MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: inverse functions and logarithms (MATH 203)

This sheet is not a summary of section 1.6 of Thomas' Calculus: you already have the course notes. It answers one question only, what makes students lose marks on inverse functions and logarithms in MATH 203 at Concordia University, and which precise gesture avoids each loss. The inverse trigonometric functions are left to their own chapter.

The idea of an inverse is rarely the problem. The marks go in the algebra inside it: an unknown that appears twice and is never collected, a logarithm law that does not exist, and a domain that nobody checked at the end. Your scientific calculator computes $\ln$ and $\log$ only, so it will not catch any of these for you.

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The thread of the chapter

When the unknown sits in two places, $x$ in a numerator and a denominator, $e^x$ and $e^{-x}$ , $x\ln 3$ and $x\ln 7$ , it must be COLLECTED on one side and FACTORED out before anything else; and since an inverse swaps inputs and outputs, every answer ends with a domain check, the piece where $f$ is one-to-one, the positive arguments of $\ln$ , the range that becomes a domain.

This chapter is part of MATH 203, Calculus I (Concordia)

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

One-to-one function
Horizontal line test
Inverse function
Inverse function domain
Restricted domain
Natural logarithm
Laws of logarithms
Change of base formula
Exponential equations
Logarithmic equations
Extraneous solutions
Logarithmic scales

The essentials

An inverse exists where f is one-to-one, and it is found by collecting the unknown

• One-to-one: $f(a) = f(b)$ forces $a = b$ ; no horizontal line meets the graph twice. One pair of inputs with the same output settles a no.
• $f^{-1}(b) = a \iff f(a) = b$ . Domain of $f^{-1}$ = range of $f$ , range of $f^{-1}$ = domain of $f$ ; the graph is reflected in $y = x$ .
• Method: $y = f(x)$ , clear the denominators, COLLECT every $x$ term on one side, FACTOR $x$ out, divide, rename. $y(x + 4) = 1 - 3x$ gives $x(y + 3) = 1 - 4y$ .
• If a $\pm$ or a square root appears, the domain of $f$ chooses: $\sqrt{x - 1} = 2 - y$ needs $y \le 2$ , so $g^{-1}(x) = 1 + (2 - x)^2$ for $x \le 2$ only.
• $f^{-1}(x)$ is never $\frac{1}{f(x)}$ : the $-1$ means undoing $f$ by composition.

g(x) = 2 - \sqrt{x - 1}

and its inverse are mirror images in

y = x

; the inverse is the solid half of the parabola, the dashed half is not

g^{-1}

The restriction is part of the formula. Without $x \le 2$ , the formula $1 + (2 - x)^2$ is a whole parabola, which is not one-to-one and is the inverse of nothing.

Logarithms: every law carries a condition

• $\log_a x = y \iff a^y = x$ , defined for $x > 0$ only; a logarithm can be negative, its argument cannot.
• $\ln(xy) = \ln x + \ln y$ , $\ln\frac{x}{y} = \ln x - \ln y$ , $\ln x^r = r\ln x$ : for $x > 0$ and $y > 0$ only.
• Change of base: $\log_a x = \frac{\ln x}{\ln a}$ , the only way to get $\log_3 20$ from a calculator that has $\ln$ and $\log$ .
• $\ln(e^x) = x$ for every $x$ , but $e^{\ln x} = x$ for $x > 0$ only: $e^{2\ln x} = x^2$ holds for $x > 0$ .
• No law for $\ln(x + y)$ , for $(\ln x)^2$ , for $(\ln x)(\ln y)$ , for $\frac{\ln x}{\ln y}$ .

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.

The laws of logarithms, and the ones students invent

Read a line as: the expression of the first column, under the condition of the second, equals the third. The red lines are rules that do not exist: each is refuted by two numbers, and the method says what to do instead.

