MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: exponential, inverse, logarithmic and inverse trigonometric functions (MATH 140)

This sheet is not a summary of sections 1.4 and 1.5 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on exponential, logarithmic and inverse functions in MATH 140 at McGill University, and which precise gesture avoids each loss.

These functions look like review, and that is the danger: the algebra is usually right, and the marks go on a domain that was never checked. Every value below is exact and done by hand, as on the midterm, and every trap comes with the sentence that earns the method mark.

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The thread of the chapter

An inverse undoes a function only where it is one-to-one, and its answers live in the RANGE of the inverse: $\ln$ eats positive numbers only, so every log equation ends by testing each candidate in the ORIGINAL equation, and $\arcsin$ returns angles of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ only, so $\arcsin(\sin x)$ is not $x$ and $\sin x = c$ has a second solution.

This chapter is part of MATH 140, Calculus 1 (McGill)

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

Exponential function
Laws of exponents
One-to-one function
Horizontal line test
Inverse function
Natural logarithm
Laws of logarithms
Change of base formula
Exponential equations
Logarithmic equations
Inverse trigonometric functions
Half-life

The essentials

An inverse exists where the function is one-to-one, and swaps domain and range

• One-to-one: $f(a) = f(b)$ forces $a = b$ ; no horizontal line meets the graph twice. One counterexample, $f(0) = f(4)$ , settles a no.
• $f^{-1}(y) = x \iff f(x) = y$ . Domain of $f^{-1}$ = range of $f$ ; range of $f^{-1}$ = domain of $f$ . The graph of $f^{-1}$ is the mirror image of $f$ in $y = x$ : $(a, b)$ becomes $(b, a)$ .
• Method: write $y = f(x)$ , solve for $x$ , rename. If a $\pm$ appears, the domain of $f$ chooses the sign: $(x - 2)^2 = y + 4$ with $x \ge 2$ gives $x = 2 + \sqrt{y + 4}$ .
• $e^x$ and $\ln x$ are inverses: $e^{\ln x} = x$ for $x > 0$ , $\ln(e^x) = x$ for every $x$ . Same for $b^x$ and $\log_b x$ .
• $f^{-1}$ is NOT $\frac{1}{f}$ : the $-1$ means the inverse for composition. $\sin^{-1} x$ is $\arcsin x$ , never $\csc x$ .

e^x

and

\ln x

are mirror images in

y = x

: the point

(0, 1)

e^x

becomes

(1, 0)

\ln x

, and

(1, e)

becomes

(e, 1)

The cancellation equations hold on the right domains only: $f(f^{-1}(x)) = x$ for $x$ in the domain of $f^{-1}$ , and $f^{-1}(f(x)) = x$ for $x$ in the domain of $f$ . Saying which domain is part of the answer.

Inverse trigonometric functions: the restricted ranges decide everything

• $\arcsin: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; $\arccos: [-1, 1] \to [0, \pi]$ ; $\arctan: \mathbb{R} \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .
• A negative input gives a negative angle for $\arcsin$ and $\arctan$ , but an angle of $\left(\frac{\pi}{2}, \pi\right]$ for $\arccos$ : $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ .
• $\sin(\arcsin x) = x$ for $x \in [-1, 1]$ ; $\arcsin(\sin x) = x$ ONLY for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
• $\cos(\arcsin x) = \sqrt{1 - x^2}$ because $\cos \ge 0$ on the range of $\arcsin$ . The triangle gives the sizes, the range gives the signs.
• $\sin x = c$ has two solutions per period, $\arcsin c$ and $\pi - \arcsin c$ ; $\tan x = c$ has one per period $\pi$ : $\arctan c + k\pi$ .

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 that do not exist

Read a line as: an expression of the first column, under the condition of the second, equals the third. The red lines are rules students invent: each is refuted by two numbers, and the method says what to write instead.

