OMPT drill — Functions

Exponential functions

Growth and decay, solving a^x = b, and recognising when an exponential model applies. Pairs directly with logarithms — study them back to back. Attempt each problem before opening the solution; the notation only becomes automatic by writing it yourself.

Tested inOMPT-A OMPT-B OMPT-D OMPT-E OMPT-F OMPT-G

Lesson

An exponential function $f (x) = b \cdot g^{x}$ describes anything that grows or shrinks by the same factor each step: interest, populations, radioactive decay, a cup of coffee cooling. The $b$ is the starting amount: what you have at $x = 0$ , since $g^{0} = 1$ . The $g$ is the growth factor per step: $g = 1.04$ means up 4% each period, $g = 0.85$ means down 15%. Growth when $g > 1$ , decay when $0 < g < 1$ .

The translation between percentage and factor is where this topic is won or lost, because the OMPT states problems in percentages and the mathematics runs on factors. "Increases by 4% per year" is multiply by 1.04; "decreases by 15%" is multiply by 0.85, not by 0.15. Once the model is written down, the remaining question types are few: evaluate it at some time, or solve $g^{x} = c$ for the time, which you do by matching bases now, and by logarithms once you have them.

Problem 1Numeric answer

A bacteria culture starts at 800 cells and doubles every 5 hours. How many cells are there after 15 hours?

Show the worked solution

Answer: 6400

In 15 hours the culture doubles $15/5 = 3$ times, so the count is $800 \cdot 2^{3} = 800 \cdot 8 = 6400$ cells. I count doubling periods first, then apply the power — never multiply by 2 once per hour.

Problem 2Multiple choice

Which of these functions describes exponential decay?

A $y = 6 \cdot (0.85)^{x}$
B $y = 0.5 \cdot (1.2)^{x}$
C $y = 6 \cdot x^{- 2}$
D $y = 85 - 6 x$

Show the worked solution

Answer: A — $y = 6 \cdot (0.85)^{x}$

Exponential decay needs the form $y = a \cdot g^{x}$ with a growth factor $0 < g < 1$ . Only the first option qualifies: $g = 0.85$ . The second grows ( $g = 1.2$ — a small starting value does not make it decay), the third is a power function of $x$ , and the fourth is linear.

Problem 3Numeric answer

Solve for $n$ : $2^{n + 3} = 32$ .

Show the worked solution

Answer: 2

I write 32 as a power of 2: $32 = 2^{5}$ . Then $2^{n + 3} = 2^{5}$ forces $n + 3 = 5$ , so $n = 2$ . Check: $2^{5} = 32$ .

Problem 4Multiple choice

A town of 2500 people grows by $4%$ per year. Which formula gives the population after $t$ years?

A $N (t) = 2500 \cdot 1.0 4^{t}$
B $N (t) = 2500 \cdot 0.0 4^{t}$
C $N (t) = 2500 + 1.04 t$
D $N (t) = 2500 \cdot 4^{t}$

Show the worked solution

Answer: A — $N (t) = 2500 \cdot 1.0 4^{t}$

Growing by $4%$ means each year the population is multiplied by $1 + 0.04 = 1.04$ . After $t$ years that multiplication has happened $t$ times: $N (t) = 2500 \cdot 1.0 4^{t}$ . The growth factor always carries the "1 plus" — the 1 is the population you already had.

Problem 5Spot the error

A sample of 3 grams doubles 4 times. A student computes the final mass. Step 1: final mass $= (3 \cdot 2)^{4}$ . Step 2: $= 6^{4}$ . Step 3: $= 1296$ grams. Which step contains the error?

Show the worked solution

Answer: Step 1

The exponent belongs to the growth factor alone, not to the starting amount. Step 1 should read $3 \cdot 2^{4}$ , which is $3 \cdot 16 = 48$ grams. Steps 2 and 3 correctly evaluate the wrong expression. Walking through the doublings by hand confirms it: $3 \to 6 \to 12 \to 24 \to 48$ .

Problem 6Numeric answer

A medicine has a half-life of 6 hours. A patient takes 160 mg. How many milligrams remain after 18 hours?

Show the worked solution

Answer: 20

Eighteen hours contain $18/6 = 3$ half-life periods, so the amount is halved three times: $160 \cdot (\frac{1}{2})^{3} = 160/8 = 20$ mg. Step by step: $160 \to 80 \to 40 \to 20$ .