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Exponential Growth and Decay: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 One recorded increase

A positive quantity rises from $240$ to $252$ in one step. It will keep the same percentage increase in every later step. Find its factor per step.
1. Hint 1
  The factor is the new amount divided by the old amount.
2. Hint 2
  The increase is only part of the new amount; include the amount already present.
Answer

$1.05$ .

Full solution

The old amount is positive, so division is valid.
$b=\frac{252}{240}$
This is $1.05$ .
The increase is $12/240=0.05$ of the old amount, confirming a $5\%$ rate and a $1.05$ factor.

Answer

$1.05$ .

Key idea

A factor is the entire new amount as a multiple of the previous amount.
Problem 1: Mark complete
Problem 2 A stationary setting

The model $A(t)=18(1+\frac c{100})^t$ uses a positive factor. Find the value of $c$ that makes the amount stay constant for every $t\ge0$ .
1. Hint 1
  A positive amount stays unchanged when each step multiplies it by one.
2. Hint 2
  Set the whole factor equal to $1$ .
Answer

$c=0$ .

Full solution

The factor must satisfy
$1+\frac c{100}=1$
giving $c=0$ .
Then $A(t)=18\cdot1^t=18$ for every allowed time.

Answer

$c=0$ .

Key idea

A zero rate corresponds to a factor of one and a constant amount.
Problem 2: Mark complete
Problem 3 A monthly statement

An account multiplies its balance by $1.0065$ each month. What nominal annual interest rate does this correspond to when the rate is compounded monthly?
1. Hint 1
  The monthly rate is the factor minus one.
2. Hint 2
  A quoted nominal annual rate is divided by $12$ to obtain the monthly rate.
Answer

$7.8\%$ per year, compounded monthly.

Full solution

The monthly rate is $0.0065$ .
Reversing the division by $12$ gives
$r=12(0.0065)$
Thus $r=0.078$ , or $7.8\%$ annually.
Dividing $0.078$ by $12$ returns $0.0065$ , verifying the schedule.

Answer

$7.8\%$ per year, compounded monthly.

Key idea

A nominal annual rate is the periodic rate multiplied by the number of periods per year.
Problem 3: Mark complete
Problem 4 A missing discount

A positive price rises by $24\%$ , then a discount is applied to the increased price. The final price is $19.4\%$ below the original price. What percentage discount was applied?
1. Hint 1
  Each percentage acts on the price present at that stage, so the changes combine by multiplying factors.
2. Hint 2
  Write the discount factor as $1-d$ and set the product of the two factors equal to the net factor.
Answer

A $35\%$ discount.

Full solution

The rise has factor $1.24$ and the net change has factor $0.806$ .
With discount $d$ ,
$1.24(1-d)=0.806$

So $1-d=0.65$ and $d=0.35$ , a $35\%$ discount.
Check: a price of $100$ rises to $124$ , and $35\%$ of $124$ is $43.4$ , leaving $80.6$ , which is $19.4\%$ below $100$ .

Answer

A $35\%$ discount.

Key idea

To find a missing discount, divide the net factor by the known nonzero factor, then subtract the resulting discount factor from one.
Problem 4: Mark complete
Problem 5 An interrupted deposit

A deposit of $1750$ dollars earns $6.8\%$ nominal annual interest compounded quarterly. After half a year, $200$ dollars is withdrawn. The remaining balance earns interest for another three-quarters of a year. Find the ending balance to the nearest cent.
1. Hint 1
  The withdrawal splits the calculation into two growth intervals.
2. Hint 2
  Convert each interval into a number of quarters, then multiply by the quarterly factor that many times.
Answer

Ending balance $\approx1693.52$ dollars.

Full solution

The quarterly rate is
$0.068/4=0.017.$
Half a year is $2$ quarters, so the balance before the withdrawal is $1750(1.017)^2=1810.00575$ dollars, and the withdrawal leaves $1610.00575$ dollars.

Three-quarters of a year is $3$ quarters.
$A=1610.00575(1.017)^3$
Thus $A\approx1693.52$ dollars, keeping the unrounded balance until the end.

Answer

Ending balance $\approx1693.52$ dollars.

Key idea

An account withdrawal changes the starting balance for the remaining compounding periods.
Problem 5: Mark complete
Problem 6 A fading signal

A signal decays exponentially. It loses $35$ units during its first half-life, then loses $9$ more units during the next $4$ hours. Find its initial amount, and its half-life exactly and to the nearest tenth of an hour.
1. Hint 1
  During one half-life a decaying quantity loses half of what it had.
2. Hint 2
  Write the amount remaining after the next four hours as a power of one half, and solve for the half-life.
Answer

Initial amount $70$ units; $h=\frac{4\log_{10}(0.5)}{\log_{10}(26/35)}\approx9.3$ hours.

Full solution

Losing $35$ units in one half-life means $35$ is half the initial amount, so the initial amount is $70$ and $35$ remain after one half-life.

Four hours later $35-9=26$ remain, so
$35\left(\tfrac12\right)^{4/h}=26$
which gives
$(\tfrac12)^{4/h}=\frac{26}{35}.$

Taking logarithms, $\frac4h\log_{10}(0.5)=\log_{10}(26/35)$ , so
$h=\frac{4\log_{10}(0.5)}{\log_{10}(26/35)}$
Both logarithms are negative, so $h$ is positive, about $9.3$ hours.

