One starting amount, one multiplier, and an exponent that counts how many times the multiplier is applied.
Which of the following is the best interpretation of 0.82 in this context?
The function v is defined by v(t) = 18,000(0.82)t/2, where v(t) is the estimated value, in dollars, of a delivery van t years after it was purchased.
Two readings are needed. The base 0.82 is what is kept, and 1 − 0.82 = 0.18 is what is lost, a drop of 18%. The exponent t2 goes up by 1 only when t goes up by 2, so that drop happens every 2 years. Check: v(2) = 18,000(0.82) = $14,760, which is 18% below $18,000. A gets the percent right but ignores the division by 2, decaying twice as fast as the model does. C and D read the base as the loss instead of the remainder — an 82% loss would leave the van worth $3,240 after two years, not $14,760.
1
Every exponential model is a starting amount times a repeated multiplier.
In y = a(b)x, a is the value at x = 0 and b is the number you multiply by once per step. Nothing else in this lesson is new.
f(x) = 800(1.12)x
a
800
b
1.12
x
steps
f(0) = 800(1.12)0 = 800 × 1 = 800
anything to the zero power is 1, so a is the y-intercept
f(1) = 800 × 1.12 = 896
one multiplication
f(2) = 896 × 1.12 = 1,003.52
the second step adds 107.52, not 96 — the rise itself grows
That is the whole difference between this and a straight line. A line adds the same amount every step; an exponential multiplies by the same factor every step, so each step is calculated on a bigger number than the last.
2
Turn the percent into a multiplier before anything else.
Up r percent → b = 1 + r100. Down r percent → b = 1 − r100. The multiplier is always what you keep.
up 6%
1 + 0.06 = 1.06
keep all of it, add 6% more
down 6%
1 − 0.06 = 0.94
94% of it survives each step
1.03 → up 3%1.25 → up 25%2 → doubled0.9 → down 10%0.6 → down 40%0.5 → halved
A city uses 7.2 billion gallons of water this year, and conservation rules cut the figure by 4% a year.
b = 1 − 0.04 = 0.96
96% of last year's use is still used
W(t) = 7.2(0.96)t
billions of gallons, t years from now
0.963 = 0.96 × 0.96 × 0.96 = 0.884736
0.9216 after two, then once more
W(3) = 7.2 × 0.884736 ≈ 6.37 billion gallons
7.2 → 6.912 → 6.636 → 6.370 the long way
0.04 and 4 are both wrong, and wrong by miles
7.2(0.04)t throws away 96% a year, leaving less than a thousandth of today's use after three years. 7.2(4)t is the percent number with the decimal point missing and quadruples the city every year. Write 1 ± r100 on the page before you write the function.
3
The base alone decides growth or decay.
With a positive a: b > 1 grows, 0 < b < 1 decays, b = 1 never moves. You do not need the graph to know which.
Pond A has 50 lily pads and the number grows 35% each month, so A(t) = 50(1.35)t. Pond B has 500 and loses 25% each month, so B(t) = 500(0.75)t.
A(4) = 50(1.35)4 = 50 × 3.3215 ≈ 166
1.35² = 1.8225, and 1.8225² = 3.3215
B(4) = 500(0.75)4 = 500 × 0.31641 ≈ 158
0.75² = 0.5625, and 0.5625² = 0.31641
Pond A passes Pond B just before t = 4
166 against 158 — by month 4 the gold curve is already on top
Look at the shapes, not just the numbers. Growth steepens forever and never turns back down. Decay flattens out, getting closer and closer to the x-axis without ever touching it, which is why a decay model never reaches 0. A quantity that falls and then rises is neither: that shape belongs to a quadratic.
4
The exponent counts periods, not clock time.
