Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: exponential and logarithmic functions (201-103)

This is the most useful chapter of the course for later studies: population growth, compound interest, the decay of memory, the spread of an innovation. Everything in it runs on the same idea, a constant relative rate, and on the same error, the power rule applied to a variable that is in the exponent.

This sheet gives the seven differentiation rules with a worked number under each, the two rules that do not exist, and the three tools of interpretation that a 201-103 exam always asks for: the relative rate, the doubling time and the reading of a semilog scale. It is written for Social Science students in the mathematics profile. Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

The whole chapter comes down to one question: is the variable in the BASE or in the EXPONENT? The power rule serves the first case, never the second, and confusing them changes the answer by a factor that grows without end.

This chapter is part of 201-103, Differential Calculus for Social Science

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Derivative of the exponential function
Derivative of the natural logarithm
Change of base
Properties of logarithms
Relative rate of change
Doubling time
Half-life
Rule of 70
Semilog scale
Logarithmic differentiation
Continuous compounding

The essentials

In the base or in the exponent: the question that decides everything

• $x^{2}$ has the variable in the BASE: power rule, $2x$ .
• $2^{x}$ has the variable in the EXPONENT: exponential rule, $2^{x}\ln 2$ .
• The two functions have nothing to do with each other: in the long run the exponential overtakes any power, and by far.
• $\left(e^{x}\right)'=e^{x}$ : it is the only nonzero function equal to its own derivative.
• $e$ is the only base whose slope equals $1$ at the point $(0, 1)$ , that is, the only base such that $\ln a=1$ .

The two curves cross at

2

and at

4

, then the exponential breaks away:

2^{5}=32

against

5^{2}=25

, and already

2^{6}=64

against

6^{2}=36

Before differentiating, circle the variable and ask where it is. This two-second reflex avoids the costliest error of the chapter.

The rules, and the factor people forget

• $\left(e^{u}\right)'=u'e^{u}$ : the factor $u'$ is never forgotten.
• $\left(a^{x}\right)'=a^{x}\ln a$ , and more generally $\left(a^{u}\right)'=u'a^{u}\ln a$ .
• $\left(\ln u\right)'=\dfrac{u'}{u}$ , on the domain where $u>0$ . The domain is found BEFORE differentiating, because the simplified derivative keeps no trace of it.
• $\log_{10}x=\dfrac{\ln x}{\ln 10}$ , so $\left(\log_{10}x\right)'=\dfrac{1}{x\ln 10}$ .
• $\ln(ab)=\ln a+\ln b$ , $\ln\dfrac{a}{b}=\ln a-\ln b$ , $\ln(a^{n})=n\ln a$ . There is nothing simple for $\ln(a+b)$ .

e^{x}

(blue) and

\ln x

(orange) are symmetric about

y=x

. At

(0, 1)

and at

(1, 0)

, the green tangents have slope

1

. The curve of

\ln

lives only for

x>0

The derivative of $\ln(3x)$ is $\dfrac{1}{x}$ : the $3$ disappears, and that is normal, since $\ln(3x)=\ln 3+\ln x$ differs from $\ln x$ only by a constant.

What the exponential tells us in the social sciences

• A model $P(t)=P_{0}e^{rt}$ satisfies $P'=rP$ : the relative rate is $r$ , CONSTANT. That is the signature of the exponential.
• Doubling time: $T=\dfrac{\ln 2}{r}$ , independent of the initial value. Half-life: the same formula with $r$ taken in absolute value.
• Rule of 70: $T\approx\dfrac{70}{\text{rate in percent}}$ , because $100\ln 2\approx 69.3$ . It holds for small rates and degrades beyond $10$ percent.
• Relative rates ADD for a product and SUBTRACT for a quotient: gross domestic product per capita grows at the difference of the two rates.
• Continuous compounding: $A(t)=A_{0}e^{rt}$ , with an annual EFFECTIVE rate of $e^{r}-1$ , always greater than the nominal rate $r$ .

With

r=1.8

percent, the population doubles every

38.5

years, whether you start from

2{,}400

or from

4{,}800

inhabitants:

T=\frac{\ln 2}{r}

does not depend on the starting level.

These five lines are the only formulas the exam expects you to know by heart in this chapter. The rest can be rederived.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Differentiating an exponential or a logarithm

The last column gives the derivative. The two red cells are rules that students invent every term: they do not exist, and two numbers are enough to show it.