Expression	Condition	Equals
$\ln(ab)$	$a > 0$ , $b > 0$	$\ln a + \ln b$
Example: $\log_2 40 - \log_2 5 = \log_2 8 = 3$ , and $\ln 12 - 2\ln 2 + \ln\frac{1}{3} = \ln 1 = 0$ .
$\ln(a^r)$	$a > 0$	$r \ln a$
Example: $e^{3\ln 2} = e^{\ln 8} = 8$ and $\ln\frac{1}{\sqrt e} = -\frac{1}{2}$ .
$\log_b a$	$a > 0$ , $b > 0$ , $b \ne 1$	$\frac{\ln a}{\ln b}$
Example: $\log_3 5 \cdot \log_5 9 = \frac{\ln 5}{\ln 3} \cdot \frac{\ln 9}{\ln 5} = 2$ .
$\ln(x^2)$	$x \ne 0$	$2\ln\|x\|$
Example: At $x = -1$ : $\ln 1 = 0 = 2\ln\|-1\|$ , while $2\ln(-1)$ does not exist.
$\ln(a + b)$	$a, b > 0$	no law rule that does not exist
Example: $\ln(1 + 1) = \ln 2 > 0$ , while $\ln 1 + \ln 1 = 0$ . What to do: Compute the sum first, then take the logarithm; or isolate the exponential before applying $\ln$ .
$\frac{\ln a}{\ln b}$	$a, b > 0$ , $b \ne 1$	not $\ln\frac{a}{b}$ rule that does not exist
Example: $\frac{\ln 8}{\ln 2} = 3$ , while $\ln\frac{8}{2} = \ln 4 = 2\ln 2 < 2$ . What to do: A quotient of logarithms is a CHANGE OF BASE: $\frac{\ln a}{\ln b} = \log_b a$ .

Every law of the first four lines carries a condition. On an equation, check it at the end on each candidate: that is where the rejected roots come from.

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. Keeping every root of a combined log equation

2 marks, and a solution that does not exist

What not to write

“ $\log_3 x + \log_3(x - 8) = 2$ , so $x(x - 8) = 9$ , so $x = 9$ or $x = -1$ .”

What to write

“Domain: $x > 8$ . Then $x(x - 8) = 9$ gives $x = 9$ or $x = -1$ ; $x = -1$ is rejected since $\log_3(-1)$ is undefined. Check: $\log_3 9 + \log_3 1 = 2$ . Solution: $x = 9$ .”

Why: Combining $\log_3 x + \log_3(x - 8)$ into $\log_3(x(x - 8))$ enlarges the domain: the product is also positive when BOTH factors are negative. The domain is written before the algebra and applied after.

2. Replacing ln of x squared by 2 ln x

1 to 2 marks, a lost solution

What not to write

“ $\ln(x^2) = \ln 9$ , so $2\ln x = \ln 9$ , so $x = 3$ .”

What to write

“ $\ln(x^2) = \ln 9$ gives $x^2 = 9$ , since $\ln$ is one-to-one, so $x = 3$ or $x = -3$ , and both work: $\ln((-3)^2) = \ln 9$ .”

y = \ln(x^2)

has two symmetric branches;

2\ln x

coincides with the right one only, so rewriting deletes every negative

x

Why: $\ln(x^2)$ is defined for every $x \ne 0$ , $2\ln x$ only for $x > 0$ : the power law needs a positive base. The correct identity is $\ln(x^2) = 2\ln|x|$ , and the figure shows the branch that $2\ln x$ deletes.

3. Taking the logarithm of a sum term by term

the whole question

What not to write

“ $3e^{2x - 1} + 4 = 10$ , so $\ln 3 + (2x - 1) + \ln 4 = \ln 10$ .”

What to write

“ $3e^{2x - 1} = 6$ , so $e^{2x - 1} = 2$ , so $2x - 1 = \ln 2$ and $x = \frac{1 + \ln 2}{2}$ .”

Why: There is no law for $\ln(a + b)$ . Isolate the exponential completely, THEN apply $\ln$ once. The same error averages two pH values: mixing pH $2$ and pH $4$ gives about $2.3$ , not $3$ .

4. Giving an arcsin answer outside its range

the whole value, on every question of this type

What not to write

“ $\arcsin\left(\sin\frac{2\pi}{3}\right) = \frac{2\pi}{3}$ , since arcsin and sin cancel.”

What to write

“ $\sin\frac{2\pi}{3} = \frac{\sqrt 3}{2}$ , and the angle of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with that sine is $\frac{\pi}{3}$ : $\arcsin\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}$ .”

Why: $\arcsin$ can only return angles of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so $\frac{2\pi}{3}$ is impossible whatever the input. The cancellation $\arcsin(\sin x) = x$ holds on that interval only.

5. Stopping at the calculator's one solution

half the question

What not to write

“ $\tan x = 2$ on $(-\pi, \pi)$ , so $x = \arctan 2$ .”

What to write

“ $\tan$ has period $\pi$ : the solutions are $\arctan 2 + k\pi$ . On $(-\pi, \pi)$ : $x = \arctan 2$ or $x = \arctan 2 - \pi$ .”

