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The Natural Base e: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two compounding counts

Let $C_n=(1+\frac1n)^n$ . Evaluate $C_6-C_3$ to the nearest thousandth.
1. Hint 1
  Keep the fraction inside the power exact until the final rounding.
2. Hint 2
  Use $(4/3)^3$ and $(7/6)^6$ for the two computations.
Answer

$C_6-C_3\approx0.151$ .

Full solution

The two exact expressions are $C_3=64/27$ and
$C_6=117649/46656.$
Evaluating gives
$C_3\approx2.370$
and
$C_6\approx2.522$
to the nearest thousandth.
Subtracting the exact fractions before rounding gives $C_6-C_3\approx0.151$ .
The positive difference also confirms that six compoundings gives the larger value.

Answer

$C_6-C_3\approx0.151$ .

Key idea

Finer compounding increases the values that approach e.
Problem 1: Mark complete
Problem 2 A natural logarithm domain

Let $g(x)=\ln(e^x-1)$ . Find the real domain of $g$ , and solve $g(x)=0$ .
1. Hint 1
  A real logarithm needs a positive argument.
2. Hint 2
  Write $1$ as a power of $e$ , and use the fact that $e^x$ increases.
Answer

Domain $x>0$ ; $g(x)=0$ at $x=\ln2$ .

Full solution

The argument must be positive: $e^x-1>0$ , that is, $e^x>1=e^0$ .
Since $e^x$ is increasing, this holds exactly when $x>0$ .

Next, $g(x)=0$ means $e^x-1=e^0=1$ , so $e^x=2$ and $x=\ln2$ .
Since $\ln2>0$ , this value lies in the domain.

Answer

Domain $x>0$ ; $g(x)=0$ at $x=\ln2$ .

Key idea

The domain of a natural logarithm comes from requiring its whole argument to be positive.
Problem 2: Mark complete
Problem 3 A continuous model

A quantity follows $A(t)=50e^{rt}$ with $t$ in hours. Find $r$ if $A(2)=50e^{-3}$ .
1. Hint 1
  The positive common coefficient can be canceled.
2. Hint 2
  Equal powers of $e$ have equal exponents.
Answer

$r=-\frac32$ per hour.

Full solution

Divide by the nonzero coefficient $50$ to get
$e^{2r}=e^{-3}.$
One-to-oneness gives
$2r=-3$
so $r=-\frac32$ per hour.
Substitution returns $50e^{-3}$ at two hours.

Answer

$r=-\frac32$ per hour.

Key idea

A continuous model's exponent combines its rate with time in the matching unit.
Problem 3: Mark complete
Problem 4 A change of time units

A signal starts at $60$ units and decays continuously at rate $0.04$ per minute. Write its amount $A(h)$ using hours $h\ge0$ and find the amount after half an hour to the nearest tenth of a unit.
1. Hint 1
  The exponent must use the same time unit as the rate.
2. Hint 2
  An hour contains $60$ minutes, and half an hour contains $30$ .
Answer

$A(h)=60e^{-2.4h}$ ; $A(0.5)\approx18.1$ units.

Full solution

At $h$ hours, the elapsed time is $60h$ minutes.
Thus
$A(h)=60e^{-0.04(60h)}$
or
$A(h)=60e^{-2.4h}.$

At half an hour the exponent is $-1.2$ , giving $A(0.5)=60e^{-1.2}\approx18.1$ units.
The negative exponent correctly gives less than the initial $60$ units.

Answer

$A(h)=60e^{-2.4h}$ ; $A(0.5)\approx18.1$ units.

Key idea

Converting a rate to a new time unit leaves the product of rate and elapsed time unchanged.
Problem 4: Mark complete
Problem 5 Two operating stages

A quantity starts at $80$ units. It grows continuously at rate $0.1$ per hour for $2$ hours, then at rate $0.2$ per hour for another $3$ hours. Find its final amount exactly and to the nearest tenth of a unit.
1. Hint 1
  The second stage acts on the amount produced by the first stage.
2. Hint 2
  Multiply the two exponential factors and add their exponents.
Answer

$80e^{0.8}\approx178.0$ units.

Full solution

The first stage multiplies by $e^{0.2}$ and the second by $e^{0.6}$ .
Their combined factor is
$e^{0.2}e^{0.6}=e^{0.8}$

The final amount is $80e^{0.8}\approx178.0$ units.
Using an averaged rate without accounting for the different durations would not preserve this exponent.

Answer

$80e^{0.8}\approx178.0$ units.

Key idea

Successive continuous growth stages combine through the sum of their rate-times-duration exponents.
Problem 5: Mark complete
Problem 6 A target window

A quantity follows $N(t)=40e^{0.12t}$ for $t\ge0$ , in hours. Find all times when $60\le N(t)\le100$ , giving exact bounds and bounds rounded to the nearest hundredth of an hour.
1. Hint 1
  Divide the entire inequality by the positive starting amount.
2. Hint 2
  The natural logarithm is increasing, so applying it preserves the order.
Answer

$\frac{\ln(3/2)}{0.12}\le t\le\frac{\ln(5/2)}{0.12}$ hours; endpoint approximations $3.38$ and $7.64$ hours.

Full solution

Dividing by $40$ gives
$1.5\le e^{0.12t}\le2.5.$
Take natural logarithms, then divide by positive $0.12$ .
$\frac{\ln(3/2)}{0.12}\le t$
and
$t\le\frac{\ln(5/2)}{0.12}$

The lower endpoint is approximately $3.38$ hours and the upper approximately $7.64$ hours.
The exact bounds give amounts $60$ and $100$ , respectively; every time between them gives an amount in the required window.

