It's divided by $5$ , so $0.84$ is the multiplier for every $5$ minutes, not every minute. Five one-minute steps have to multiply out to $0.84$ , so the one-minute multiplier is the fifth root:

0.84^{1/5}.

Type 0.84^(1/5) into Desmos:

0.84^{1/5}\approx0.9657.

That's closest to $0.97$ , so the answer is C.

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Each minute, 0.84^(1/5) multiplies only the amount above the baseline of 7, not the whole concentration.

Choose the job before the tool

First ask what the question hands you: a model you can use as it is, or data you still have to fit a model to.

The question gives you the model

When it asks for an output, a multiplier, or a percent, evaluate the model directly, like typing f(3) for $f(x)=200(1.05)^x$ .
When the unknown sits in the exponent, as in $120(0.85)^{t/4}=50$ , and the powers don’t match up easily, graph both sides and read where they meet.
When the answer choices are possible equivalent forms of an expression in $x$ , graph the original and find the choice whose graph lies right on top of it along the whole visible curve. That’s full-curve overlap.

The question gives you data

When a table gives several inputs and outputs, and either the question points to an exponential model or each output is about the same multiple of the one before, run exponential regression.
When the values were measured and rounded, trust the base the regression finds, because it uses the whole table and not just one pair of rounded numbers.

Sometimes a line of algebra beats Desmos: when both sides already share a base, as in $2^{x+1}=2^5$ , when a ratio you can see gives the answer, or when the question wants an exact form and has no answer choices to test.

Regression finds the unknown numbers in a model, its parameters, from a table or a few given facts. If the question already gives you the whole function, there’s nothing left to find.

Check your understanding:

One question gives $P(t)=300(1.04)^t$ and asks for $P(6)$ . Another gives five measured pairs and says an exponential model is appropriate. Which Desmos method would you use for each?

Read every layer of the model

The general form of these models has five letters, and that's a lot to take in at once. So let's start with numbers. Here's the gas model from the SAT example again:

y=48(0.84)^{\frac{x-10}{5}}+7.

Take it one number at a time:

The $10$ . At $x=10$ , the exponent is $\frac{10-10}{5}=0$ , and any base to the power $0$ is $1$ . So the model starts counting at minute $10$ .
The $+7$ . It's added after the power, so no multiplier ever touches it. Think of it as a fixed level the rest of the model sits on.
The $48$ . At $x=10$ , the concentration is $7+48=55$ parts per million, which is $48$ above that level.
The $5$ . The exponent goes up by $1$ each time $x$ goes up by $5$ . So the model takes steps of $5$ minutes.
The $0.84$ . Each $5$ -minute step multiplies the amount above $7$ by $0.84$ .

Now the same thing with letters. The general form is

f(x)=c+ab^{\frac{x-h}{k}},

where $k>0$ , $b>0$ , and $b\ne1$ . Each letter does the job you just saw:

$h$ is the reference input ( $10$ above). At $x=h$ , the exponent is $0$ .
$c$ is the baseline ( $7$ above), a fixed amount added to the exponential part.
$a$ tells you how far the graph is from the baseline at $x=h$ ( $48$ above), because $f(h)=c+a$ . A positive $a$ puts the graph above the baseline there. A negative $a$ puts it below, and the distance is $\lvert a\rvert$ .
$k$ is the interval ( $5$ above). Each time $x$ grows by $k$ , the exponent grows by $1$ .
$b$ is the multiplier for each interval ( $0.84$ above). It multiplies $f(x)-c$ , the gap between the graph and the baseline.

The signs of $a$ and $b$ tell you which way the graph goes. In the gas model, $a=48$ is positive and $b=0.84$ is less than $1$ . So the part above $7$ keeps shrinking, and the graph levels off toward the line $y=7$ , its horizontal asymptote. In general, when $a>0$ and $0<b<1$ , the graph levels off toward $y=c$ .

Change the base to $1.2$ and keep $a=48$ . Now the part above the baseline, $f(x)-c$ , grows each step instead. That's the case $a>0$ and $b>1$ .

Now make $a$ negative too, as in $7-48(1.2)^{\frac{x-10}{5}}$ . The graph starts $48$ below the baseline, at $-41$ , and each step multiplies that distance by $1.2$ . So when $a<0$ and $b>1$ , the distance from the baseline, $\lvert f(x)-c\rvert$ , still grows, but the graph heads downward.

Check your understanding:

The function $p(t)=50(0.6)^{t/5}$ models a mass in milligrams after $t$ days. What does the factor $0.6$ mean?