Form of $f$	What we apply	$f'$
$e^{x}$	it is its own derivative	$e^{x}$
Example: At $x=0$ the slope is $1$ ; at $x=2$ it is $e^{2}\approx 7.389$ .
$e^{u}$	the chain rule	$u'e^{u}$
Example: $e^{3x}$ has derivative $3e^{3x}$ , which equals $3$ at $x=0$ .
$a^{x}$	the exponential with base $a$	$a^{x}\ln a$
Example: $2^{x}$ has derivative $2^{x}\ln 2$ , which is about $0.693$ at $x=0$ .
$\ln u$	$\dfrac{u'}{u}$ , on $u>0$	$\dfrac{u'}{u}$
Example: $\ln(3x)$ has derivative $\dfrac{1}{x}$ , which equals $0.5$ at $x=2$ .
$\log_{10}x$	the change of base	$\dfrac{1}{x\ln 10}$
Example: At $x=2$ : $\dfrac{1}{2\ln 10}\approx 0.217$ .
$\left(e^{x}\right)'=xe^{x-1}$	the power rule	does not exist nonexistent rule
Example: At $x=2$ , the true derivative is $e^{2}\approx 7.389$ , whereas this expression would give $2e\approx 5.437$ . What to do: The variable is in the EXPONENT: the derivative is $e^{x}$ , never $nx^{n-1}$ .
$\ln(a+b)=\ln a+\ln b$	nothing at all	does not split nonexistent rule
Example: $\ln 5\approx 1.609$ against $\ln 2+\ln 3\approx 1.792$ . What to do: Compute the sum FIRST, then the logarithm. Only products and quotients split.

The first five rows can be recited in thirty seconds and cover the whole exam. The last two cost four marks per paper on average.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Forgetting the inner factor on a composite exponential

2 marks, and a growth rate announced fifty times too large

What not to write

“ $f(t)=e^{0.018t}$ , so $f'(t)=e^{0.018t}$ .”

What to write

“ $f'(t)=0.018\,e^{0.018t}$ : the chain rule multiplies by the derivative of the exponent.”

t=10

, the true tangent (green) has slope

0.018\times 2{,}873\approx 51.7

inhabitants per year. Without the factor

0.018

, one announces a slope of

2{,}873

: the red line, almost vertical.

Why: The exponential is its own derivative only when the exponent is $x$ alone. As soon as the exponent contains a coefficient, that coefficient comes out as a factor.

2. Applying the power rule to an exponential

the whole question, and the error comes back in the curve-sketching study

What not to write

“ $f(x)=2^{x}$ , so $f'(x)=x\,2^{x-1}$ .”

What to write

“The variable is in the exponent: $f'(x)=2^{x}\ln 2$ , which is about $0.693$ at $x=0$ .”

x=0

, the power rule would give

0\times 2^{-1}=0

, a horizontal tangent (red) on a curve that rises. The correct slope is

\ln 2\approx 0.693

(green).

Why: $x^{n}$ and $n^{x}$ are two unrelated families. The power rule assumes the exponent is a constant, which is exactly what is false here.

3. Splitting the logarithm of a sum

2 marks, and an entirely wrong derivative

What not to write

“ $\ln(x+3)=\ln x+\ln 3$ .”

What to write

“ $\ln(x+3)$ does not split. Its derivative is computed by the chain rule: $\dfrac{1}{x+3}$ .”

\ln(x+3)

(green) and

\ln x+\ln 3=\ln(3x)

(red) are two different functions: different domains, and they cross only at

x=1.5

. The logarithm of a sum does not split.

Why: The logarithm turns a PRODUCT into a sum, not a sum into a sum. Two numbers prove it: $\ln 5\approx 1.609$ and $\ln 2+\ln 3\approx 1.792$ .

4. Differentiating a logarithm without having stated its domain

1.5 marks, and a function study carried out on a half-line that does not exist

What not to write

“ $f(x)=\ln(5-x)$ , so $f'(x)=\dfrac{-1}{5-x}$ , defined everywhere except at $5$ .”

What to write

“ $f$ exists only for $x<5$ . On that domain, $f'(x)=\dfrac{-1}{5-x}$ .”

\ln(5-x)

exists only to the left of

x=5

. On the right (red), the formula

\frac{-1}{5-x}

still gives numbers,

0.5

x=7

, but there is no curve whose slope it would be.

