Exponential Growth and Decay: Practice

Quarterly compounding means $n = 4$ periods a year, so divide the rate and multiply the time.

$\frac{r}{n} = \frac{0.08}{4} = 0.02, \qquad nt = 4 \cdot 6 = 24$

Now apply the model with those values.

$A = 1500(1.02)^{24} \approx 1500(1.60844) \approx 2412.66$

Compounding once a year instead would give $1500(1.08)^6 \approx 2380.31$ dollars, which is less.

Chained percent changes multiply their factors; they do not add their rates.

$1.20 \times 0.80 = 0.96$

The final price is $96\%$ of the original, a $4\%$ decrease. The two $20\%$ figures are percentages of different amounts, so the fall is taken from a larger price than the rise was.

Divide both sides by the principal first, so the equation involves only the multiplier.

$5000(1.07)^t = 8000 \quad \Longrightarrow \quad (1.07)^t = 1.6$

Take logarithms and bring the exponent down.

$t = \frac{\log 1.6}{\log 1.07} \approx \frac{0.20412}{0.029384} \approx 6.95 \text{ years}$

Dividing $1.6$ by $0.07$ gives about $22.9$, but that treats the growth as if it were linear.

The yearly factor $b$ must produce a halving after eight years, so it satisfies an equation you can undo with a rational exponent.

$b^{8} = \tfrac{1}{2} \quad \Longrightarrow \quad b = \left(\tfrac{1}{2}\right)^{1/8} \approx 0.917$

The value $0.875$ comes from subtracting $\tfrac{1}{8}$ of the sample each year, but decay removes a fixed fraction of what is left, not of the original.

In $t$ years the number of triplings that have taken place is $t/4$, and each tripling multiplies by $3$.

$A = P \cdot 3^{t/4}$

Check it at $t = 4$, where exactly one tripling should have happened: $P \cdot 3^{4/4} = 3P$. The version $3^{4t}$ would triple four times a year.

First convert the time into doubling periods: four hours contains eight half-hours, so eight doublings occur.

$A = 500 \cdot 2^{8} = 500 \cdot 256 = 128000$

Counting the four hours as four doublings would give $500 \cdot 16 = 8000$, far too few. The period, not the hour, is the unit of growth here.

The growth factor is $b = 1.5$, and doubling means the multiplier reaches $2$ (the starting amount cancels).

$(1.5)^t = 2$

Take logarithms and solve for the exponent.

$t = \frac{\log 2}{\log 1.5} \approx \frac{0.30103}{0.176091} \approx 1.71 \text{ years}$

Two years would over-double, since $(1.5)^2 = 2.25$, so the true answer must be under $2$.

Semiannual compounding means $n = 2$ periods a year, so each period pays half the annual rate and there are $2$ periods per year.

$\frac{r}{n} = \frac{0.09}{2} = 0.045, \qquad nt = 2 \cdot 3 = 6$

Apply the model.

$A = 800(1.045)^{6} \approx 800(1.302260) \approx 1041.81$

Halving the rate but forgetting to double the periods gives $800(1.045)^3 \approx 912.93$ dollars, which is far too low.

Being $50\%$ larger means the multiplier has reached $1.5$, and the starting population cancels out of the equation.

$(1.15)^t = 1.5$

Take logarithms and use the power law.

$t = \frac{\log 1.5}{\log 1.15} \approx \frac{0.176091}{0.060698} \approx 2.90 \text{ years}$

Dividing $50$ by $15$ gives about $3.33$, but that assumes each year adds a flat $15\%$ of the original population.

Multiply the two factors and look for a product of exactly $1$.

$1.25 \times 0.80 = 1.00$

The others fall short: $1.20 \times 0.80 = 0.96$, $1.30 \times 0.70 = 0.91$, and $1.50 \times 0.50 = 0.75$. A rise and fall of the same percent never cancels, because $(1+r)(1-r) = 1 - r^2 < 1$.

Use the half-life form of the model, where the exponent counts how many half-lives have elapsed.

$A(t) = 240\left(\tfrac{1}{2}\right)^{t/15}$

At $t = 20$ the exponent is $\tfrac{20}{15} = \tfrac{4}{3}$, so a little more than one half-life has passed.

$A(20) = 240\left(\tfrac{1}{2}\right)^{4/3} = \frac{240}{2^{4/3}} \approx \frac{240}{2.5198} \approx 95.2 \text{ grams}$

That is sensible: after a full $15$ years there would be exactly $120$ grams, and $20$ years leaves somewhat less.

The decay factor is $b = 0.97$. Divide the target by the starting population so only the multiplier is left.

$(0.97)^t = \frac{40000}{50000} = 0.8$

Take logarithms of both sides.

$t = \frac{\log 0.8}{\log 0.97} \approx \frac{-0.09691}{-0.013228} \approx 7.33 \text{ years}$

A linear estimate ($20\%$ lost at $3\%$ a year) suggests about $6.7$ years, but each year's loss shrinks along with the population, so it actually takes longer.