Common mistake:

In $c+ab^{(x-h)/k}$ , the base multiplies only the exponential part, $f(x)-c$ . The baseline $c$ never gets multiplied. In the gas model, $5$ minutes after $x=10$ , the amount above $7$ goes from $48$ to $48(0.84)=40.32$ , so $y=47.32$ . Multiplying the whole $55$ by $0.84$ would give $46.2$ instead, which is wrong.

Convert an interval factor into a per-unit factor

Say $b$ is the multiplier for every $k$ input units. What multiplier $r$ , used once per unit, gives the same change? After $k$ one-unit steps, you've multiplied by $r$ a total of $k$ times, which is $r^k$ , and that has to equal $b$ . So $r$ is the $k$ th root of $b$ :

r=b^{1/k}.

You can check that it works:

r^k=\left(b^{1/k}\right)^k=b,

so the two agree over the whole interval.

Back to the mass from the quick check,

p(t)=50(0.6)^{t/5}.

The per-day multiplier is

0.6^{1/5}\approx0.9029.

So about $90.29\%$ of the mass is left after each day, and the mass drops by about

100(1-0.9029)\%=9.71\%

per day.

To turn any multiplier $r$ into a percent change, compare it with $1$ :

\begin{aligned} \text{growth percent}&=100(r-1)\%, && r>1,\\[1.4em] \text{decay percent}&=100(1-r)\%, && 0<r<1. \end{aligned}

Check your understanding:

A population is multiplied by $1.44$ every 2 hours. What is the hourly multiplier, and what is the hourly percent increase?

Common mistake:

Say a population grows $21\%$ every $2$ hours. Splitting that into $10.5\%$ an hour sounds reasonable, but growth compounds: two hours at $10.5\%$ gives $1.105^2\approx1.221$ , more than $1.21$ . Turn the percent into a multiplier first, then take the root that matches the interval. Here $1.21^{1/2}=1.1$ , so the growth is $10\%$ an hour.

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The fifth root turns the five-day factor into a one-day factor. The second line turns that into the daily percent decrease.

Choose an equivalent form by full-curve overlap

The same exponential can be written in different forms, and each form shows you something different. The transformed form

a b^{\frac{x-h}{k}}+c

shows the reference input, the interval length, and the baseline. The standard form

A\cdot B^x+c

shows the per-unit multiplier $B$ .

When a multiple-choice question in one variable asks which form is equivalent, let the answer choices do the work. Put the original on one line and leave it alone. Put one choice on a second line, and edit only that line to try each choice in turn. The choice whose graph lies on top of the original along the whole visible curve is the answer. That's full-curve overlap, and it's the fastest reliable way through these questions.

Rewrite with exponent rules instead when there are no answer choices, when the question wants the exact form written out, or when you'd like a quick confirmation.

Here's how the rewrite goes in general:

\begin{aligned} c+ab^{\frac{x-h}{k}} &=c+ab^{-h/k}\left(b^{1/k}\right)^x\\[1.4em] &=c+A\cdot B^x, \end{aligned}

where

A=ab^{-h/k} \qquad\text{and}\qquad B=b^{1/k}.

The shift $c$ doesn't change, and the new base $B$ is the per-unit multiplier.

Worked example

Consider

F(t)=72\cdot9^{\frac{t-2}{2}}.

Which choice is equivalent to $F(t)$ ?

A
$8\cdot3^t$
B
$8\cdot9^t$
C
$24\cdot3^t$
D
$72\cdot3^t$

Desmos draws $y=$ graphs against $x$ , so first swap the problem's $t$ for $x$ , in the original and in every choice. Keep y=72*9^((x-2)/2) on line 1. Type choice A as y=8*3^x on line 2, then edit only line 2 to try the others. Choice A lies on the original along the whole visible curve, so the answer is A.

Now try choice B, y=8*9^x. It meets the original at $x=0$ , where both equal $8$ , but to the right of that point it climbs much faster. That's why you check the whole curve: two different expressions can meet at one point without being equivalent.

The algebra is short here, so you can also confirm it exactly:

\begin{aligned} 72\cdot9^{\frac{t-2}{2}} &=72\cdot\left(3^2\right)^{\frac{t-2}{2}}\\[1.4em] &=72\cdot3^{t-2}\\[1.4em] &=72\cdot\frac{3^t}{9}\\[1.4em] &=8\cdot3^t. \end{aligned}

Both forms describe the same curve in different ways. The transformed form says the output is multiplied by $9$ every $2$ input units. The standard form says it's multiplied by $3$ every unit, and two steps of $3$ make $9$ .