Why: The simplified derivative keeps no trace of the condition $u>0$ : at $x=7$ it gives a perfectly computable number, whereas the function is not defined there.

5. Believing that the doubling time depends on the starting population

2 marks of interpretation

What not to write

“A city of $50{,}000$ inhabitants takes longer to double than a village of $2{,}400$ .”

What to write

“With the same rate $r$ , both double in $T=\dfrac{\ln 2}{r}$ : the initial value cancels and plays no role.”

The three vertical lines are equally spaced and the curve is halved each time: the length of the half-life depends neither on the starting level nor on where you start counting.

Why: The equation $P_{0}e^{rT}=2P_{0}$ is divided by $P_{0}$ on both sides. That is the whole proof, and it fits in one line.

6. Confusing the nominal rate and the effective rate under continuous compounding

1.5 marks, and a skewed comparison of investments

What not to write

“An investment at $6$ percent compounded continuously earns $6$ percent per year.”

What to write

“The nominal rate is $6$ percent, the effective rate is $e^{0.06}-1\approx 6.18$ percent: continuous compounding always earns a little more than the advertised rate.”

At the same nominal rate of

6

percent, continuous compounding (green) stays ahead of annual compounding (orange): its effective rate is

e^{0.06}-1\approx 6.18

percent, and the gap widens with time.

Why: Between any two instants, interest earns interest itself. The gap is small over one year and becomes very visible over twenty.

7. Applying the rule of 70 to a high rate

1 mark, and above all a false statement of method

What not to write

“At $30$ percent per year, doubling takes $\dfrac{70}{30}\approx 2.3$ years.”

What to write

“The rule of 70 is an approximation valid for small rates. At $30$ percent, the exact calculation gives $\dfrac{\ln 2}{0.30}\approx 2.31$ years if the rate is continuous, but $2.64$ years if it is compounded annually.”

At small rates,

\frac{70}{r}

(red) and the exact doubling time at an annually compounded rate (green) coincide. At

30

percent, the gap reaches three tenths of a year:

2.33

against

2.64

Why: The rule of 70 comes from $100\ln 2\approx 69.3$ , an approximation that assumes a continuous and small rate. Knowing where it comes from is knowing when it stops holding.

8. Reading a straight line on the ordinary scale as proof of an exponential

2 marks, and a poorly chosen model for the rest of the problem

What not to write

“The points are almost aligned, so the growth is exponential.”

What to write

“Points aligned on the ORDINARY scale signal linear growth. It is on the semilog scale that an exponential becomes a straight line.”

On this scale the exponential is a perfect straight line, while the linear model rises fast and then flattens: the shape of the trace is enough to choose the model.

Why: The slope of $\ln P(t)$ is $\dfrac{P'}{P}$ : a straight line on the semilog scale means a constant relative rate, which is exactly the definition of an exponential.

9. Adding the quantities instead of the rates in a product

2 marks, and sometimes a result with the sign reversed

What not to write

“Gross domestic product grows by $3$ percent and the population by $1.2$ percent, so the product per capita grows by $4.2$ percent.”

What to write

“The product per capita is a QUOTIENT: its relative rate is the DIFFERENCE of the two, about $1.8$ percent per year.”

Why: Logarithmic differentiation proves it in two lines: $\ln\dfrac{A}{B}=\ln A-\ln B$ , and differentiating, $\dfrac{(A/B)'}{A/B}=\dfrac{A'}{A}-\dfrac{B'}{B}$ .

Which method to choose

Which rule, depending on where the variable is

Locate the variable before anything else: in the base, in the exponent, or in the argument of a logarithm. Three positions, three rules, none interchangeable.

If the variable is in the BASE and the exponent is a number → power rule, $nx^{n-1}$

Example: $x^{5}$ gives $5x^{4}$
If the variable is in the EXPONENT, base $e$ → $u'e^{u}$

Example: $e^{0.018t}$ gives $0.018e^{0.018t}$
If the variable is in the EXPONENT, any base → $u'a^{u}\ln a$

Example: $2^{3x}$ gives $3\times 2^{3x}\ln 2$
If the variable is in the argument of a $\ln$ → $\dfrac{u'}{u}$ , after writing the condition $u>0$

Example: $\ln(5-x)$ gives $\dfrac{-1}{5-x}$ on $x<5$
If the variable is in the base AND in the exponent → logarithmic differentiation, the only possible method

Example: $x^{x}$ gives $x^{x}(\ln x+1)$

this case is rare in 201-103, but it turns up as a reasoning question
If a product or quotient of three or more factors → logarithmic differentiation: take $\ln$ , split, then differentiate

Example: relative rates add and subtract

If the variable appears in two places, no simple rule applies and you must go through the logarithm. That is the signal for logarithmic differentiation.