The line

y = 2

meets

y = \tan x

twice on

(-\pi, \pi)

, one period apart;

\arctan

only ever returns the intersection between

-\frac{\pi}{2}

and

\frac{\pi}{2}

Why: An inverse function returns ONE value, the one in its range; the equation asks for all of them. Add the period, $\pi$ for tangent, and use $\pi - \arcsin c$ or $-\arccos c$ for the symmetric solution of sine or cosine.

6. Trusting the triangle for the sign

1 mark, the sign

What not to write

“With a $3$ - $4$ - $5$ triangle, $\tan\left(\arccos\left(-\frac{3}{5}\right)\right) = \frac{4}{3}$ .”

What to write

“ $\theta = \arccos\left(-\frac{3}{5}\right) \in \left(\frac{\pi}{2}, \pi\right)$ , so $\sin\theta = \frac{4}{5}$ and $\tan\theta = \frac{4/5}{-3/5} = -\frac{4}{3}$ .”

Why: A triangle has positive sides: it gives the size of the answer, never its sign. The sign comes from the range of the inverse function: $[0, \pi]$ for $\arccos$ , where the sine is positive and the tangent takes the sign of the cosine.

7. Writing the reciprocal for the inverse

the whole question

What not to write

“ $f(x) = 2x + 6$ , so $f^{-1}(x) = \frac{1}{2x + 6}$ .”

What to write

“ $y = 2x + 6$ gives $x = \frac{y - 6}{2}$ , so $f^{-1}(x) = \frac{x - 6}{2}$ . Check: $f(1) = 8$ and $f^{-1}(8) = 1$ .”

Why: The exponent $-1$ of $f^{-1}$ means undoing $f$ by composition, not dividing by it. The one-value check, $f^{-1}(f(1)) = 1$ , catches the error at once: $\frac{1}{8} \ne 1$ .

8. Inverting a function that is not one-to-one

2 marks

What not to write

“The inverse of $g(x) = x^2 - 4x$ is $2 \pm \sqrt{x + 4}$ .”

What to write

“ $g(0) = g(4)$ , so $g$ has no inverse on $\mathbb{R}$ . On $[2, \infty)$ its inverse is $2 + \sqrt{x + 4}$ ; on $(-\infty, 2]$ it is $2 - \sqrt{x + 4}$ .”

Why: A $\pm$ gives two outputs for one input, so it is not a function. Restrict the domain first, as Stewart does for $\sin$ to build $\arcsin$ ; the restriction then chooses the sign.

9. Accepting a negative base

1 mark

What not to write

“ $Cb = 6$ and $Cb^3 = 24$ give $b^2 = 4$ , so $b = \pm 2$ .”

What to write

“ $b^2 = 4$ and the base of an exponential function is positive, so $b = 2$ and $C = 3$ ; $b = -2$ is rejected.”

Why: $(-2)^x$ is not defined for $x = \frac{1}{2}$ , so it is not an exponential function. The rejection must be written with its reason.

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 both sides are powers of the same base → rewrite with that base and equate the exponents (one-to-one)

Example: $9^{x+1} = 27^{2x - 1}$ : $2x + 2 = 6x - 3$ , $x = \frac{5}{4}$
If one exponential, plus constants → isolate it completely, then apply $\ln$ once

Example: $3e^{2x - 1} + 4 = 10$ : $e^{2x - 1} = 2$ , $x = \frac{1 + \ln 2}{2}$
If $e^{2x}$ and $e^x$ together, or $4^x$ and $2^x$ → set $u = e^x$ (or $2^x$ ), solve the quadratic, reject $u \le 0$

Example: $e^{2x} - 2e^x - 3 = 0$ : $u = 3$ kept, $u = -1$ rejected, $x = \ln 3$
If exponentials with unrelated bases → take $\ln$ of both sides and collect the $x$ terms

Example: $5^x = 2^{x+1}$ : $x = \frac{\ln 2}{\ln 5 - \ln 2}$
If a sum or difference of logarithms → write the domain, combine into one log, exponentiate, test each candidate

Example: $\log_3 x + \log_3(x - 8) = 2$ : $x = 9$ , $x = -1$ rejected
If $\sin x = c$ , $\cos x = c$ or $\tan x = c$ on an interval → one value from the inverse function, the others by symmetry and period

Example: $\sin x = \frac{1}{3}$ on $[0, 2\pi]$ : $\arcsin\frac{1}{3}$ and $\pi - \arcsin\frac{1}{3}$

The last step of every branch is the same: check each candidate against the domain of the ORIGINAL equation, and say why a rejected one is rejected.

