Exponential Functions

Adding versus multiplying

Two people offer you a deal for the next twelve days. The first gives you $ $50$ today and another $ $50$ every day after. The second gives you one cent today, and every day promises to double whatever they gave you yesterday.

For the first week the first deal is obviously better — you’re up hundreds of dollars while the second person is still handing you pocket change. Then the doubling catches up, and by day twelve it isn’t close. Nothing about the second deal ever got faster. The rule never changed. It was multiply by two, every single day, from the start.

That is the whole idea of this module. A5 gave you a machine that adds the same amount at every step — a linear function, with a rate you can read off a table. This one multiplies by the same amount at every step. F7 already taught you what $b^x$ means when $x$ was a number you were handed; now $x$ is the input, and the exponent counts the steps.

The table gives it away

Here is the skill the SAT asks for over and over, usually without warning you which kind of function you’re looking at. You get a table, and you have to decide.

The test is two columns. Subtract each row from the next — if that difference is the same every time, something is being added over and over, and it’s linear. If it isn’t, divide each row by the one before it. If that is the same every time, something is being multiplied over and over, and it’s exponential.

try subtracting

The table reads

3,\ 6,\ 12,\ 24

. The differences are

3

, then

6

, then

12

— not equal, so nothing is being added over and over. One column down.

try dividing

Now divide:

6 \div 3 = 2

12 \div 6 = 2

24 \div 12 = 2

. The same factor every step.

name it

Constant ratio

2

, so it’s exponential — and the rule is “start at

3

, double each step.”

Before you touch the columns, take the two offers at the board — five dollars up front, or one dollar that doubles — and pick one before the fourth day lands.

Transcript

Two offers. A: $ $5$ now, plus $ $1$ every day. B: a single $ $1$ — but it doubles every day. Four days. Pick.

$A : $ 5 + $ 1 a day$

$B : $ 1 \times 2 a day$

Day $1$ : A has $ $6$ , B has $ $2$ . Day $2$ : A $ $7$ , B $ $4$ . A is winning, and it is not close.

$$ 6$

$$ 7$

Day $3$ : A has $ $8$ . B has $ $8$ . Dead tie — the first and last there will ever be.

Day $4$ : A adds one more: $ $9$ . B doubles: $ $16$ — off this chart. Then $32$ . Then $64$ .

$16, 32, 64, \dots$

$\times 2$

One word does all the work: doubles means $\times 2$ , not $+ 2$ . B is $1 \cdot 2 \cdot 2 \cdot 2 = 8$ by day $3$ .

$1 \cdot 2 \cdot 2 \cdot 2 = 8$

Nothing sped up. B ran one rule from day $1$ : multiply by $2$ . That is all an exponential is — the same multiplication, repeated.

On the test: subtract neighbors — same gap? The adder: linear. Divide neighbors — same factor? The multiplier: exponential.

$6 - 5 = 1$

$32 \div 16 = 2$

Two offers: adding versus multiplying · 1:30

Both columns live side by side in Ratio Table, with a badge on whichever one has gone constant.

Two columns decide everything: the difference between rows, and the ratio between them. Exactly one of them goes constant — and which one names the model.

start ateach stepfactor

x	y	difference	ratioconstant
0	$3$
1	$6$	$3$	$2$
2	$12$	$6$	$2$
3	$24$	$12$	$2$
4	$48$	$24$	$2$

The verdictThe ratios are all

2

, so this table is exponential:

y = 3 (2)^{x}

. Every step multiplies by the same amount.

Switch the rule between adds and multiplies, and watch the constant badge jump columns.

Which column goes constant?

Predict before you switch: set the rule to adds with an amount of $100$ and look at the table — it climbs by a hundred every row, which feels dramatic. Which column has the badge? Now flip the rule to multiplies by and watch the badge jump. Then try Name that table and call the second one before you read the columns; it is the trap below, wearing a table.

What each letter is doing

Every function in this module has the same two-part shape:

f (x) = a \cdot b^{x}

$a$ is the starting amount — the value before anything has happened. That falls straight out of F7’s rule that $b^0 = 1$ : put $x = 0$ in and the whole power collapses to one, leaving $f(0) = a$ . So when a question shows you $f(x) = 500(1.08)^x$ and asks what the $500$ represents, you’re not calculating anything. It’s the value at the start.