Which tool, depending on what the question asks

The wording of growth problems is stable. Each phrasing calls for one formula and only one.

If “by what percent per year” → relative rate $\dfrac{P'}{P}$ , which equals $r$ for an exponential

Example: $P(t)=2400e^{0.018t}$ grows by $1.8$ percent per year
If “how long until the population doubles” → $T=\dfrac{\ln 2}{r}$

Example: with $r=0.018$ : $T\approx 38.5$ years
If “half-life”, “half of it remains” → same formula, with $|r|$

Example: with $r=-0.15$ : half-life of $4.62$ units
If “estimate in your head”, “order of magnitude” → rule of 70, valid for small rates

Example: $\dfrac{70}{1.8}\approx 38.9$ , very close to $38.5$
If “does this data follow an exponential model” → plot on a semilog scale and look for a straight line

Example: a straight line means a constant relative rate
If “the rate of the product, of the quotient, of the product per capita” → add or subtract the relative rates

Example: $3$ percent minus $1.2$ percent gives $1.8$ percent

None of these branches requires solving a complicated equation: they all come down to $\ln$ on both sides, then a division.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Computing and writing up a doubling time

When to use it: Whenever the question asks when a quantity doubles, triples, reaches a threshold, or falls to half.

1 Write the equation that the word translates: $P_{0}e^{rT}=2P_{0}$ for a doubling.
2 Divide by $P_{0}$ and SAY that the initial value cancels, so the result does not depend on it.
3 Take the natural logarithm of both sides: $rT=\ln 2$ .
4 Solve for $T$ and give the value with its units.
5 Check with the rule of 70 and conclude with a sentence in words.

Concluding sentence

“The equation $2400e^{0.018T}=4800$ gives $e^{0.018T}=2$ , so $0.018T=\ln 2$ and $T=\dfrac{\ln 2}{0.018}\approx 38.5$ years. The population therefore doubles in a little under forty years, whatever its starting size.”

The trap: Keeping $2400$ in the calculation to the end. It is not wrong, but the cancellation is exactly what the question wants to see written.

Marking: 1 mark for the equation set up, 1 mark for the cancellation, 1 mark for the logarithm, 1 mark for the value and its unit.

Carrying out a logarithmic differentiation

When to use it: Whenever there are three or more factors, a quotient of products, or a variable in both the base and the exponent. Also whenever the question asks for the relative rate of a ratio.

1 Write $\ln F$ and SPLIT it using the logarithm properties, products into sums and powers into coefficients.
2 Differentiate both sides: on the left $\dfrac{F'}{F}$ appears, on the right a sum of simple terms.
3 Read the result directly when the question is about the relative rate: $\dfrac{F'}{F}$ is already isolated.
4 Multiply by $F$ only if the question asks for $F'$ itself.
5 Interpret as a percentage, since $\dfrac{F'}{F}$ is a relative rate.

Concluding sentence

“Let $F=\dfrac{A}{B}$ . Then $\ln F=\ln A-\ln B$ , and differentiating, $\dfrac{F'}{F}=\dfrac{A'}{A}-\dfrac{B'}{B}=0.030-0.012=0.018$ . Gross domestic product per capita therefore grows by about $1.8$ percent per year.”

The trap: Differentiating $\ln F$ as if it were $F$ . On the left, the chain rule requires $\dfrac{F'}{F}$ , and it is precisely this factor that makes the method useful.

Marking: 1 mark for the splitting, 1 mark for differentiating both sides, 1 mark for reading the rate, 1 mark for the interpretation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The rule of 70 as an order of magnitude

After computing a doubling time, divide $70$ by the rate in percent. The two results should be close for a small rate.

$\dfrac{\ln 2}{0.018}\approx 38.5$ and $\dfrac{70}{1.8}\approx 38.9$ . A gap of several years would signal a decimal-point error on the rate.
The relative rate must give back $r$

Divide the derivative you found by the function. On a pure exponential, the result must be exactly the coefficient of the exponent.

