The Natural Base e: Additional Practice (Set 2)

The two models agree for every $t$ exactly when their one-year growth factors agree. Monthly compounding at $6\%$ multiplies by $\left(1 + \frac{0.06}{12}\right)^{12} = (1.005)^{12}$ each year, and the continuous model multiplies by $e^{r}$.

$e^{r} = (1.005)^{12} \;\Longrightarrow\; r = \ln\left((1.005)^{12}\right) = 12\ln(1.005)$

$r \approx 12(0.0049875) \approx 0.059850 = 5.985\%$

The continuous rate is slightly below $6\%$, which is what we expect: continuous compounding is more efficient, so it needs a smaller rate to match. The value $\ln(1.06)$ would match yearly compounding, not monthly.

Write the radical as a rational exponent, then use the power rule of logarithms.

$\ln\left(\sqrt{x}\right) = \ln\left(x^{1/2}\right) = \frac{1}{2}\ln x$

The exponent $\frac{1}{2}$ comes out front as a coefficient. It does not put a square root on the logarithm, which is a different operation.

Compute each balance separately. Bank A uses the discrete formula with $n = 4$ and $r = 0.07$.

$A = 10{,}000\left(1 + \frac{0.07}{4}\right)^{20} = 10{,}000(1.0175)^{20} \approx 14{,}147.78$

Bank B uses the continuous model with $rt = (0.069)(5) = 0.345$.

$A = 10{,}000e^{0.345} \approx 14{,}119.90$

Bank A ends up ahead by about $27.88$ dollars. Continuous compounding is the more efficient schedule, but it cannot make up for a rate that is a tenth of a percentage point lower. Upgrading bank A from quarterly to continuous compounding at the same $7\%$ would add only about $43$ dollars, while the missing tenth of a percentage point is worth about $71$ dollars, so the rate matters more here than the schedule does.

Simplify the numerator and denominator separately with $\ln\left(e^{x}\right) = x$, and only then divide.

$\frac{\ln\left(e^{6}\right)}{\ln\left(e^{2}\right)} = \frac{6}{2} = 3$

Do not cancel the logarithms as if they were factors. A quotient of logarithms is a change of base, and here it is just the plain number $3$.

The balance $\left(1 + \frac{1}{n}\right)^{n}$ increases with $n$, so the finest schedule listed wins.

$n = 1 \Rightarrow 2.000000, \quad n = 4 \Rightarrow 2.441406, \quad n = 12 \Rightarrow 2.613035, \quad n = 31{,}536{,}000 \Rightarrow 2.718282$

Compounding every second pays the most, though the whole point of this lesson is how little "the most" is worth here: it beats monthly compounding by only about ten cents on the dollar, and it can never reach $e$ dollars.

Use $A = Pe^{rt}$ with $P = 5000$, $r = 0.09$ and $t = 3$, so the exponent is $rt = 0.27$.

$A = 5000e^{0.27} \approx 5000(1.309964) = 6549.82$

Each distractor is a specific error: $5470.87$ leaves out the $3$ years, $6475.15$ compounds once a year instead of continuously, and $6350.00$ applies $27\%$ as simple interest.

Compute both balances. The continuous account uses $rt = (0.05)(20) = 1$, so its factor is just $e$.

$1000e^{1} \approx 2718.28 \qquad 1000(1.05)^{20} \approx 2653.30$

The continuous account leads by about $64.98$ dollars, roughly $65$. Continuous compounding always pays at least as much as any discrete schedule, and over twenty years even a small per-year edge compounds into a visible gap.

Bring everything to one side and factor, treating $e^{x}$ as the common factor.

$e^{2x} - 4e^{x} = 0 \;\Longrightarrow\; e^{x}\left(e^{x} - 4\right) = 0$

A product is zero when a factor is zero, but $e^{x}$ is positive for every real $x$, so $e^{x} = 0$ has no solution and that factor can be discarded.

$e^{x} = 4 \;\Longrightarrow\; x = \ln 4 \approx 1.386$

The extra root $x = 0$ is exactly the mistake of forgetting that an exponential is never zero.

The year-end balance on 1 dollar at $100\%$ compounded $n$ times is exactly $\left(1 + \frac{1}{n}\right)^{n}$, and that expression is trapped below $3$ for every $n$.

$2 \le \left(1 + \frac{1}{n}\right)^{n} < e = 2.718281828\ldots < 3 < 4$

No compounding schedule, however fine, can reach even $e$ dollars, let alone $4$. Compounding more often does pay more, so it certainly beats simple interest, but the gains shrink toward nothing.

Compute both balances. Monthly compounding uses $n = 12$, so the monthly rate is $\frac{0.12}{12} = 0.01$.

$1000(1.01)^{12} \approx 1126.83 \qquad 1000e^{0.12} \approx 1127.50$

The difference is about $0.67$ dollars, sixty-seven cents. Continuous compounding is the ceiling and always pays a little more, but even at a hefty $12\%$ the edge over monthly compounding is under a tenth of a percent.

The fraction remaining is whatever multiplies $N_{0}$, so the starting value of $80$ never matters. Put $t = 10$ into the exponential.

$e^{(-0.15)(10)} = e^{-1.5} \approx 0.223130$

So about $22\%$ of the original amount is left, and roughly $78\%$ has decayed away. Reading the exponent as the answer would suggest $15\%$, but $-0.15$ is the rate per year, not the fraction remaining after ten of them.

The name records where the formula comes from. Substituting $m = \frac{n}{r}$ into the discrete formula rewrites it as

$A = P\left[\left(1 + \frac{1}{m}\right)^{m}\right]^{rt}$

and as the compounding gets finer, $m$ grows and the bracket closes in on $e$, leaving $A = Pe^{rt}$. The rate $r$ stays fixed throughout, and the irrationality of $e$, though true, has nothing to do with the name.