Check your understanding:

Rewrite $40\cdot25^{(x-2)/2}$ in the form $A\cdot B^x$ .

Common mistake:

The subtraction and the division in $b^{(x-h)/k}$ both live inside the exponent. So $9^{(t-2)/2}$ isn’t $9^t-2$ divided by $2$ . Split the exponent first, $\frac{t-2}{2}=\frac t2-1$ , then apply exponent rules to the whole power: $9^{t/2-1}=\frac{3^t}{9}$ .

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Choice A lies right on top of the original. Leave line 1 alone and edit only line 2 to try choices B, C, and D.

Solve an exponential target by graphing both sides

When the unknown is in an exponent, graph the model and the target on separate lines. Where they meet, the model's output equals the target, and the input there is your answer.

Say

M(t)=120(0.85)^{t/4}

models a mass in grams after $t$ hours, and you want to know when the mass reaches $50$ grams. Enter

M(t)=120(0.85)^(t/4)

and

y=50

The graphs meet near

(21.547,50).

So the mass reaches $50$ grams after about

\boxed{21.5}

hours, to the nearest tenth.

You never needed a logarithm. The point Desmos shows is a decimal, though, so round it and give the units the way the question asks.

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The graphs meet where the mass first reaches the target of 50 grams.

Try it yourself:

Change y=50 to y=40. Before you click the new point, predict: will it move left or right? Then reset the example.

Here's where a correct graph can still lead to a wrong answer. When a question asks for the least whole number that meets or passes a target, don't just round the intersection to the nearest integer. Look at the direction of the inequality, and test the next whole number that's allowed.

Check your understanding:

Look at $200(0.5)^{t/6}=25$ before you reach for a graph. Is graphing the fastest way here, and what is $t$ ?

Use automatic Log Mode for exponential data

Now for data. Say a table has only positive outputs, and the question tells you, or the numbers show, that the model has the form

y=ab^x.

Put the inputs in x_1 and the outputs in y_1, then type

y_1~a*b^(x_1)

Desmos finds the $a$ and $b$ that fit the table. In the current College Board testing calculator, an exponential regression like this one runs in Log Mode on its own. You don't switch it on, which is why it's called automatic.

So what is Log Mode? It's a way of fitting, not a separate tool. A straight-line pattern is simpler to fit than a curve, so Desmos takes logarithms, which turns the exponential pattern into a straight-line one. It fits that line, then hands you back $a$ and $b$ . You never type a logarithm or change the table yourself.

Try it on this exact table:

$x_1$	$0$	$1$	$2$	$3$
$y_1$	$90$	$72$	$57.6$	$46.08$

Log Mode gives

a=90 \qquad\text{and}\qquad b=0.8.

So the model is

y=90(0.8)^x.

A base of $0.8$ means $80\%$ remains at each step, so the output drops by $20\%$ per input interval. You can see it in the table, too: each output is $0.8$ times the one before.

Try it yourself:

Change the last output from $46.08$ to $45$ . Will the fitted base land above or below $0.8$ ? Predict first, check the result, then reset the table.

Keep in mind that Log Mode only does the fitting. It won't tell you whether an exponential model is the right choice. That call is yours, and it comes from the question, the context, and outputs that keep multiplying by about the same factor.

Log Mode also has limits, and most of them come from those logarithms:

Logarithms work only on positive numbers, so every output has to be positive.
If even one output is zero or negative, Log Mode can't run. Desmos may still show a fit, but it comes from ordinary nonlinear regression, which adjusts $a$ and $b$ directly until the curve sits as close to the points as it can.
Taking logarithms of a model like $ab^x+c$ with an unknown $c$ doesn't give a straight line, so Desmos uses ordinary nonlinear regression for it too.
Rounded measurements give you estimates of $a$ and $b$ , not exact values.

Later in the course, regression failure modes looks closer at residuals, fit statistics, and exact versus best-fit data.

Check your understanding:

A table has an output of $0$ , and a student wants to fit $y=ab^x$ with Log Mode. What should they notice first?

Common mistake:

Log Mode fits a model to table data. It isn’t a command for solving an exponential equation. When you’re given an equation with the unknown in the exponent, graph both sides or match the powers.

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Log Mode fits the positive data with a starting value of 90 and a multiplier of 0.8.

Example: Read the model, graph the target, then find the first whole minute

Worked example

The temperature $T(t)$ , in degrees Celsius, of a liquid $t$ minutes after cooling begins is modeled by

T(t)=18+72(0.75)^{t/6}.

