Exponential Growth and Decay: Additional Practice (Set 1)

The decay factor is $0.96$. Divide the target by the starting population to isolate the multiplier.

$(0.96)^t = \frac{8000}{12000} = \frac{2}{3}$

Take logarithms to find the crossing time.

$t = \frac{\log \frac{2}{3}}{\log 0.96} \approx \frac{-0.176091}{-0.017729} \approx 9.93$

The population crosses $8000$ partway through the tenth year, so the first whole year below the target is $t = 10$. Checking: after $9$ years there are about $8310$ left, and after $10$ years about $7978$.

Compare the final mass with the starting mass to see how many halvings occurred.

$\frac{5}{80} = \frac{1}{16} = \left(\tfrac{1}{2}\right)^{4}$

Four half-lives fit into the $24$ days, so each one lasts

$h = \frac{24}{4} = 6 \text{ days}.$

The masses run $80 \to 40 \to 20 \to 10 \to 5$ grams, one step every $6$ days.

Monthly compounding gives $n = 12$, so divide the rate and multiply the time.

$\frac{r}{n} = \frac{0.048}{12} = 0.004, \qquad nt = 12 \cdot 8 = 96$

Apply the model with those values.

$A = 2500(1.004)^{96} \approx 2500(1.46702) \approx 3667.55$

Compounding once a year would give about $3637.73$ dollars, and simple interest only $3460$ dollars.

Tripling means the multiplier reaches $3$, and the principal cancels out of the equation.

$(1.06)^t = 3$

Take logarithms and bring the exponent down.

$t = \frac{\log 3}{\log 1.06} \approx \frac{0.477121}{0.025306} \approx 18.85 \text{ years}$

The doubling time at the same rate is about $11.9$ years, and tripling reasonably takes longer than that but far less than the linear guess of $300 \div 6 = 50$ years.

Three years of decay leave a multiplier of $\tfrac{18000}{30000} = 0.6$, so the annual factor satisfies a cube equation.

$b^{3} = 0.6 \quad \Longrightarrow \quad b = 0.6^{1/3} \approx 0.8434$

The rate is $r = b - 1 \approx -0.1566$, a loss of about $15.7\%$ a year. Spreading the $40\%$ loss evenly to get about $13.3\%$ a year is too gentle, since each year's loss is taken from a smaller value than the last.

Five years of growth multiply the balance by $\tfrac{6000}{4000} = 1.5$, so the annual factor satisfies a fifth-power equation.

$b^{5} = 1.5 \quad \Longrightarrow \quad b = 1.5^{1/5} \approx 1.0845$

The rate is about $8.45\%$ a year. Dividing the $50\%$ gain by $5$ years to get $10\%$ overshoots, because $(1.10)^5 = 1.61$, well past $1.5$.

After $n$ half-lives the fraction remaining is $\left(\tfrac{1}{2}\right)^{n}$, and the condition is that this drops below $0.01$.

$n > \frac{\log 0.01}{\log 0.5} = \frac{-2}{-0.30103} \approx 6.64$

The first whole number past $6.64$ is $n = 7$. Checking: six half-lives leave $\tfrac{1}{64} \approx 1.56\%$, still above $1\%$, while seven leave $\tfrac{1}{128} \approx 0.78\%$.

The monthly factor is $0.95$, applied twelve times.

$(0.95)^{12} \approx 0.5404$

About $54\%$ of the quantity survives, so the decrease is $1 - 0.5404 \approx 0.46$, about $46\%$. Multiplying $5\%$ by $12$ to get $60\%$ overstates the loss, because each month's $5\%$ is taken from a smaller amount than the month before.

Write the model in doubling-time form, with the exponent counting doublings.

$A(t) = 4 \cdot 2^{t/3}$

At $t = 10$ the exponent is $\tfrac{10}{3} \approx 3.33$, a little more than three doublings.

$A(10) = 4 \cdot 2^{10/3} \approx 4(10.079) \approx 40$

Three full doublings ($9$ hours) would reach $32$ people, so $10$ hours lands a bit above that.

Doubling means the multiplier reaches $2$, and the coefficient $5000$ cancels before you do anything else.

$(1.045)^t = 2$

Take logarithms and solve for the exponent.

$t = \frac{\log 2}{\log 1.045} \approx \frac{0.30103}{0.019116} \approx 15.75 \text{ years}$

The linear estimate $100 \div 4.5 \approx 22.2$ years is far too long, since it ignores that the growth compounds.

Losing $25\%$ each hour keeps $75\%$, so the hourly factor is $0.75$ and six hours apply it six times.

$(0.75)^{6} \approx 0.178$

About $17.8\%$ of the dose is still present. Subtracting $25\%$ of the original dose six times would remove $150\%$ and predict nothing left, which is impossible: a fixed fraction of a positive amount is still positive, so the drug never fully disappears in this model.

Half the original size means the multiplier has dropped below $0.5$, and the coefficient $2000$ cancels.

$(0.9)^t < 0.5$

Solve the equality to find the crossing time.

$t = \frac{\log 0.5}{\log 0.9} \approx \frac{-0.30103}{-0.045757} \approx 6.58$

The population crosses the halfway mark partway through year $7$, so year $7$ is the first whole year below it. Checking: $(0.9)^6 \approx 0.531$, still above half, while $(0.9)^7 \approx 0.478$.