Expression	Condition	Equals
$\log_a(xy)$	$x > 0$ , $y > 0$	$\log_a x + \log_a y$
Example: $\log_6 4 + \log_6 9 = \log_6 36 = 2$ .
$\ln\frac{x}{y}$	$x > 0$ , $y > 0$	$\ln x - \ln y$
Example: $\ln 8 - \ln 2 = \ln 4$ , which checks $x = 4$ in $\ln(x + 4) - \ln(x - 2) = \ln x$ .
$\ln x^r$	$x > 0$	$r\ln x$
Example: $e^{-2\ln 5} = e^{\ln 5^{-2}} = \frac{1}{25}$ .
$\log_a x$	$x > 0$ , $a > 0$ , $a \ne 1$	$\frac{\ln x}{\ln a}$
Example: $\log_3 20 = \frac{\ln 20}{\ln 3} \approx 2.7268$ , between $2$ and $3$ since $9 < 20 < 27$ .
$\ln(x + y)$	$x, y > 0$	no law rule that does not exist
Example: $\ln(1 + 1) = \ln 2 \approx 0.69$ , while $\ln 1 + \ln 1 = 0$ . What to do: Isolate the exponential first, or multiply by $e^x$ to get a quadratic: $e^x + 6e^{-x} = 5$ becomes $u^2 - 5u + 6 = 0$ .
$(\ln x)^2$	$x > 0$	not $2\ln x$ rule that does not exist
Example: At $x = e$ : $(\ln e)^2 = 1$ , while $2\ln e = 2$ . What to do: Set $t = \ln x$ and solve the quadratic in $t$ ; only $\ln(x^2)$ equals $2\ln x$ .
$\frac{\ln x}{\ln y}$	$x, y > 0$ , $y \ne 1$	not $\ln\frac{x}{y}$ rule that does not exist
Example: $\frac{\ln 8}{\ln 2} = 3$ , while $\ln\frac{8}{2} = \ln 4 \approx 1.39$ . What to do: A quotient of logarithms is a CHANGE OF BASE: $\frac{\ln x}{\ln y} = \log_y x$ .

The conditions of the first four lines are what produces the rejected roots: combining two logarithms into one enlarges the domain, and the test at the end restores it.

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. Leaving x on both sides of an inverse formula

the whole question

What not to write

“ $y(x + 4) = 1 - 3x$ , so $x = \frac{1 - 3x - 4y}{y}$ .”

What to write

“ $xy + 4y = 1 - 3x$ , so $xy + 3x = 1 - 4y$ , $x(y + 3) = 1 - 4y$ and $x = \frac{1 - 4y}{y + 3}$ . So $f^{-1}(x) = \frac{1 - 4x}{x + 3}$ .”

Why: Nothing is solved while $x$ is on both sides. Collect every term containing $x$ , factor $x$ out, and only then divide.

2. Forgetting the restriction or leaving a plus or minus

2 marks

What not to write

“ $\sqrt{x - 1} = 2 - y$ , so $x = 1 + (2 - y)^2$ and $g^{-1}(x) = 1 + (2 - x)^2$ .”

What to write

“ $\sqrt{x - 1} = 2 - y$ requires $y \le 2$ , so $g^{-1}(x) = 1 + (2 - x)^2$ for $x \le 2$ . Check: $g(5) = 0$ and $g^{-1}(0) = 5$ .”

Why: Squaring loses the sign condition. The full parabola gives $g^{-1}(3) = g^{-1}(1) = 2$ , two inputs for one output, so it is not an inverse; the figure of the first block shows the half that is.

3. Turning the square of a logarithm into twice the logarithm

the whole question

What not to write

“ $(\ln x)^2 = \ln(x^3) + 4$ , so $2\ln x = 3\ln x + 4$ and $x = e^{-4}$ .”

What to write

“With $t = \ln x$ : $t^2 - 3t - 4 = 0$ , $t = 4$ or $t = -1$ , so $x = e^4$ or $x = e^{-1}$ , both positive, both kept.”

(\ln x)^2

and

2\ln x

are two different curves that cross only at

(1, 0)

and

(e^2, 4)

; everywhere else, replacing one by the other changes the equation.

Why: The power law moves an exponent INSIDE the logarithm. In $(\ln x)^2$ the square is outside; the two functions only agree at $x = 1$ and $x = e^2$ , as the figure shows.

4. Keeping a root that makes a logarithm negative

2 marks, and a solution that does not exist

What not to write

“ $\frac{x + 4}{x - 2} = x$ gives $x^2 - 3x - 4 = 0$ , so $x = 4$ or $x = -1$ .”

What to write

“Domain: $x > 2$ . The candidates are $4$ and $-1$ ; $x = -1$ is rejected because $\ln(-1 - 2)$ is undefined. Solution: $x = 4$ .”

Why: The quotient law and clearing the denominator both enlarge the domain. The domain is written BEFORE the algebra and applied after it, with the reason.

