The Natural Base e: Free Response

Write the base as a single decimal first: $1 + \frac{1}{50} = 1.02$. Raise it to the $50$th power.

$(1.02)^{50} \approx 2.6916$

Compare that to $e \approx 2.718281828$: $2.6916$ is smaller, consistent with $n=50$ being a finite value of $n$, not the unreachable limit.

Two facts govern every term of this sequence regardless of how large $n$ gets.

$\left(1+\frac{1}{50}\right)^{50} < \left(1+\frac{1}{10{,}000}\right)^{10{,}000} < 3$

First, the values increase with $n$, so trading $n=50$ for $n=10{,}000$ can only raise the balance, never lower it. Second, the values never reach $3$, whatever $n$ is. Neither fact requires recomputing the expression.

The lesson asserts the borrowed theorem as a plain fact: a column of numbers that keeps rising but never passes a fixed ceiling has to settle on one single value.

$2 < \left(1+\tfrac{1}{n}\right)^{n} < 3 \text{ for every } n$

Checking that this particular column satisfies both conditions, always rising and always under $3$, only confirms that the theorem's hypotheses hold here. It does not reprove the theorem itself, which is a general statement about every column of real numbers with those two properties, not only this one built from compounding. That general statement is proved using a rigorous treatment of limits, the machinery of a calculus course, which is exactly why the lesson calls it a loan rather than deriving it from the algebra already in hand.

The first expression starts with $e$ raised to a power and applies $\ln$ to it, which is exactly the identity $\ln\left(e^{x}\right)=x$, valid for every real $x$:

$\ln\left(e^{7}\right) = 7.$

The second expression starts with $\ln$ applied to a positive number and then raises $e$ to that result, which is the other identity, $e^{\ln x}=x$, valid for $x>0$:

$e^{\ln 9} = 9.$

Rearrange $20 = 4 \cdot 5$ into a quotient instead: $5 = \frac{20}{4}$. The quotient law turns the logarithm of that quotient into a difference of the two given logarithms:

$\ln 5 = \ln\!\left(\frac{20}{4}\right) = \ln 20 - \ln 4 \approx 2.9957 - 1.3863 = 1.6094.$

Each of the three logarithm laws was proved for an arbitrary allowed base $b$, using only the definition of a logarithm and the laws of exponents, neither of which cares what particular number the base happens to be.

$\log_b(MN)=\log_b M+\log_b N \;\xrightarrow{\,b=e\,}\; \ln(MN)=\ln M+\ln N$

The natural logarithm is not a different kind of object built from scratch; it is exactly $\log_e$, the same logarithm with the base pinned at the specific number $e$. Since the proofs never singled out a particular base, substituting $b=e$ into an already-proved statement is all that is needed, so every law transfers with no new derivation.

Monthly compounding means $n=12$, so the formula uses a monthly rate of $\frac{0.07}{12}$ over $12 \cdot 6 = 72$ periods:

$A = 4800\left(1+\frac{0.07}{12}\right)^{72} \approx 4800(1.520105) \approx 7296.51.$

The deposit grows to about $7296.51$ dollars.

Continuous compounding uses $A=Pe^{rt}$ with the same $P=4800$, $r=0.07$, $t=6$, so $rt = 0.42$:

$A = 4800e^{0.42} \approx 4800(1.521962) \approx 7305.42.$

The claim conflates "compounding very often" with "compounding continuously," but those name two different kinds of object. Daily compounding is a specific, finite schedule: it plugs $n=365$ into the very same formula used in part A, and it produces a specific finite balance, a little more than the monthly figure from part A but still strictly below the continuous figure from part B.

$A_{\text{daily}} < A_{\text{continuous}} = 4800e^{0.42}$

Continuous compounding is not one large value of $n$ among many; the model $A=Pe^{rt}$ has no $n$ in it at all, because it is defined as the number every finite schedule approaches while none of them, including daily, ever reaches it. So "daily" names one point on the increasing list of finite schedules, while "continuous" names the ceiling those points climb toward.

Substitute $C_{0}=210$, $k=-0.17$, and $t=6$ directly into the model:

$C = 210e^{-0.17 \cdot 6} = 210e^{-1.02} \approx 210(0.360595) \approx 75.72.$

About $75.72$ ppm remain after $6$ hours.

Set $C=45$ and isolate the exponential first, dividing by the initial concentration:

$210e^{-0.17t} = 45 \quad \Longrightarrow \quad e^{-0.17t} = \frac{45}{210} \approx 0.214286.$

Take $\ln$ of both sides, which strips the base $e$ away immediately:

$-0.17t = \ln(0.214286) \approx -1.540445 \quad \Longrightarrow \quad t \approx \frac{1.540445}{0.17} \approx 9.06.$

Solve the general equation the same way part B did, without committing to specific numbers yet. Divide by $C_{0}$ and take $\ln$:

$C_{0}e^{kt} = C \quad \Longrightarrow \quad e^{kt} = \frac{C}{C_{0}} \quad \Longrightarrow \quad kt = \ln\!\left(\frac{C}{C_{0}}\right) \quad \Longrightarrow \quad t = \frac{\ln(C/C_{0})}{k}.$

The numerator $\ln(C/C_{0})$ depends only on the starting concentration and the target, neither of which changed. The fading constant $k$ sits alone in the denominator, so replacing $k=-0.17$ with $k=-0.34$ (a magnitude twice as large) divides the same numerator by a denominator with a magnitude twice as large, which halves the quotient. That relies on one condition: $C_0>0$ and $0<C<C_0$, so that $C/C_0$ is strictly between $0$ and $1$, making $\ln(C/C_0)$ a fixed negative number rather than zero or undefined. With that condition in place, doubling the magnitude of $k$ always halves the time to reach that target, for any initial concentration and any smaller positive target, not only the numbers in this problem.

At every fixed time $t$, the amount fact says the continuously compounded balance is at least as large as the monthly compounded balance, and in fact strictly larger for any finite schedule.

$A_Y(t) \ge A_X(t) \text{ for every } t \quad\Longrightarrow\quad t_Y \le t_X$

So at the moment account X first reaches the target, account Y has already grown at least as much, meaning it reached that same target no later. A balance that is always ahead reaches any fixed target no later than one that is always behind, so account Y gets there first.

The principal cancels, since only the growth factor matters here: set $e^{rt}=2.6$ with $r=0.038$.

$e^{0.038t} = 2.6 \quad \Longrightarrow \quad 0.038t = \ln(2.6) \approx 0.955511 \quad \Longrightarrow \quad t \approx \frac{0.955511}{0.038} \approx 25.15.$

The continuous growth model buries the elapsed time inside an exponent, so a statement about how much the quantity has grown is really a statement about the value of that exponential.

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

The natural logarithm exists precisely as the function that undoes $e$ raised to a power, so applying it to both sides pulls the exponent $rt$ down out of hiding. That is exactly why the amount fact from part A could be read backward: the same inverse relationship that lets you compute a growth ratio from a given time also lets you compute the time from a given growth ratio.