Which choice is the least integer value of $t$ for which $T(t)\le40$ ?

A
$24$
B
$25$
C
$30$
D
$31$

Step 1

Read the model

The baseline is $18$ . At $t=0$ , the exponent is $0$ , so the liquid starts at

18+72=90

degrees Celsius. Every $6$ minutes, the amount above $18$ is multiplied by $0.75$ .

Step 2

Graph the model and the target

Enter the model and the target line:

T(t)=18+72(0.75)^(t/6)

y=40

The graphs meet at about

(24.728,40).

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The graphs meet after minute 24 but before minute 25.

Step 3

Find the first whole minute that works

At $t=24$ , the liquid is still a little too warm:

T(24)=40.78125.

At $t=25$ , it has cooled past the target:

T(25)\approx39.715,

so the least integer time with $T(t)\le40$ is

\boxed{25}.

The answer is B.

Common mistake:

The graphs meet at about $24.7$ , so $24$ is tempting. But at minute $24$ the liquid is still above $40$ degrees. It’s cooling, and it has to be at or below the target, so the first whole minute that works is $25$ .

Finish the solution

The measured data are already in the calculator. Add the exponential regression, then use the fitted base to find the percent decrease per interval.

Turn the fitted base into a percent

Finish the solution

A scientist measures a sample once every $4$ hours. Let $n$ be the number of $4$ -hour intervals after the first measurement.

$n$	$0$	$1$	$2$	$3$
Mass (milligrams)	$900$	$744$	$620$	$511$

An exponential model of the form

M(n)=ab^n

is used to fit the measured data. According to the fitted model, by what percent does the sample's mass decrease every $4$ hours, rounded to the nearest tenth of a percent?

First steps

The table is already loaded, with the interval number in x_1 and the mass in y_1.
Enter y_1~a*b^(x_1).
Use the fitted $b$ directly. Don't retype the decimal Desmos shows.

Finish it

Your answer

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Add the regression, then turn its fitted base into the percent decrease per 4-hour interval.

Practice problems

Before you calculate, ask what the exponent and the base mean, and whether the question gives you a model or data.

Read an interval factor

Practice problem

The concentration $m(t)$ , in parts per million, of a substance $t$ days after a treatment begins is modeled by

m(t)=12+80(0.72)^{t/6}.

Which choice best describes the meaning of the factor $0.72$ ?

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Evaluate the model at inputs 6 apart and compare each output with the baseline of 12.

Choose an equivalent exponential

Practice problem

A company models the amount of data, in gigabytes, stored in an archive after $r$ weeks with

D(r)=96\cdot4^{\frac{3r-2}{2}}.

Which choice gives an equivalent function in the form

D(r)=A\cdot B^r,

where $A$ and $B$ are constants?

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Test the choices one at a time against the original by full-curve overlap.

Find the least allowable input

Practice problem

The function

S(t)=600(1.08)^{t/3}

models the number of subscribers on a channel $t$ days after a promotion begins. Which choice is the least integer value of $t$ for which

S(t)\ge900?

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Graph the model and the target, then find the first whole day that works.

The next problem adds an unknown vertical shift, $q$ , so Log Mode can't fit it. Use custom regression with several conditions instead. Put the three given points in one table and fit the model to all of them at once, so Desmos finds the three unknowns together.

Fit a shifted exponential model

Practice problem

The function $m$ is defined by

m(x)=p\,r^x+q,

where $p$ , $q$ , and $r$ are positive constants. The graph of $y=m(x)$ passes through $(0,5)$ , $(1,9)$ , and $(3,57)$ . Which choice gives the value of $m(2)$ ?

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Fit the three points with the given shifted form, then evaluate at x = 2.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Read the exponent. In $c+ab^{(x-h)/k}$ , the base applies once every $k$ input units.
Watch the baseline. The base multiplies the distance from $c$ , not the whole output.
Convert the interval. The per-unit factor is $b^{1/k}$ .
Turn it into a percent. Growth is $100(r-1)\%$ , and decay is $100(1-r)\%$ .
Pick equivalent forms by full-curve overlap. Test one choice at a time against a fixed original, then use exponent rules when you need an exact form or a quick check.
Hit a target by graphing. When algebra isn't quick, graph the model against the target.
Fit a table with Log Mode. It fits positive data of the form $ab^x$ . A zero or negative output, or an unknown shift, means ordinary regression instead.
Choose the job. Use a model you're given as it is, and run a regression only when there are parameters left to find.

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