The Natural Base e: Practice

Combine the left side with the product rule, then compare the inputs, since a logarithm is one-to-one.

$\ln\left(x(x + 3)\right) = \ln 10 \;\Longrightarrow\; x^{2} + 3x = 10$

$x^{2} + 3x - 10 = 0 \;\Longrightarrow\; (x + 5)(x - 2) = 0$

The candidates are $x = 2$ and $x = -5$, but $\ln x$ requires $x > 0$, so $x = -5$ is not in the domain and is rejected. Checking the survivor: $\ln 5 + \ln 2 = \ln 10$, which is true.

Isolate the exponential first by dividing both sides by $P$, and only then apply the natural logarithm.

$e^{rt} = \frac{A}{P} \;\Longrightarrow\; rt = \ln\left(\frac{A}{P}\right)$

$t = \frac{1}{r}\ln\left(\frac{A}{P}\right)$

The logarithm can only be taken once the exponential stands alone, which is why $\ln(A - P)$ never appears: nothing was ever subtracted.

Set the final amount to twice the principal and divide it out, so the answer cannot depend on how much was invested.

$Pe^{rt} = 2P \;\Longrightarrow\; e^{rt} = 2 \;\Longrightarrow\; rt = \ln 2$

$t = \frac{\ln 2}{r}$

Since $\ln 2 \approx 0.693$, this is close to the familiar rule of thumb that divides $0.72$ by the rate, but $\frac{\ln 2}{r}$ is the exact answer.

Notice that $e^{2x} = \left(e^{x}\right)^{2}$, so the equation is a quadratic in the quantity $u = e^{x}$.

$u^{2} - 7u + 12 = 0 \;\Longrightarrow\; (u - 3)(u - 4) = 0$

So $e^{x} = 3$ or $e^{x} = 4$. Both are positive, so both are attainable, and taking the natural logarithm of each gives

$x = \ln 3 \approx 1.099 \qquad x = \ln 4 \approx 1.386$

The answers $x = 3$ and $x = 4$ skip the last step: those are the values of $e^{x}$, not of $x$.

Compute both balances to the cent. Daily compounding uses $n = 365$, so there are $3650$ compoundings.

$1000\left(1 + \frac{0.06}{365}\right)^{3650} \approx 1822.03 \qquad 1000e^{0.6} \approx 1822.12$

The difference is about nine cents across a full decade. Continuous compounding is genuinely larger, since it is the ceiling no finite schedule reaches, but on realistic rates the gap over daily compounding is negligible.

Two facts govern the expression: the values increase with $n$, and they never reach $3$. Together they force the values to close in on a single number, which is $e$.

$2 = \left(1 + \tfrac{1}{1}\right)^{1} < \cdots < \left(1 + \tfrac{1}{365}\right)^{365} \approx 2.714567 < e < 3$

No finite $n$ reaches $e$ exactly, so it does not equal $e$ at $n = 1000$. And it does not shrink toward $1$: the base does shrink toward $1$, but the exponent grows at the same time, and the growth of the exponent wins.

Tripling gives $A = 3P$, so divide out $P$ and take the natural logarithm with $t = 12$.

$e^{12r} = 3 \;\Longrightarrow\; 12r = \ln 3 \approx 1.098612$

$r = \frac{1.098612}{12} \approx 0.091551 = 9.2\%$

The value $5.8\%$ is what you get by using $\ln 2$, which answers the doubling question instead of the tripling one.

Take the natural logarithm of both sides. On the left, $\ln\left(e^{x}\right) = x$; on the right, the power rule brings the exponent down.

$x = (x - 1)\ln 5$

Expand and gather the $x$ terms on one side.

$x = x\ln 5 - \ln 5 \;\Longrightarrow\; x(1 - \ln 5) = -\ln 5$

$x = \frac{-\ln 5}{1 - \ln 5} = \frac{\ln 5}{\ln 5 - 1} \approx \frac{1.609438}{0.609438} \approx 2.64$

Since $\ln 5 > 1$, the denominator $\ln 5 - 1$ is positive and the answer is positive, which is a useful sign check.

Apply $\ln\left(e^{x}\right) = x$ to each term separately, then subtract.

$\ln\left(e^{a + b}\right) - \ln\left(e^{a}\right) = (a + b) - a = b$

The quotient rule gives the same thing in one step: the difference is $\ln\left(\frac{e^{a + b}}{e^{a}}\right) = \ln\left(e^{b}\right) = b$.

Doubling in $5$ hours means that putting $t = 5$ into the model multiplies $N_{0}$ by $2$.

$N_{0}e^{5k} = 2N_{0} \;\Longrightarrow\; e^{5k} = 2 \;\Longrightarrow\; 5k = \ln 2$

$k = \frac{\ln 2}{5} \approx \frac{0.693147}{5} \approx 0.1386 \text{ per hour}$

The constant $k$ is positive because the colony is growing. The reciprocal $\frac{5}{\ln 2}$ inverts the fraction and has the wrong units entirely.

Pull the coefficient back inside the logarithm with the power rule, then use the undoing rule $e^{\ln u} = u$.

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

A quick sanity check with $x = 2$: $3\ln 2 \approx 2.0794$, and $e^{2.0794} \approx 8 = 2^{3}$.

There is no rule that breaks up the logarithm of a sum, and multiplying logarithms is never the result of any property. Test it with $x = y = 1$.

$\ln(1 + 1) = \ln 2 \approx 0.693 \qquad (\ln 1)(\ln 1) = 0 \cdot 0 = 0$

Since $0.693 \neq 0$, the claim is false. The other three are all true: the domain of $\ln$ is the positive numbers, $\ln e = 1$ because $e^{1} = e$, and the product rule holds for every base, $e$ included.