The Natural Base e: Additional Practice (Set 1)

Rewrite in exponential form, using $\ln u = y$ exactly when $e^{y} = u$.

$2x - 1 = e^{2} \approx 7.389056$

$2x = 8.389056 \;\Longrightarrow\; x = 4.194528 \approx 4.19$

The answer $50.50$ comes from treating $\ln$ as base $10$ and using $10^{2}$, and $3.69$ forgets to add the $1$ back before halving.

Evaluate $\left(1 + \frac{1}{n}\right)^{n}$ at each listed value, remembering that the balance rises with $n$.

$n = 12 \Rightarrow 2.613035, \quad n = 50 \Rightarrow 2.691588, \quad n = 100 \Rightarrow 2.704814, \quad n = 365 \Rightarrow 2.714567$

The first two fall short of $2.7$. Both $n = 100$ and $n = 365$ clear it, but $100$ is the smaller, so it is the answer.

Use the power-of-a-power law to rewrite $e^{3x}$ in terms of $e^{x}$, which you already know.

$e^{3x} = \left(e^{x}\right)^{3} = 3^{3} = 27$

No logarithm is needed at all. Tripling the exponent cubes the value; it does not triple it.

Isolate the exponential before taking any logarithm: divide both sides by $4$ first.

$e^{0.2t} = 5 \;\Longrightarrow\; 0.2t = \ln 5 \approx 1.609438$

$t = \frac{1.609438}{0.2} \approx 8.05$

Taking $\ln$ of both sides while the $4$ was still attached would give $\frac{\ln 20}{0.2} \approx 14.98$, which is wrong: $\ln(4e^{0.2t})$ is not $0.2t$.

Decay means the constant in the exponent is negative, so use $N = N_{0}e^{kt}$ with $k = -0.03$ and $t = 8$, giving $kt = -0.24$.

$N = 12{,}000e^{-0.24} \approx 12{,}000(0.786628) \approx 9439.5$

That rounds to $9440$ people. Dropping the minus sign gives $15{,}255$, a growing town, and $9120$ comes from subtracting $24\%$ all at once with no compounding.

After $30$ years half the sample remains, so substitute $t = 30$ and $N = \frac{1}{2}N_{0}$.

$e^{30k} = \frac{1}{2} \;\Longrightarrow\; 30k = \ln\left(\frac{1}{2}\right) = -\ln 2 \approx -0.693147$

$k = \frac{-0.693147}{30} \approx -0.0231 \text{ per year}$

The sign matters: $k$ must be negative for decay. The positive version $\frac{\ln 2}{30}$ would describe a substance that doubles every $30$ years.

Divide out the principal to isolate the exponential, then take the natural logarithm.

$e^{4r} = \frac{700}{500} = 1.4 \;\Longrightarrow\; 4r = \ln 1.4 \approx 0.336472$

$r = \frac{0.336472}{4} \approx 0.084118 = 8.4\%$

The answer $10.0\%$ divides the total $40\%$ gain by $4$ years, which ignores compounding altogether. The answer $3.7\%$ comes from reaching for the base-$10$ key, since $\frac{\log 1.4}{4} \approx 0.0365$, but the base of the exponential here is $e$.

Evaluate all four with $e \approx 2.718282$.

$e^{2} \approx 7.389 \qquad 2^{e} \approx 6.581 \qquad 2e \approx 5.437 \qquad e + 2 \approx 4.718$

So $e^{2}$ is the largest. The comparison of $e^{2}$ with $2^{e}$ is the interesting one: swapping a base and an exponent changes the value, and here the larger base in the exponent position loses.

Factor $12$ into powers of the numbers you know, then split the logarithm with the product and power rules.

$\ln 12 = \ln\left(2^{2} \cdot 3\right) = 2\ln 2 + \ln 3$

$\approx 2(0.693147) + 1.098612 = 1.386294 + 1.098612 = 2.484906$

So $\ln 12 \approx 2.4849$. The distractor $1.7918$ is $\ln 6 = \ln 2 + \ln 3$, which forgets that $12$ carries two factors of $2$, and $0.7615$ is the product $(\ln 2)(\ln 3)$, which no logarithm rule ever produces.

Growing by $40\%$ means the final amount is $1.4$ times the original, so the growth factor over $5$ years is $1.4$.

$e^{5r} = 1.4 \;\Longrightarrow\; 5r = \ln 1.4 \approx 0.336472$

$r = \frac{0.336472}{5} \approx 0.067294 = 6.7\%$

The near-miss $7.0\%$ is the equivalent rate compounded once a year, since $1.4^{1/5} - 1 \approx 0.0696$. It answers a different question: the problem specified continuous compounding, so the continuous rate is the one to report.

Combine the left side with the product rule, then equate the inputs.

$\ln\left(x(x - 3)\right) = \ln 4 \;\Longrightarrow\; x^{2} - 3x - 4 = 0$

$(x - 4)(x + 1) = 0 \;\Longrightarrow\; x = 4 \text{ or } x = -1$

Now check the domain. Both $\ln x$ and $\ln(x - 3)$ need positive inputs, so $x$ must exceed $3$. That rejects $x = -1$ and leaves $x = 4$, which checks out: $\ln 4 + \ln 1 = \ln 4$.

For a positive exponent, a bigger base gives a bigger power, and $e$ sits between $2$ and $3$.

$2 < e = 2.718281828\ldots < 3 \;\Longrightarrow\; 2^{x} < e^{x} < 3^{x} \quad (x > 0)$

Check at $x = 2$: $4 < 7.389 < 9$. This is the graphical meaning of $e$ lying between $2$ and $3$, that the curve $y = e^{x}$ is sandwiched between $y = 2^{x}$ and $y = 3^{x}$ to the right of the $y$-axis.