The Natural Base e: Practice

Use $A = Pe^{rt}$ with $P = 1500$, $r = 0.04$ and $t = 5$, so the exponent is $rt = 0.20$.

$A = 1500e^{0.20} \approx 1500(1.221403) = 1832.10$

The distractors are the usual slips: $1824.98$ compounds once a year, $1800.00$ uses simple interest, and $1561.22$ forgets to multiply the rate by the $5$ years.

The unknown is stuck in the exponent, so take the natural logarithm of both sides and use $\ln\left(e^{x}\right) = x$.

$0.03t = \ln 4 \approx 1.386294$

$t = \frac{1.386294}{0.03} \approx 46.21$

Using the base-$10$ key by mistake gives $\frac{\log 4}{0.03} \approx 20.07$, which is wrong because the base of the exponential is $e$, not $10$.

The quotient rule holds for $\ln$ exactly as for any logarithm, since $\ln$ is $\log_{e}$.

$\ln 5 - \ln 2 = \ln\left(\frac{5}{2}\right)$

A difference of logarithms becomes the logarithm of a quotient. It is not $\ln 3$: subtracting the logarithms divides the inputs, it does not subtract them.

The power rule of logarithms brings the exponent out in front as a multiplier.

$\ln\left(x^{3}\right) = 3\ln x$

The exponent becomes a coefficient of the logarithm, not an exponent on the logarithm, so $(\ln x)^{3}$ is a different quantity entirely.

Compute both growth factors. Monthly compounding uses $n = 12$, and continuous compounding uses $e^{rt}$.

$\left(1 + \frac{0.10}{12}\right)^{12} \approx 1.104713 \qquad e^{0.10} \approx 1.105171$

Continuous compounding wins, as it must: it is the ceiling that every finite compounding schedule falls short of. The gap is tiny, less than a twentieth of a cent on the dollar, and the principal does not affect which is larger, since both are just multipliers.

Start from $A = Pe^{rt}$ and solve for $P$ by dividing, with $rt = (0.045)(6) = 0.27$.

$P = \frac{A}{e^{rt}} = \frac{1047.97}{e^{0.27}} \approx \frac{1047.97}{1.309964} \approx 800.00$

As a check, run it forward: $800e^{0.27} \approx 1047.97$ dollars.

This is the change-of-base rule with the natural logarithm as the common base.

$\log_{b} x = \frac{\ln x}{\ln b} \;\Longrightarrow\; \frac{\ln 20}{\ln 4} = \log_{4} 20$

Dividing two logarithms is not the same as subtracting them, so the answer is not $\ln 5$: that would be $\ln 20 - \ln 4$.

Rewrite the logarithmic equation in exponential form, using $\ln x = y$ exactly when $e^{y} = x$.

$\ln x = 3 \;\Longrightarrow\; x = e^{3} \approx 20.085537$

The answer $1000$ comes from treating $\ln$ as the base-$10$ logarithm, which would give $10^{3}$. The base here is $e$.

Set $A = 8000$ and $P = 5000$, then isolate the exponential before taking the logarithm.

$5000e^{0.03t} = 8000 \;\Longrightarrow\; e^{0.03t} = 1.6$

$0.03t = \ln 1.6 \approx 0.470004 \;\Longrightarrow\; t = \frac{0.470004}{0.03} \approx 15.67 \text{ years}$

Using $\log 1.6$ instead of $\ln 1.6$ produces the wrong value $6.80$.

Half-life means the amount left is $\frac{1}{2}N_{0}$, so divide by $N_{0}$ and take the natural logarithm.

$e^{-0.05t} = 0.5 \;\Longrightarrow\; -0.05t = \ln 0.5 \approx -0.693147$

$t = \frac{-0.693147}{-0.05} \approx 13.86 \text{ years}$

Both negatives cancel, which they must, since a half-life is a positive amount of time.

Any positive base can be written as a power of $e$, since $e^{\ln a} = a$. Start with $2 = e^{\ln 2}$.

$2^{x} = \left(e^{\ln 2}\right)^{x} = e^{x\ln 2}$

The power-of-a-power law multiplies the exponents. This is why every exponential function can be written with base $e$, which is what makes $e$ the standard base for growth models.

The balance climbs with $n$ but is squeezed against the ceiling $e$, and by the per-second row it already agrees with $e$ to six decimal places.

$e = 2.718281828\ldots$

A thousand times more compoundings changes nothing you could see on a dollar: the answer is still $2.718282$ dollars to six places. This is the whole moral of the lesson, that compounding faster does not pay unboundedly more.