Exponential Growth and Decay: Additional Practice (Set 2)

Each quarter multiplies by $1 + \tfrac{0.06}{4} = 1.015$, and a year contains four quarters. The effective annual factor is what those four multiplications amount to.

$(1.015)^4 \approx 1.0614$

So $6\%$ compounded quarterly behaves like $6.14\%$ compounded once a year. Compounding semiannually instead would give $(1.03)^2 = 1.0609$, slightly less.

Each quarter multiplies by $1 + \tfrac{0.08}{4} = 1.02$, and a year holds four quarters.

$(1.02)^4 \approx 1.0824$

Subtract $1$ to read off the effective rate: about $8.24\%$. Compounding semiannually would give $(1.04)^2 = 1.0816$, an effective $8.16\%$, so the more often you compound, the more you earn.

Let $b$ be the hourly decay factor. The $12$-hour reading gives $b^{12} = 0.3$, so $\log b = \tfrac{1}{12}\log 0.3$.

The half-life $h$ satisfies $b^{h} = 0.5$, so $h \log b = \log 0.5$. Substituting the expression for $\log b$ removes $b$ entirely.

$h = \frac{\log 0.5}{\tfrac{1}{12}\log 0.3} = 12 \cdot \frac{\log 0.5}{\log 0.3} \approx 12 \cdot \frac{-0.30103}{-0.52288} \approx 6.91 \text{ hours}$

The answer must be under $12$ hours, since more than half the sample is already gone by then.

The annual factor applied twice must give the two-year factor.

$b^{2} = 1.44 \quad \Longrightarrow \quad b = \sqrt{1.44} = 1.2$

Subtract $1$ to get the rate: $r = 0.2$, a $20\%$ increase each year. Halving the $44\%$ to get $22\%$ ignores compounding, and indeed $(1.22)^2 = 1.4884$, which overshoots.

Multiply the two factors rather than adding the rates.

$1.30 \times 0.70 = 0.91$

The stock is worth $91\%$ of its starting value, a $9\%$ loss. This is the identity $(1+r)(1-r) = 1 - r^2$ with $r = 0.3$, and $1 - 0.09 = 0.91$. A matched rise and fall always loses, because the fall is applied to a bigger number than the rise was.

The three-year factor is $1 + 0.331 = 1.331$, and the annual factor applied three times must produce it.

$b^{3} = 1.331 \quad \Longrightarrow \quad b = \sqrt[3]{1.331} = 1.1$

So the quantity grows $10\%$ a year, and $(1.1)^3 = 1.331$ confirms it. Splitting the $33.1\%$ into three equal parts gives $1.1103$, which is wrong because compounding contributes some of the growth for free.

Each year contributes a factor, and the factors multiply in sequence.

$1.10 \times 0.90 \times 1.10 = 0.99 \times 1.10 = 1.089$

The net result is an $8.9\%$ gain, not the $10\%$ that a naive "the gain and loss cancel" argument suggests. The first gain and loss already leave $0.99$, a $1\%$ hole that the third year has to climb out of.

Use the half-life form of the model and divide out the starting amount.

$\left(\tfrac{1}{2}\right)^{t/30} = 0.20$

Take logarithms of both sides and use the power law.

$\frac{t}{30} = \frac{\log 0.2}{\log 0.5} \approx \frac{-0.69897}{-0.30103} \approx 2.32$

Multiplying by $30$ gives $t \approx 69.7$ years. That is sensible: two half-lives ($60$ years) leave $25\%$, and three ($90$ years) leave $12.5\%$, so $20\%$ falls between them.

Each half-life multiplies by $\tfrac{1}{2}$, so $2.5$ of them multiply by $\left(\tfrac{1}{2}\right)^{2.5}$, which rational exponents give a meaning to.

$\left(\tfrac{1}{2}\right)^{2.5} = \frac{1}{2^{5/2}} = \frac{1}{4\sqrt{2}} \approx 0.1768$

So about $17.7\%$ remains. It has to land between the $25\%$ left after two half-lives and the $12.5\%$ left after three.

Evaluate both after one year. Annual compounding is a single multiplication.

$1000(1.05) = 1050.00$

Monthly compounding applies one twelfth of the rate twelve times.

$1000\left(1 + \frac{0.049}{12}\right)^{12} \approx 1000(1.050115) \approx 1050.12$

The monthly account wins by about $0.12$ dollars, or $12$ cents. Compounding more often is worth a little extra, just enough here to overcome the lower quoted rate.

Evaluate each model at $t = 10$.

$1000(1.06)^{10} \approx 1790.85, \qquad 900(1.07)^{10} \approx 1770.44$

The difference is about $20.41$ dollars in favour of the larger principal. Ten years is not yet long enough for the extra $1\%$ to overcome a $100$-dollar head start, though eventually it would: the higher factor always wins in the long run.

Let $P$ be the deposit. Twelve years of growth must turn it into $10000$ dollars.

$P(1.06)^{12} = 10000$

Solve for $P$ by dividing by the twelve-year factor.

$P = \frac{10000}{(1.06)^{12}} \approx \frac{10000}{2.012196} \approx 4969.69$

Roughly half the target is enough, because the money doubles in about twelve years at this rate. Multiplying rather than dividing gives $20122$ dollars, which runs the clock the wrong way.