With $P(t)=2400e^{0.018t}$ and $P'(t)=43.2e^{0.018t}$ : $\dfrac{P'}{P}=\dfrac{43.2}{2400}=0.018$ . Any other number signals a forgotten factor.
The sign of the derivative of a logarithm

The derivative of $\ln u$ has the sign of $u'$ , since $u$ is positive on the domain. This check catches chain-rule sign errors.

$\ln(5-x)$ is decreasing, so its derivative $\dfrac{-1}{5-x}$ must be negative for $x<5$ : at $x=4$ it equals $-1$ , which is consistent.
The numerical difference on a calculator

Compare the derivative you found with $\dfrac{f(a+0.0001)-f(a-0.0001)}{0.0002}$ . The two should agree to three decimal places.

For $2^{x}$ at $x=0$ , the numerical calculation gives $0.6931$ , which confirms $\ln 2$ and rules out the power rule.
The unit of the rate

A relative rate has NO unit of quantity: it is expressed per unit of time, in percent per year. A rate in dollars per year is an absolute rate, not a relative one.

$P'(0)=43.2$ inhabitants per year is absolute; $1.8$ percent per year is relative. They answer different questions.

The typical problem, taken apart

A population in continuous growth: rate, doubling and check

The population of a municipality is modelled by $P(t)=2400\,e^{0.018t}$ , where $t$ is the number of years since 2020.

Compute $P'(t)$ , give the relative growth rate, the population and its growth rate in 2030, then the doubling time.

Step 1

The variable is in the EXPONENT, with a coefficient: this is $\left(e^{u}\right)'=u'e^{u}$ with $u=0.018t$ , so $u'=0.018$ .

Why

Identifying $u$ and $u'$ before writing the formula is what prevents you from forgetting the factor, which is the number one error of the chapter.

Step 2

$P'(t)=2400\times 0.018\,e^{0.018t}=43.2\,e^{0.018t}$ inhabitants per year.

Why

The constant $2400$ stays, the coefficient $0.018$ comes down. The units read off immediately: inhabitants per year.

Step 3

Relative rate: $\dfrac{P'(t)}{P(t)}=\dfrac{43.2\,e^{0.018t}}{2400\,e^{0.018t}}=0.018$ , that is $1.8$ percent per year, for every $t$ .

Why

The exponential cancels entirely: this is the signature of the model, a CONSTANT relative rate. A function for which this ratio depends on $t$ is not exponential.

Step 4

In 2030, $t=10$ : $P(10)=2400\,e^{0.18}\approx 2400\times 1.1972\approx 2873$ inhabitants.

Why

We substitute into the function, not into the derivative. We round to the unit, since these are inhabitants.

Step 5

$P'(10)=43.2\,e^{0.18}\approx 51.7$ inhabitants per year.

Why

The absolute rate increases with time, while the relative rate stays at $1.8$ percent. Both statements are true at the same time.

Step 6

Doubling: $2400\,e^{0.018T}=4800$ , so $e^{0.018T}=2$ , then $0.018T=\ln 2$ and $T=\dfrac{\ln 2}{0.018}\approx 38.5$ years.

Why

The division by $2400$ shows that the result does not depend on the starting population. This is the line the marking scheme pays for.

Step 7

Check with the rule of 70: $\dfrac{70}{1.8}\approx 38.9$ years, very close to $38.5$ .

Why

The approximation comes from $100\ln 2\approx 69.3$ . It confirms the order of magnitude and catches a decimal-point error on the rate.

Step 8

Caveat: the model was fitted on recent data. Extended over two centuries, it would give an absurd population.

Why

An exponential has no ceiling, whereas reality does. Stating the window of validity is part of the answer in the social sciences.

The conclusion, written out

“ $P'(t)=43.2\,e^{0.018t}$ inhabitants per year, and the relative rate is $1.8$ percent per year whatever $t$ is. In 2030 the municipality will have about $2873$ inhabitants and will gain about $52$ inhabitants per year. The population doubles in $T=\dfrac{\ln 2}{0.018}\approx 38.5$ years, regardless of its initial size.”

The classic mistake on this problem: Announcing that the population grows by $43.2$ percent per year. The number $43.2$ is an ABSOLUTE rate, in inhabitants per year; the relative rate is the ratio $\dfrac{P'}{P}$ , and it equals $1.8$ percent.