Answer

Initial amount $70$ units; $h=\frac{4\log_{10}(0.5)}{\log_{10}(26/35)}\approx9.3$ hours.

Key idea

Two losses measured over known times can determine a decaying quantity's starting amount and half-life.
Problem 6: Mark complete
Problem 7 A repeated cycle

In each two-day cycle a quantity rises by $25\%$ on the first day and falls by $10\%$ on the second. It starts at $80$ units. Find a model $C(n)$ for the amount after $n$ complete cycles, then the real value of $n$ with $C(n)=160$ . Convert that $n$ to days, exactly and to the nearest tenth of a day.
1. Hint 1
  Combine the two daily factors into the factor for one whole cycle.
2. Hint 2
  Find the cycle count for a factor of two, then convert cycles to days.
Answer

$C(n)=80(1.125)^n$ ; $t=\frac{2\log_{10}2}{\log_{10}(1.125)}\approx11.8$ days.

Full solution

One full cycle multiplies the amount by
$1.25(0.9)=1.125.$
Therefore
$C(n)=80(1.125)^n$

The doubled target requires $1.125^n=2$ , so
$n=\log_{10}2/\log_{10}(1.125).$
Each cycle lasts two days, giving the stated time, approximately $11.8$ days.
The model is exact only at whole cycles: day by day, the quantity itself first passes $160$ at the end of day $9$ , since $80(1.125)^4(1.25)\approx160.18$ .

Answer

$C(n)=80(1.125)^n$ ; $t=\frac{2\log_{10}2}{\log_{10}(1.125)}\approx11.8$ days.

Key idea

A repeating cycle of fixed positive multipliers has a cycle factor equal to their product.
Problem 7: Mark complete
Problem 8 Order of two changes

Two tanks each start with $P>0$ liters. Tank A has $14\%$ of its contents removed and then receives $6$ liters. Tank B receives $6$ liters and then has $14\%$ of its contents removed. Do they finish with equal volumes? If not, which contains more, and by how much?
1. Hint 1
  A percentage acts on the amount present at that moment.
2. Hint 2
  Write each tank's final volume in terms of $P$ and subtract.
Answer

No; tank A contains $0.84$ liters more, whatever $P$ is.

Full solution

Tank A: removing $14\%$ leaves $0.86P$ , then adding $6$ gives $0.86P+6$ .

Tank B: adding $6$ gives $P+6$ , then removing $14\%$ leaves
$0.86(P+6)=0.86P+5.16.$

The difference is $(0.86P+6)-(0.86P+5.16)=0.84$ liters, for every $P>0$ , so tank A contains more.
In tank B the percentage also takes $14\%$ of the added $6$ liters.

Answer

No; tank A contains $0.84$ liters more, whatever $P$ is.

Key idea

A nonzero percentage change and a nonzero fixed addition do not commute: the percentage also acts on whatever was added before it.
Problem 8: Mark complete
Problem 9 Three factor settings

For $P>0$ , consider $A(t)=Pb^t$ at $t\ge0$ . A student says that $b=1.02$ , $b=1$ , and $b=0.98$ produce growth, standstill, and decay, respectively. Is the classification correct? Explain what happens after each unit step.
1. Hint 1
  Compare each factor with one: above one, equal to one, or below one.
2. Hint 2
  The ratio of the amount after a step to the amount before it is $b$ .
Answer

Yes: $b=1.02$ gives a $2\%$ increase each step; $b=1$ leaves the amount unchanged; $b=0.98$ gives a $2\%$ decrease each step.

Full solution

Every current amount is positive.
A step with $b=1.02$ retains it and adds $2\%$ ; a step with $b=1$ leaves it unchanged; a step with $b=0.98$ removes $2\%$ .
$\frac{A(t+1)}{A(t)}=b$
The denominator is positive because $P>0$ and each listed base is positive.

Answer

Yes: $b=1.02$ gives a $2\%$ increase each step; $b=1$ leaves the amount unchanged; $b=0.98$ gives a $2\%$ decrease each step.

Key idea

Comparing a positive factor with one determines whether positive exponential amounts rise, stay constant, or fall.
Problem 9: Mark complete
Problem 10 Two schedules

Two positive quantities grow exponentially. Quantity A doubles every $7$ hours; quantity B triples every $11$ hours. Give both doubling times exactly and determine which doubles sooner. You may use $2^{11}=2048$ and $3^7=2187$ .
1. Hint 1
  Write B's doubling time as the time $d$ with $3^{d/11}=2$ .
2. Hint 2
  Compare the two doubling times by comparing powers of $2$ and $3$ .
Answer

$d_A=7$ hours and $d_B=\frac{11\log_{10}2}{\log_{10}3}$ hours; B doubles sooner.

Full solution

A doubles every $7$ hours, so $d_A=7$ .
For B, $3^{d_B/11}=2$ , so
$d_B=\frac{11\log_{10}2}{\log_{10}3}$

B doubles sooner exactly when $d_B<7$ , that is, when $11\log_{10}2<7\log_{10}3$ , which says
$2^{11}<3^7.$
Since $2048<2187$ , it does, so B doubles sooner (in about $6.94$ hours).

Answer

$d_A=7$ hours and $d_B=\frac{11\log_{10}2}{\log_{10}3}$ hours; B doubles sooner.

Key idea

Doubling times of different schedules compare through the powers their growth factors reach.
Problem 10: Mark complete

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