Answer

$\frac{\ln(3/2)}{0.12}\le t\le\frac{\ln(5/2)}{0.12}$ hours; endpoint approximations $3.38$ and $7.64$ hours.

Key idea

Taking natural logarithms can convert a continuous-growth amount interval into a time interval.
Problem 6: Mark complete
Problem 7 Two deposits

An account earns interest compounded continuously at $4.7\%$ per year. It receives $2000$ dollars at $t=0$ and another $3000$ dollars at $t=5$ , with $t$ in years, and nothing is withdrawn. Find, exactly and to the nearest hundredth of a year, when the balance first reaches $10000$ dollars.
1. Hint 1
  First decide whether the target is reached before or after the second deposit.
2. Hint 2
  After $t=5$ , both deposits grow by the same factor $e^{0.047t}$ up to a constant, so factor it out.
Answer

$t=\frac{1000}{47}\ln\frac{10}{2+3e^{-0.235}}\approx17.60$ years.

Full solution

Before the second deposit the balance is $2000e^{0.047t}$ , which at $t=5$ is $2000e^{0.235}\approx2529.82$ dollars, short of the target.
So the target is reached after $t=5$ .

For $t\ge5$ the balance $2000e^{0.047t}+3000e^{0.047(t-5)}$ counts both deposits.
Since $e^{0.047(t-5)}=e^{0.047t}e^{-0.235}$ , factoring out $e^{0.047t}$ and dividing by $1000$ gives
$e^{0.047t}(2+3e^{-0.235})=10$

Dividing and taking $\ln$ gives
$0.047t=\ln\frac{10}{2+3e^{-0.235}}$
so $t=\frac{1000}{47}\ln\frac{10}{2+3e^{-0.235}}\approx17.60$ years, which is after $t=5$ as required.

Answer

$t=\frac{1000}{47}\ln\frac{10}{2+3e^{-0.235}}\approx17.60$ years.

Key idea

Deposits growing at the same continuous rate share one exponential factor, so the natural logarithm can solve for the time.
Problem 7: Mark complete
Problem 8 A ceiling argument

For positive integers $n$ , let $C_n=(1+\frac1n)^n$ . A student argues: each $C_n$ is a fraction and $e$ is irrational, so $C_n<e$ for every $n$ . Is that reasoning sufficient? Name the facts from this lesson that do establish $C_n<e$ .
1. Hint 1
  Ask what the irrationality of $e$ rules out, and what it leaves open.
2. Hint 2
  Revisit the lesson's description of how the compounding values behave as n grows.
Answer

No. The facts are that $C_n$ strictly increases with $n$ and closes in on $e$ .

Full solution

Each $C_n=\frac{(n+1)^n}{n^n}$ is a fraction, and $e$ is irrational, so $C_n\ne e$ .
That says nothing about order: the fraction $3$ is also unequal to $e$ , yet $3>e$ .

The order comes from the lesson's two facts about the values: they strictly increase with $n$ , and they close in on $e$ .
If some $C_m$ were at least $e$ , then every value from $C_{m+1}$ on would be at least $C_{m+1}$ , a number greater than $e$ , so the values could not close in on $e$ .
Therefore $C_n<e$ for every positive integer $n$ .

Answer

No. The facts are that $C_n$ strictly increases with $n$ and closes in on $e$ .

Key idea

Two numbers being unequal does not decide which is larger; a strictly increasing list that closes in on a value stays below it.
Problem 8: Mark complete
Problem 9 A rate interpretation

A model is $A(t)=Pe^{0.07t}$ with $P>0$ and $t$ in years. A student calls $7\%$ its exact percentage increase over one full year. Is that correct? Give the actual one-year percentage increase to the nearest hundredth of a percent.
1. Hint 1
  A one-year percentage change compares the gain with the amount at the start of that year.
2. Hint 2
  Compare $A(1)$ with $A(0)$ through their ratio.
Answer

No; the one-year increase is approximately $7.25\%$ .

Full solution

The yearly multiplier is
$\frac{A(1)}{A(0)}=e^{0.07}$
The denominator is $P>0$ .
Therefore the yearly percentage increase is $100(e^{0.07}-1)\%$ , approximately $7.25\%$ .
The continuous rate $0.07$ and the effective yearly rate describe different quantities.

Answer

No; the one-year increase is approximately $7.25\%$ .

Key idea

A continuous rate $r$ gives an effective one-period rate of $e^r-1$ .
Problem 9: Mark complete
Problem 10 A shared factor claim

For $x>0$ , a student claims that $\ln(\frac{x^3}{e^2})=3\ln x-2$ . Is the claim correct? Explain using the meanings of the terms.
1. Hint 1
  The ordinary logarithm rules also hold in base $e$ .
2. Hint 2
  Use the quotient and power rules, then evaluate the logarithm of $e^2$ .
Answer

Yes; the identity holds for every $x>0$ .

Full solution

All arguments are positive and $e^2\ne0$ .
The quotient rule subtracts $\ln(e^2)$ , the power rule gives $\ln(x^3)=3\ln x$ , and the inverse law gives
$\ln(e^2)=2.$
Therefore
$\ln\left(\frac{x^3}{e^2}\right)=3\ln x-2$
which proves the claim on the stated domain.

Answer

Yes; the identity holds for every $x>0$ .

Key idea

Natural logarithms obey the same product, quotient, and power rules as logarithms in other valid bases.
Problem 10: Mark complete

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