Exponents, Exponentials and Logarithms · Lesson 2

Exponential growth and decay

An exponential model multiplies by the same factor each step. In y=a⋅bty = a \cdot b^t, aa is the starting amount (the value at t=0t = 0) and bb is the growth factor per step. A linear model adds the same amount each step; an exponential model multiplies by the same number each step.

Turn a percent into a factor. Growth by a rate rr per step uses b=1+rb = 1 + r, and decay uses b=1−rb = 1 - r, with rr written as a decimal. Growth of 6% per year gives a(1.06)ta(1.06)^t; decay of 8% per year gives a(0.92)ta(0.92)^t. Read it backward too: a factor of 0.92 means 92% remains and 8% is lost each step.

Doubling and half-life. If an amount doubles every dd units of time, it is a⋅2t/da \cdot 2^{t/d}; if it halves every dd units, it is a(12)t/da\left(\frac{1}{2}\right)^{t/d}. The exponent td\frac{t}{d} counts how many doublings or halvings have happened, so convert the time units first.

Spot exponential data by ratios. In a table with equally spaced inputs, divide each output by the one before. A constant ratio means exponential; a constant difference means linear.

The graphs of y = 3 \cdot 2^x and y = 3\left(\frac{1}{2}\right)^x−4−3−2−11234246810120xy(0, 3)(2, 12)(1, 1.5)
The graphs of y=3⋅2xy = 3 \cdot 2^x and y=3(12)xy = 3\left(\frac{1}{2}\right)^x Two curves that both cross the y-axis at (0, 3), the starting amount. The growth curve y = 3 · 2^x rises from left to right through (1, 6) and (2, 12). The decay curve y = 3(1/2)^x falls from left to right through (1, 1.5) and (2, 0.75). Each curve gets closer and closer to the x-axis on one side, the horizontal asymptote y = 0, and never touches it.
Reading the factor bb in y=a⋅bty = a \cdot b^t
DescriptionFactor per stepModel
Grows 6% per year1.061.06a(1.06)ta(1.06)^t
Shrinks 8% per year0.920.92a(0.92)ta(0.92)^t
Doubles every 5 hours22 per 5 hoursa⋅2t/5a \cdot 2^{t/5}
Half remains every 12 days12\frac{1}{2} per 12 daysa(12)t/12a\left(\frac{1}{2}\right)^{t/12}

Worked example Easy

The population of a small town is 12,000 people and is growing by 3% per year. Which function gives the population P(t)P(t) after tt years?

  1. P(t)=12,000(1.3)tP(t) = 12{,}000(1.3)^t
  2. P(t)=12,000(1.03)tP(t) = 12{,}000(1.03)^tAnswer
  3. P(t)=12,000(0.97)tP(t) = 12{,}000(0.97)^t
  4. P(t)=12,000(0.03)tP(t) = 12{,}000(0.03)^t

How to solve it

  1. Growth by 3% means each year's population is 100%+3%=103%100\% + 3\% = 103\% of the year before.
  2. As a decimal, the growth factor is 1+0.03=1.031 + 0.03 = 1.03.
  3. Start at 12,000 and multiply by 1.03 once per year: P(t)=12,000(1.03)tP(t) = 12{,}000(1.03)^t.

Why each choice is right or wrong

  • A. Incorrect. A factor of 1.3 writes 3% as 0.3 instead of 0.03, which would model 30% growth per year.
  • B. Correct. The starting value is 12,000 and the growth factor is 1+0.03=1.031 + 0.03 = 1.03, so P(t)=12,000(1.03)tP(t) = 12{,}000(1.03)^t.
  • C. Incorrect. A factor of 0.97 is 1−0.031 - 0.03, a 3% decrease each year. A growing population needs a factor greater than 1.
  • D. Incorrect. A factor of 0.03 uses the rate rr instead of 1+r1 + r; this model would keep only 3% of the population each year.

Worked example Medium

A lab sample of a radioactive substance has a mass of 640 grams. The substance has a half-life of 5 days. What mass of the substance, in grams, remains after 20 days?

  1. 4040Answer
  2. 8080
  3. 160160
  4. 320320

How to solve it

  1. A half-life of 5 days means the mass halves once every 5 days.
  2. In 20 days there are 205=4\frac{20}{5} = 4 half-lives.
  3. M=640(12)4=64016=40M = 640\left(\frac{1}{2}\right)^4 = \frac{640}{16} = 40.

Why each choice is right or wrong

  • A. Correct. Four half-lives multiply by 12\frac{1}{2} four times: 640⋅116=40640 \cdot \frac{1}{16} = 40. The halvings go 320, 160, 80, 40.
  • B. Incorrect. 80 divides by 2⋅4=82 \cdot 4 = 8 instead of 24=162^4 = 16. Four halvings multiply by 12\frac{1}{2} four times.
  • C. Incorrect. 160 divides 640 by 4, the number of half-lives. Each half-life halves the mass, so divide by 2 four times.
  • D. Incorrect. 320 is the mass after one half-life, 5 days. Twenty days is 4 half-lives, so keep halving.

Trap Using $r$ instead of $1 + r$

A 4% growth rate is a factor of 1.04, not 0.04. A factor below 1 shrinks the amount, so a(0.04)ta(0.04)^t would keep only 4% each year. Write the factor as 1 plus the rate.

Trap Decay: subtract the rate from 1

A decrease of 8% keeps 92%, so the factor is 1−0.08=0.921 - 0.08 = 0.92, not 0.08. In reverse, a model with factor 0.92 loses 8% per step, not 92%.

Calculator When a calculator is faster

Exponential models are a good place for a calculator. Type the model, such as y = 640(0.5)^(x/5), into the built-in Desmos calculator if you test online and read values from the graph, or type 640(0.5)^(20/5) to evaluate it directly. On paper, a scientific calculator handles the powers just as well.

To pick a model from a table, enter the points and see which choice's graph passes through all of them.