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.

Solving a logarithmic equation

When to use it: Any equation with the unknown inside one or several logarithms

1 Write the domain of the ORIGINAL equation first: each argument of a logarithm strictly positive.
2 Combine the logarithms with the laws, naming them, until one logarithm equals one logarithm or a constant.
3 Remove the logarithm by exponentiating, or by the one-to-one property of $\ln$ , and solve the algebraic equation.
4 Test every candidate against the domain of step 1; reject the ones outside it, with the reason.
5 Substitute the kept solution into the original equation as a final check.

Concluding sentence

“Domain: $x > 8$ . By the product law, $\log_3(x(x - 8)) = 2$ , so $x^2 - 8x - 9 = 0$ and $x = 9$ or $x = -1$ . Since $-1 \le 8$ , $x = -1$ is rejected. Check: $\log_3 9 + \log_3 1 = 2$ . The solution is $x = 9$ .”

The trap: Writing the domain at the end, from the combined equation: $\log_3(x(x - 8))$ accepts $x = -1$ , the original does not.

Marking: Typically 2 marks for the domain, 3 for combining and exponentiating, 3 for solving, 2 for the rejection with its reason.

Check before you hand in

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

The answer is in the range of the inverse

An arcsin or arctan lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , an arccos in $[0, \pi]$ , a logarithm's argument is positive. Compare before you box it.

$\arcsin(\sin 5) = 5$ is impossible, since $5 > \frac{\pi}{2}$ ; the answer is $5 - 2\pi$ .
Substitute each solution back

Put every kept solution into the ORIGINAL equation, not the combined one. It takes ten seconds and catches both a wrong root and a forgotten rejection.

$x = -2$ satisfies $x^2 = x + 6$ , but $2\ln(-2)$ does not exist: the original equation refuses it.
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.

$f(x) = \frac{2x + 1}{x - 3}$ : $f(4) = 9$ , and $f^{-1}(9) = \frac{28}{7} = 4$ .
The sign of a logarithm

$\ln x < 0$ exactly when $0 < x < 1$ . Bracket a logarithm by powers of its base to know its size without a calculator.

$\log_2 5$ lies between $2$ and $3$ since $4 < 5 < 8$ ; an answer of $0.7$ is wrong at a glance.

The typical problem, taken apart

A log equation with a root that the merged logarithm invents

Solve $\ln(x - 2) + \ln(x + 1) = 2\ln 2$ .

No calculator. Every step must be justified as on a MATH 140 midterm.

The original left side exists for

x > 2

only and meets

y = \ln 4

once, at

x = 3

; the merged

\ln((x - 2)(x + 1))

has a second branch that meets it again at

x = -2

Step 1

Domain: $x - 2 > 0$ and $x + 1 > 0$ , so $x > 2$ .

Why

The domain comes from the ORIGINAL equation, before any law is used; it is the test that the last step will need, and it is worth a mark on its own.

Step 2

By the product law, legal since both arguments are positive on the domain: $\ln((x - 2)(x + 1)) = \ln 4$ , because $2\ln 2 = \ln 2^2 = \ln 4$ .

Why

Turning the constant into a logarithm lets the one-to-one property of $\ln$ remove both logarithms at once. The merged left side is defined on a LARGER set, $x < -1$ or $x > 2$ , and that is where an extra root can hide.

Step 3

$\ln$ is one-to-one, so $(x - 2)(x + 1) = 4$ , that is $x^2 - x - 6 = 0$ , $(x - 3)(x + 2) = 0$ : $x = 3$ or $x = -2$ .

Why

Naming the one-to-one property justifies dropping the logarithms. The algebra is correct and gives two candidates; it cannot know which one the original equation accepts.

Step 4

$x = 3 > 2$ is kept. $x = -2$ is rejected: $\ln(-2 - 2) = \ln(-4)$ is undefined.

Why

The rejection sentence, with its reason, is the mark most students lose. $x = -2$ does solve the merged equation, $\ln((-4)(-1)) = \ln 4$ : the figure shows it on the dashed branch.

Step 5

Check: $\ln(3 - 2) + \ln(3 + 1) = \ln 1 + \ln 4 = \ln 4 = 2\ln 2$ .

Why

Substituting into the ORIGINAL equation closes the question and catches any slip in the factorization.

The conclusion, written out

“On the domain $x > 2$ , the equation becomes $(x - 2)(x + 1) = 4$ , whose roots are $3$ and $-2$ . Since $-2 \le 2$ , it is rejected; the only solution is $x = 3$ .”