$b$ is the growth factor — what one step multiplies by. If $b > 1$ the quantity grows; if $b$ is between $0$ and $1$ it decays, shrinking by the same proportion each step. And the exponent binds to $b$ alone: $a \cdot b^x$ is not $(a \cdot b)^x$ . The starting amount is multiplied by the power at the end, never dragged inside it.

Percent in disguise

Here is where the SAT hides this topic, and where F5 hands you the answer already made. You learned there that raising something by $p\%$ means multiplying by $1 + \tfrac{p}{100}$ — a single multiplier. That multiplier is $b$ .

So “grows by 8% a year” is $b = 1.08$ , and “falls by 15% a year” is $b = 1 - 0.15 = 0.85$ . A population of $500$ growing at 8% a year is $f(x) = 500(1.08)^x$ , and that’s the entire translation.

Watching the curve

Numbers in a table are one view; the shape is another. In Growth Dial the two letters are two dots you can drag: the start dot sets $a$ , and the one-step dot sets how high the first step lands — which fixes $b$ , since $b$ is just that height divided by the start.

Drag the start dot to set where the curve begins, and the one step dot to set how much one step multiplies by. The factor and the percent follow your hand.

f (x) = 4 (1.5)^{x}

Each step multiplies by

1.5

growth · +50% a step

$b = 1.5$ is bigger than $1$ , so every step ADDS. The part past $1$ is the rate: $50%$ more each step.

Drag the one-step dot BELOW the start dot and watch growth become decay.

The dashed line starts at the same place and moves by the same amount on the FIRST step. Look right: the line keeps that pace for ever, and the curve does not.

Set the factor by hand

Predict before you drag: put the start at $4$ and the one-step dot at $6$ . What percent is that per step? Now drag the one-step dot below the start dot and watch the read-out flip from growth to decay. Finally, follow a decaying curve to the right — it keeps halving, or keeps taking a fifth off, and never lands on zero. Multiplying by a positive number can shrink something forever without ever finishing the job.

That is why an exponential graph flattens toward the axis without touching it, and why “it decays to nothing” is never quite true.

When you can solve one

You can only solve an exponential equation exactly when both sides can be written as powers of the same base — then the exponents must match:

one base

2^{x+1} = 8

, and

8

2^3

, so both sides are powers of

2

equate

x + 1 = 3

, so

x = 2

The general question — “how many years until this reaches $1000$ ?” — needs logarithms, which this course doesn’t cover yet. On the SAT that question is answered from a table or a graph instead, so if the bases won’t match, look for the value rather than solving for it.

The one thing to remember

Linear adds, exponential multiplies. Off a table, subtract to test the first and divide to test the second — exactly one column goes constant. In $a \cdot b^x$ the $a$ is where you started and the $b$ is what one step does, with $b = 1 \pm \tfrac{r}{100}$ turning any percent into that multiplier. And when $b < 1$ , remember that the factor is what survived, not what was lost.

Which model is it?

Test	Do this	Constant means
linear	subtract each row from the next	same difference → adds the same amount
exponential	divide each row by the one before	same ratio → multiplies by the same amount

f (x) = a \cdot b^{x} a = f (0) (the start), b = one step’s multiplier

Growth or decay: $b > 1$ grows; $0 < b < 1$ decays; $b = 1$ is flat.

Percent to factor: up $r\%$ → $b = 1 + \tfrac{r}{100}$ ; down $r\%$ → $b = 1 - \tfrac{r}{100}$ . So $+8\% \to 1.08$ , $-15\% \to 0.85$ , doubling $\to 2$ , halving $\to 0.5$ .

Factor back to percent: $b = 1.2$ is $+20\%$ ; $b = 0.9$ is $-10\%$ . Compare to $1$ .

Landmarks: $f(0) = a$ always (since $b^0 = 1$ ). A negative exponent steps backwards: $b^{-2} = \tfrac{1}{b^2}$ .

Solving: only by matching bases — $2^{x+1} = 8 \Rightarrow x + 1 = 3$ . Anything else needs logarithms (not on this course yet, and not required by the SAT).