5. Rejecting a valid negative root by reflex

half the question

What not to write

“ $\log_4(x^2) = \log_4(3x + 10)$ gives $x = 5$ or $x = -2$ ; $-2$ is negative, so it is rejected.”

What to write

“At $x = -2$ : $x^2 = 4 > 0$ and $3x + 10 = 4 > 0$ , so both logarithms exist. Solutions: $x = 5$ and $x = -2$ .”

Why: A candidate is rejected only when it makes an ARGUMENT zero or negative in the original equation. Its own sign does not matter.

6. Taking the logarithm of a sum term by term

the whole question

What not to write

“ $e^x + 6e^{-x} = 5$ , so $x + \ln 6 - x = \ln 5$ .”

What to write

“Multiply by $e^x > 0$ : $e^{2x} - 5e^x + 6 = 0$ , $(e^x - 2)(e^x - 3) = 0$ , so $x = \ln 2$ or $x = \ln 3$ .”

Why: There is no law for $\ln(a + b)$ . When the unknown sits in $e^x$ and $e^{-x}$ , multiplying by $e^x$ gathers it into one quadratic in $u = e^x$ .

7. Writing the change of base upside down

1 to 2 marks

What not to write

“ $\log_3 20 = \frac{\ln 3}{\ln 20} \approx 0.3667$ .”

What to write

“ $\log_3 20 = \frac{\ln 20}{\ln 3} \approx 2.7268$ , consistent with $3^2 < 20 < 3^3$ .”

Why: The argument goes on top: $\log_a x = \frac{\ln x}{\ln a}$ . Bracketing by powers of the base catches the inversion in five seconds.

8. Reflecting the graph in the wrong line

the sketch, 2 to 3 marks

What not to write

“The graph of $\ln x$ is the graph of $e^x$ reflected in the $y$ -axis.”

What to write

“The graph of $\ln x$ is the reflection of the graph of $e^x$ in the line $y = x$ : $(0, 1)$ becomes $(1, 0)$ .”

Reflected in the

y

-axis,

e^x

becomes the dashed

e^{-x}

; reflected in

y = x

, it becomes

\ln x

, the inverse.

Why: The $y$ -axis reflection replaces $x$ by $-x$ and gives $e^{-x}$ ; the inverse swaps the COORDINATES, which is the mirror $y = x$ .

Which method to choose

Which method, by the FORM of the equation

Look at where the unknown sits before writing anything: the form picks the method

If powers with unrelated bases, the unknown in both exponents → take $\ln$ of both sides, distribute, collect the $x$ terms, factor

Example: $3^{2x + 1} = 7^x$ : $x(2\ln 3 - \ln 7) = -\ln 3$ , $x \approx -4.3715$
If $e^x$ and $e^{-x}$ together → multiply by $e^x$ , set $u = e^x$ , solve the quadratic, reject $u \le 0$

Example: $e^x + 6e^{-x} = 5$ : $u = 2$ or $u = 3$ , $x = \ln 2$ or $\ln 3$
If a sum or difference of logarithms → write the domain, combine into one logarithm, remove it, test each candidate

Example: $\ln(x + 4) - \ln(x - 2) = \ln x$ : $x = 4$ , $x = -1$ rejected
If logarithms in different bases → change of base to a single base first

Example: $\log_2 x + \log_4 x = 6$ : $\frac{3}{2}\log_2 x = 6$ , $x = 16$
If $\ln x$ squared, or $\ln x$ in a quadratic → set $t = \ln x$ , solve for $t$ , then $x = e^t$

Example: $(\ln x)^2 = \ln(x^3) + 4$ : $t = 4$ or $-1$ , $x = e^4$ or $e^{-1}$
If the unknown in the denominator of an exponent → isolate the power, take $\ln$ , solve for the reciprocal, then invert

Example: $80 \cdot 2^{-9/h} = 50$ : $h = \frac{9\ln 2}{\ln(8/5)} \approx 13.27$

The last step of every branch is the same: test each candidate in the ORIGINAL equation, and write the reason for each rejection.

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.