The classic mistake on this problem: Answering $x = 3$ or $x = -2$ ; or rejecting $-2$ with no reason given, which earns the answer but not the method mark.

Learn by heart

• Only a one-to-one function has an inverse; domain of $f^{-1}$ = range of $f$ , and the graph reflects in $y = x$ .
• $\ln(ab) = \ln a + \ln b$ , $\ln a^r = r\ln a$ : for POSITIVE $a, b$ only. No law for $\ln(a + b)$ .
• $\ln(x^2) = 2\ln|x|$ ; $\frac{\ln a}{\ln b} = \log_b a$ ; $e^{\ln x} = x$ for $x > 0$ .
• Equation: domain first, algebra, then test every candidate in the ORIGINAL equation.
• $\arcsin \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , $\arccos \in [0, \pi]$ , $\arctan \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .
• $\arcsin(\sin x) = x$ ONLY on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; $\cos(\arcsin x) = \sqrt{1 - x^2}$ .
• $\sin x = c$ : $\arcsin c$ and $\pi - \arcsin c$ ; $\tan x = c$ : $\arctan c + k\pi$ .

Frequently asked questions

Why do I have to reject some solutions of a logarithmic equation?

Because combining logarithms enlarges the domain. The sum ln of x plus ln of x minus 8 exists only when x is greater than 8, but the single logarithm of their product also exists when both factors are negative. The algebra then produces a candidate like x equals minus 1 that solves the combined equation and not the original one. Write the domain first and test each candidate against it.

Is arcsin of sin x always equal to x?

No. Arcsin always returns an angle between minus pi over 2 and pi over 2, so arcsin of sin x equals x only when x is already in that interval. For x between pi over 2 and 3 pi over 2 it equals pi minus x, so arcsin of sin of 2 pi over 3 is pi over 3. In the other order, sin of arcsin x equals x for every x between minus 1 and 1.

How do I find the inverse of a function in MATH 140?

First check that the function is one-to-one, by algebra or with the horizontal line test, and restrict its domain if it is not. Then write y equals f of x, solve for x, and rename the variables. The domain of the inverse is the range of the original function. Finish by checking one value: f of an easy input, then the inverse of that output, must give the input back.

What is the difference between sin inverse of x and 1 over sin x?

Sin inverse of x is arcsin x, the angle between minus pi over 2 and pi over 2 whose sine is x, defined only for x between minus 1 and 1. One over sin x is the cosecant, defined wherever sin x is not zero. At x equals 2, arcsin 2 does not exist while one over sin 2 does. The exponent minus one on a function name means the inverse, never the reciprocal.

Can I use a calculator for logarithms on the MATH 140 exam?

No, the midterm and final are written without a calculator, so answers stay exact: ln 3, log base 2 of 5, pi over 3. When a question asks for a size, bracket the value by powers of the base: log base 2 of 5 is between 2 and 3 because 5 is between 4 and 8. Known values such as ln 2 close to 0.69 only serve to check an order of magnitude.

Practise it

Corrected exercises: Exponential, inverse, log and inverse trig functions, MATH 140 at McGill

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, graphs and transformations Next sheet Limits and the limit laws

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-inverse-exponential-log. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: exponential, inverse, logarithmic and inverse trigonometric functions (MATH 140)

The essentials

An inverse exists where the function is one-to-one, and swaps domain and range

Inverse trigonometric functions: the restricted ranges decide everything

The rules in table form

The laws of logarithms, and the ones that do not exist

The mistakes that cost marks

1. Keeping every root of a combined log equation

2. Replacing ln of x squared by 2 ln x

3. Taking the logarithm of a sum term by term

4. Giving an arcsin answer outside its range

5. Stopping at the calculator's one solution

6. Trusting the triangle for the sign

7. Writing the reciprocal for the inverse

8. Inverting a function that is not one-to-one

9. Accepting a negative base

Which method to choose

Which method, by the FORM of the equation

How the answer is expected to be written

Solving a logarithmic equation

Check before you hand in

The typical problem, taken apart

A log equation with a root that the merged logarithm invents

Learn by heart

Frequently asked questions

Why do I have to reject some solutions of a logarithmic equation?

Is arcsin of sin x always equal to x?

How do I find the inverse of a function in MATH 140?

What is the difference between sin inverse of x and 1 over sin x?

Can I use a calculator for logarithms on the MATH 140 exam?

Corrected exercises: Exponential, inverse, log and inverse trig functions, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?