Exponential Growth and Decay: Practice

Write the model in doubling-time form, where the exponent counts doublings.

$A(t) = P \cdot 2^{t/10}$

Tripling means the multiplier reaches $3$, and the starting amount cancels.

$2^{t/10} = 3 \quad \Longrightarrow \quad \frac{t}{10} = \frac{\log 3}{\log 2} \approx 1.585$

So $t \approx 15.85$ years. Tripling takes more than one and a half doubling times, not one and a half times $10$ scaled by $3/2$.

Group the exponent so the model shows one factor per year. Since $(1.03)^{4t} = \left[(1.03)^4\right]^t$, the annual factor is $(1.03)^4$.

$(1.03)^4 \approx 1.1255$

Subtract $1$ to get the rate: $r \approx 0.1255$, or about $12.55\%$ a year. The quoted $3\%$ is the rate per quarter, and multiplying it by four to get $12\%$ ignores the compounding within the year.

Set the two models equal and collect the exponentials on one side.

$1000(1.05)^t = 1500(1.02)^t \quad \Longrightarrow \quad \left(\frac{1.05}{1.02}\right)^t = 1.5$

The ratio of factors is itself a growth factor, about $1.029412$, so take logarithms.

$t = \frac{\log 1.5}{\log 1.029412} \approx \frac{0.176091}{0.012589} \approx 14.0 \text{ years}$

Town A always catches up eventually: its larger factor beats any head start, given enough time.

The annual factor must produce a doubling after nine years, so undo the ninth power with a rational exponent.

$b^{9} = 2 \quad \Longrightarrow \quad b = 2^{1/9} \approx 1.0801$

Subtracting $1$ gives $r \approx 0.0801$, about $8.01\%$ a year. The linear guess $100 \div 9 \approx 11.1\%$ is far too high, because it forgets that the interest starts earning interest.

Monthly compounding gives a period factor of $1 + \tfrac{0.05}{12}$ and an exponent of $12t$. Divide by the principal to isolate the multiplier.

$\left(1 + \frac{0.05}{12}\right)^{12t} = \frac{5000}{3000} = \frac{5}{3}$

Take logarithms and bring the exponent down.

$12t = \frac{\log \frac{5}{3}}{\log 1.0041667} \approx \frac{0.221849}{0.0018058} \approx 122.85$

Dividing by $12$ gives $t \approx 10.24$ years. Compounding annually instead would take about $10.47$ years, slightly longer.

The decay factor is $b = 0.75$, and the condition is $(0.75)^t < 0.05$. Solve the equality first to locate the crossing.

$t = \frac{\log 0.05}{\log 0.75} \approx \frac{-1.30103}{-0.124939} \approx 10.41$

The value drops below $5\%$ partway through the eleventh year, so the first whole year that satisfies the strict inequality is $t = 11$. Checking confirms it: $(0.75)^{10} \approx 0.0563$, still above $5\%$, while $(0.75)^{11} \approx 0.0422$, comfortably below.

The annual factor $b$ must compound to $2$ over six years, so undo the sixth power with a rational exponent.

$b^{6} = 2 \quad \Longrightarrow \quad b = 2^{1/6} \approx 1.1225$

The rate is $r = b - 1 \approx 0.1225$, about $12.25\%$ a year. Splitting the $100\%$ increase into six equal pieces gives about $16.7\%$, which is too big, since compounding does part of the work for you.

Divide both sides by $500$ so only the exponential remains.

$(0.92)^t = \frac{200}{500} = 0.4$

Take logarithms of both sides and use the power law.

$t = \frac{\log 0.4}{\log 0.92} \approx \frac{-0.39794}{-0.036212} \approx 10.99$

Both logarithms are negative because both numbers are below $1$, so the quotient is positive.

Each month multiplies the balance by $1 + \tfrac{0.12}{12} = 1.01$, and twelve of those multiplications make a year.

$(1.01)^{12} \approx 1.1268$

Subtracting $1$ gives an effective annual rate of about $12.68\%$, more than the $12\%$ advertised. Compounding semiannually would give $(1.06)^2 = 1.1236$, an effective $12.36\%$, so more frequent compounding costs the borrower more.

Divide the later reading by the earlier one. The unknown $P$ cancels, and the exponents subtract.

$\frac{P b^{7}}{P b^{3}} = b^{4} = \frac{640}{250} = 2.56$

Undo the fourth power with a fourth root, taking it in two square-root steps.

$b = 2.56^{1/4} = \sqrt{\sqrt{2.56}} = \sqrt{1.6} \approx 1.2649$

The number $2.56$ is the factor for the whole four-year gap, not for one year.

Divide the readings to find the factor first, since $P$ cancels in the ratio.

$b^{3} = \frac{151.875}{45} = 3.375 \quad \Longrightarrow \quad b = \sqrt[3]{3.375} = 1.5$

Now substitute $b = 1.5$ back into the earlier reading and solve for $P$.

$P(1.5)^{2} = 45 \quad \Longrightarrow \quad P = \frac{45}{2.25} = 20$

Check: $20(1.5)^5 = 20(7.59375) = 151.875$, as required.

Four years of decay leave a multiplier of $0.6$, so the annual factor satisfies a fourth-power equation.

$b^{4} = 0.6 \quad \Longrightarrow \quad b = 0.6^{1/4} \approx 0.8801$

The rate is $r = b - 1 \approx -0.1199$, a decrease of about $12.0\%$ a year. Splitting the $40\%$ loss into four equal $10\%$ pieces is wrong: $(0.9)^4 = 0.6561$, which leaves too much behind.