Learn by heart

• $\left(e^{u}\right)'=u'e^{u}$ and $\left(a^{u}\right)'=u'a^{u}\ln a$ . The factor $u'$ is never forgotten.
• $\left(\ln u\right)'=\dfrac{u'}{u}$ , on the domain $u>0$ , written BEFORE differentiating.
• A variable in the BASE calls for $nx^{n-1}$ ; in the EXPONENT, it calls for $a^{x}\ln a$ . Never the other way around.
• $\ln(ab)=\ln a+\ln b$ and $\ln(a^{n})=n\ln a$ . NOTHING for $\ln(a+b)$ .
• An exponential $P_{0}e^{rt}$ has a CONSTANT relative rate equal to $r$ : that is its signature.
• Doubling time $T=\dfrac{\ln 2}{r}$ , independent of the initial value. The rule of 70 serves as a check.
• Relative rates ADD for a product and SUBTRACT for a quotient.
• On a semilog scale, an exponential becomes a STRAIGHT LINE, and its slope is the relative rate.

Frequently asked questions

What is the derivative of 2 to the power x?

It is 2 to the power x multiplied by the natural logarithm of 2, that is, about 0.693 times the function itself. The power rule does not apply here, because it assumes a constant exponent, whereas it is precisely the exponent that varies. For base e, this factor equals 1, which is why the natural exponential is its own derivative.

How do you calculate a doubling time?

Write that the function reaches twice its initial value, divide by that initial value, which cancels, then take the natural logarithm of both sides. What remains is the rate times the time equals the logarithm of 2, so the time is the logarithm of 2 divided by the rate. The result never depends on the starting quantity.

Where does the rule of 70 come from?

From the fact that one hundred times the natural logarithm of 2 is about 69.3, which is rounded to 70 because it divides nicely. Dividing 70 by the rate expressed as a percentage therefore gives the approximate doubling time. The approximation is excellent for small rates and degrades noticeably beyond ten percent.

Why does the logarithm of a sum not split?

Because the logarithm turns products into sums, not sums into sums. Two numbers are enough to check: the logarithm of 5 is about 1.609, whereas the logarithm of 2 plus the logarithm of 3 is about 1.792. To differentiate the logarithm of a sum, use the chain rule, not a splitting.

What is a semilog scale used for?

To recognize exponential growth at a glance. On this scale, you plot the logarithm of the quantity, and its slope is exactly the relative rate. An exponential therefore becomes a perfect straight line, and linear growth becomes a curve that flattens out. If the points line up, the rate is constant.

How do you get the growth rate of gross domestic product per capita?

By subtracting the population growth rate from that of gross domestic product. It is a quotient, and relative rates subtract for a quotient just as they add for a product. With three percent and one point two percent, the product per capita therefore grows by about one point eight percent per year.

Practise it

Corrected exercises: Exponential and logarithmic functions, 201-103

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: exponential and logarithmic functions (201-103)

The essentials

In the base or in the exponent: the question that decides everything

The rules, and the factor people forget

What the exponential tells us in the social sciences

The rules in table form

Differentiating an exponential or a logarithm

The mistakes that cost marks

1. Forgetting the inner factor on a composite exponential

2. Applying the power rule to an exponential

3. Splitting the logarithm of a sum

4. Differentiating a logarithm without having stated its domain

5. Believing that the doubling time depends on the starting population

6. Confusing the nominal rate and the effective rate under continuous compounding

7. Applying the rule of 70 to a high rate

8. Reading a straight line on the ordinary scale as proof of an exponential

9. Adding the quantities instead of the rates in a product

Which method to choose

Which rule, depending on where the variable is

Which tool, depending on what the question asks

How the answer is expected to be written

Computing and writing up a doubling time

Carrying out a logarithmic differentiation

Check before you hand in

The typical problem, taken apart

A population in continuous growth: rate, doubling and check

Learn by heart

Frequently asked questions

What is the derivative of 2 to the power x?

How do you calculate a doubling time?

Where does the rule of 70 come from?

Why does the logarithm of a sum not split?

What is a semilog scale used for?

How do you get the growth rate of gross domestic product per capita?

Corrected exercises: Exponential and logarithmic functions, 201-103

See also

Looking for a 201-103 tutor in Montreal?