Finding an inverse function

When to use it: Any question that asks for $f^{-1}$ , its formula, its domain or one of its values

1 Prove that $f$ is one-to-one on its domain, or restrict the domain until it is, and say which piece you keep.
2 Find the range of $f$ : it will be the domain of $f^{-1}$ .
3 Write $y = f(x)$ , clear the denominators, collect every term containing $x$ on one side and factor $x$ out.
4 Divide, stating what must not be zero; if a $\pm$ appears, choose the sign with the restricted domain.
5 Rename $x$ and $y$ , write the domain next to the formula, and check one value through $f$ and back.

Concluding sentence

“ $f$ is one-to-one. From $y(x + 4) = 1 - 3x$ we get $x(y + 3) = 1 - 4y$ , so $f^{-1}(x) = \frac{1 - 4x}{x + 3}$ for $x \ne -3$ . Check: $f(0) = \frac{1}{4}$ and $f^{-1}\left(\frac{1}{4}\right) = 0$ .”

The trap: Stopping at a line with $x$ on both sides, or writing the formula without its domain.

Marking: Typically 2 marks for one-to-one and the range, 4 for the algebra, 2 for the domain, 2 for the check.

Check before you hand in

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

One value through f and back

After finding $f^{-1}$ , compute $f$ at an easy input, then $f^{-1}$ at the output: you must get the input back.

$f(-\frac{3}{4}) = 1$ and $f^{-1}(1) = \frac{1 - 4}{1 + 3} = -\frac{3}{4}$ . A swapped fraction would give $-\frac{4}{3}$ and fail at once.
Bracket a logarithm by powers of its base

Before trusting a calculator value, place the argument between two powers of the base.

$9 < 20 < 27$ , so $\log_3 20$ is between $2$ and $3$ : $2.7268$ is plausible, $0.3667$ is not.
Substitute into the ORIGINAL equation

Put every kept solution into the equation as it was given, not the combined one. It catches both a wrong root and a forgotten rejection.

$x = -1$ satisfies $\frac{x + 4}{x - 2} = x$ , but $\ln(x - 2) = \ln(-3)$ does not exist.
The domain of the inverse is the range

Compare the domain you wrote for $f^{-1}$ with the values $f$ actually takes. They must be the same set.

$m(x) = \frac{2^x + 1}{2^x - 1}$ never takes the value $0$ , so $m^{-1}(0)$ cannot exist, and indeed it would be $\log_2(-1)$ .

The typical problem, taken apart

An inverse where the unknown is e to the x, twice

Let $f(x) = \frac{3e^x - 1}{e^x + 2}$ . Show that $f$ is one-to-one, find its range, and find $f^{-1}$ with its domain.

Every step must be justified as on a MATH 203 final.

The graph rises between the lines

y = -\frac{1}{2}

and

y = 3

without touching them; the algebra below proves what the picture suggests.

Step 1

Rewrite $f(x) = \frac{3(e^x + 2) - 7}{e^x + 2} = 3 - \frac{7}{e^x + 2}$ . As $x$ increases, $e^x + 2$ increases, $\frac{7}{e^x + 2}$ decreases, so $f$ increases: $f$ is one-to-one.

Why

Splitting off the constant puts the unknown in ONE place, which makes both the monotonicity and the range readable without any derivative.

Step 2

$e^x + 2$ takes every value in $(2, \infty)$ , so $\frac{7}{e^x + 2}$ takes every value in $\left(0, \frac{7}{2}\right)$ and $f$ every value in $\left(-\frac{1}{2}, 3\right)$ .

Why

The range is announced before the formula: it will be the domain of $f^{-1}$ , and the next steps must find it again.

Step 3

$y(e^x + 2) = 3e^x - 1$ , so $ye^x + 2y = 3e^x - 1$ . Collect: $ye^x - 3e^x = -1 - 2y$ , and factor: $e^x(y - 3) = -(1 + 2y)$ , so $e^x = \frac{1 + 2y}{3 - y}$ .

Why

This is the gesture of the chapter: $e^x$ appears in two terms, and only collecting and factoring isolates it. Dividing needs $y \ne 3$ , which the range guarantees.

Step 4

$e^x > 0$ requires $\frac{1 + 2y}{3 - y} > 0$ , that is $-\frac{1}{2} < y < 3$ . Then $x = \ln\frac{1 + 2y}{3 - y}$ , so $f^{-1}(x) = \ln\frac{1 + 2x}{3 - x}$ on $\left(-\frac{1}{2}, 3\right)$ .