If the multiplier is applied once every k units of time, the exponent is tk. The exponent must equal the number of times b has been used.
v(t) = 18,000(0.82)t/2
t = 2: exponent = 22 = 1
one 18% drop has happened
v(2) = 18,000 × 0.82 = $14,760
18,000 − 3,240, exactly 18% off
t = 4: exponent = 2, so v(4) = 18,000 × 0.822
0.82² = 0.6724
v(4) = 18,000 × 0.6724 = $12,103.20
14,760 × 0.82 gives the same figure
The sentence says
Exponent
Why
every year, t in years
t
one period per unit
every 2 years, t in years
t2
fewer periods than years
every 3 months, t in years
4t
four periods per year
every 6 hours, t in days
4t
four periods per day
half-life of 9 days, t in days
t9
count the halvings
One sanity check settles every one of these: put in one whole period's worth of time and confirm the exponent comes out to exactly 1.
5
Several periods at once: raise the base, then subtract 1.
A hard SAT item hands you a model and asks for the percent change over a stretch longer than one period.
f(x) = 400(1.3)x/2, and x increases by 4
exponent goes up by 42 = 2
two full periods, not four
f is multiplied by 1.32 = 1.69
the start value 400 cancels and never matters
1.69 − 1 = 0.69 → an increase of 69%
a multiplier becomes a percent by subtracting 1
Decay works the same way. If g(t) = 900(0.85)t/5 and t rises by 15, the exponent rises by 3, so g is multiplied by 0.85³ = 0.614125 — about 61.4% of what it was, a decrease of about 38.6%.
×1.69 is up 69% · ×0.614 is down 38.6%
69% is not 2 × 30%
Two 30% rises do not make a 60% rise, because the second 30% is taken on the already larger amount. Adding percentages is the same mistake as adding multipliers; multiply them instead.
6
Say what each part means in the situation's own words.
Interpretation items hand you the finished model and ask what one number represents. Name the part, then translate it.
N(t) = 6,400(1.045)t
N(t) is the estimated number of trees in a reserve t years after 2015.
Part
Its job
In this context
6,400
value at t = 0
there were 6,400 trees in 2015
1.045
multiplier per year
the number of trees increases by 4.5% each year
1.045 − 1
the growth rate
0.045, which is where the 4.5% comes from
t
number of periods
years since 2015
The base is never the percent change itself. Subtract 1 first, then read the decimal as a percent — and if the exponent is tk, attach "every k years" to that percent.
The choice that keeps the number and loses the meaning
"The value of the van decreases by 82% each year" in the opening problem is built from the right digits and the wrong reading, twice over: 0.82 is what remains rather than what is lost, and the exponent t2 means the change happens every second year. Two wrong turns, one very tempting sentence.
Desmos is on your screen — use it
Type y = 18000(0.82)x/2 and read the table, or type y = 50(1.35)^x and y = 500(0.75)^x together and click the intersection to see exactly when Pond A overtakes Pond B (about x = 3.92). Desmos is also the fastest check on an interpretation item: if the stem says "decreases by 18% every 2 years", the value at x = 2 should be 82% of the value at x = 0.
✓
Check: does one whole period behave like the sentence?
the van18,000 × 0.82 = 14,76018% gone after 2 years ✓
the pond50 → 67.5 → 91.125×1.35 each month ✓
the city7.2 → 6.912 → 6.636 → 6.370each year keeps 96% ✓
Remember the flow
a is the start · b = 1 ± r · the exponent counts periods
Multiplier to percent: subtract 1. Percent to multiplier: add to 1.
Now you try
Practice problem
What is the estimated population 9 weeks from now?
A field contains an estimated 5,000 insects of a certain species, and the population is expected to decrease by 20% every 3 weeks.
The multiplier is 1 − 0.20 = 0.8 and it is applied once every 3 weeks, so P(w) = 5,000(0.8)w/3. Nine weeks is 93 = 3 periods: 0.8³ = 0.512 and 5,000 × 0.512 = 2,560. The long way agrees: 5,000 → 4,000 → 3,200 → 2,560. A uses 0.2 instead of 0.8, the part that disappears rather than the part that remains, and wipes out 99.2% of the field. B subtracts three lots of 20% of the original 5,000, which is the linear reading of a percent decrease. D stops after two periods — that is the population at 6 weeks, not 9.
Hint: answer two questions in order before you compute anything. What is the multiplier? And how many times is it applied? The percent settles the first; the period named in the sentence settles the second.