Why

The condition of $\ln$ gives back exactly the range of step 2: two independent roads to the same set is the best check of the domain.

Step 5

Check: $f(0) = \frac{2}{3}$ , and $f^{-1}\left(\frac{2}{3}\right) = \ln\frac{7/3}{7/3} = \ln 1 = 0$ .

Why

One value through $f$ and back takes twenty seconds and catches a swapped fraction or a sign lost while collecting.

The conclusion, written out

“ $f$ is increasing, hence one-to-one, with range $\left(-\frac{1}{2}, 3\right)$ . Its inverse is $f^{-1}(x) = \ln\frac{1 + 2x}{3 - x}$ for $-\frac{1}{2} < x < 3$ .”

The classic mistake on this problem: Taking $\ln$ of the numerator and the denominator separately, $\ln(3e^x - 1) - \ln(e^x + 2)$ , which leaves $x$ trapped in two logarithms; or giving the formula with no domain.

Learn by heart

• Only a one-to-one function has an inverse; domain of $f^{-1}$ = range of $f$ , and the graph reflects in $y = x$ .
• Unknown in two places: collect it on one side, factor it out, THEN divide or take $\ln$ .
• $\ln(xy) = \ln x + \ln y$ , $\ln x^r = r\ln x$ : for POSITIVE $x, y$ only. No law for $\ln(x + y)$ or $(\ln x)^2$ .
• $\log_a x = \frac{\ln x}{\ln a}$ : the argument on top.
• $\ln(e^x) = x$ for every $x$ ; $e^{\ln x} = x$ for $x > 0$ only.
• Equation: domain first, algebra, then test every candidate in the ORIGINAL equation; reject only for a reason.

Frequently asked questions

How do I find the inverse of a function in MATH 203?

Check that the function is one-to-one, restricting its domain if needed. Write y equals f of x, clear the denominators, move every term that contains x to one side, factor x out and divide. Rename x and y, and write the domain of the inverse, which is the range of the original function. Finish by checking one value through f and back.

Why do some solutions of a logarithmic equation have to be rejected?

Combining logarithms and clearing denominators enlarge the domain. The difference ln of x plus 4 minus ln of x minus 2 only exists when x is greater than 2, but the combined equation also accepts x equals minus 1. Reject a candidate only when it makes the argument of a logarithm zero or negative in the original equation; a negative candidate can be a perfectly valid solution.

How do I compute log base 3 of 20 on a calculator that only has ln and log?

Use the change of base formula: log base 3 of 20 equals ln 20 divided by ln 3, about 2.7268. The argument goes on top. Check it by bracketing: 20 is between 3 squared and 3 cubed, so the answer must be between 2 and 3. Getting 0.3667 means the fraction was typed upside down.

Is the square of ln x the same as 2 ln x?

No. Two ln x equals ln of x squared, because the power law moves an exponent that is inside the logarithm. The square of ln x squares the value of the logarithm, and it has no simpler form. At x equals e the first gives 1 and the second 2. In an equation, substitute t for ln x and solve the quadratic in t.

Practise it

Corrected exercises: Inverse functions and logarithms, MATH 203 at Concordia

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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Functions, trigonometric and exponential functions Next sheet Rates of change and limit laws

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-inverse-functions-logarithms. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: inverse functions and logarithms (MATH 203)

The essentials

An inverse exists where f is one-to-one, and it is found by collecting the unknown

Logarithms: every law carries a condition

The rules in table form

The laws of logarithms, and the ones students invent

The mistakes that cost marks

1. Leaving x on both sides of an inverse formula

2. Forgetting the restriction or leaving a plus or minus

3. Turning the square of a logarithm into twice the logarithm

4. Keeping a root that makes a logarithm negative

5. Rejecting a valid negative root by reflex

6. Taking the logarithm of a sum term by term

7. Writing the change of base upside down

8. Reflecting the graph in the wrong line

Which method to choose

Which method, by the FORM of the equation

How the answer is expected to be written

Finding an inverse function

Check before you hand in

The typical problem, taken apart

An inverse where the unknown is e to the x, twice

Learn by heart

Frequently asked questions

How do I find the inverse of a function in MATH 203?

Why do some solutions of a logarithmic equation have to be rejected?

How do I compute log base 3 of 20 on a calculator that only has ln and log?

Is the square of ln x the same as 2 ln x?

Corrected exercises: Inverse functions and logarithms, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?