Exercise 1: Exponential functions: the laws of exponents and a curve read from two points

An exponential function is $f(x) = b^x$ with a base $b > 0$ , $b \ne 1$ : the VARIABLE is in the exponent. For $b > 1$ it increases, for $0 < b < 1$ it decreases, it is defined for every real $x$ and its values are always positive. The laws, for $b > 0$ and all real $x, y$ : $b^{x+y} = b^x b^y$ , $b^{x-y} = \frac{b^x}{b^y}$ , $(b^x)^y = b^{xy}$ , $(ab)^x = a^x b^x$ .

The figure shows the graph of an exponential function $f(x) = C b^x$ passing through the two marked points. No calculator: every answer is exact.

a) Simplify $\frac{8^{x} \cdot 4^{1-x}}{2^{x+3}}$ , then $\frac{(e^{2x})^3\, e^{-x}}{\sqrt{e^{4x}}}$ and $\left(\frac{1}{9}\right)^{-x/2}$ .
b) Starting from the graph of $y = 3^x$ , describe in order the transformations that give $g(x) = 2 - 3^{-x}$ . State the domain and range of $g$ , its $y$ -intercept, and the horizontal line its graph approaches.
c) Using the figure, find $C$ and $b$ , then $f(-1)$ .
d) Which of these are exponential functions: $(-2)^x$ , $1^x$ , $\left(\frac{1}{2}\right)^x$ , $x^2$ , $\pi^x$ , $e^{-x}$ ? Justify each answer in one line.
e) True for every real $x$ , or false? (i) $2^x + 2^x = 4^x$ ; (ii) $(2^x)^3 = 2^{x^3}$ ; (iii) $3^x \cdot 3^x = 9^x$ ; (iv) $e^{-x} = -e^x$ .

Show the solution

Answers

a) $\frac{1}{2}$ (a constant); $e^{3x}$ ; $3^x$
b) Reflect in the $y$ -axis, then in the $x$ -axis, then shift up $2$ . Domain $\mathbb{R}$ , range $(-\infty, 2)$ , $y$ -intercept $1$ , line $y = 2$ .
c) $C = 3$ , $b = 2$ ( $b = -2$ rejected), $f(-1) = \frac{3}{2}$
d) Yes: $\left(\frac{1}{2}\right)^x$ , $\pi^x$ , $e^{-x}$ . No: $(-2)^x$ (negative base), $1^x$ (constant), $x^2$ (variable in the base).
e) (i) false, $2^x + 2^x = 2^{x+1}$ ; (ii) false, $2^{3x}$ ; (iii) true; (iv) false, $e^{-x} > 0 > -e^x$ .

a) Write every base as a power of $2$ BEFORE applying a law: $8^x = 2^{3x}$ and $4^{1-x} = 2^{2-2x}$ . Then $\frac{2^{3x} \cdot 2^{2-2x}}{2^{x+3}} = 2^{3x + 2 - 2x - x - 3} = 2^{-1} = \frac{1}{2}$ . The $x$ disappears completely: the expression is the constant $\frac{1}{2}$ , which you can check at $x = 0$ : $\frac{1 \cdot 4}{8} = \frac{1}{2}$ . Next, $(e^{2x})^3 = e^{6x}$ and $\sqrt{e^{4x}} = (e^{4x})^{1/2} = e^{2x}$ , so the quotient is $e^{6x - x - 2x} = e^{3x}$ . Finally $\left(\frac{1}{9}\right)^{-x/2} = 9^{x/2} = (3^2)^{x/2} = 3^x$ . The error to avoid is adding exponents of DIFFERENT bases: $8^x \cdot 4^{1-x}$ is not $32^{1}$ , the laws only combine powers of the same base.

b) $3^{-x}$ is $3^x$ with $x$ replaced by $-x$ : a reflection in the $y$ -axis, giving the decreasing curve $\left(\frac{1}{3}\right)^x$ . Then $-3^{-x}$ reflects that in the $x$ -axis, and $2 - 3^{-x}$ shifts it up by $2$ . The order matters for the last two: shifting first and reflecting second would give $-(3^{-x} + 2) = -2 - 3^{-x}$ , another function. Domain: all reals, since $3^{-x}$ is defined everywhere. Range: $3^{-x}$ takes every value in $(0, \infty)$ , so $-3^{-x}$ takes every value in $(-\infty, 0)$ and $g$ every value in $(-\infty, 2)$ ; the value $2$ itself is never reached, because $3^{-x}$ is never $0$ . The $y$ -intercept is $g(0) = 2 - 1 = 1$ . As $x$ grows, $3^{-x}$ becomes as small as we like, so the graph approaches the horizontal line $y = 2$ from below without ever touching it. Check two points: $g(-1) = 2 - 3 = -1$ and $g(1) = 2 - \frac{1}{3} = \frac{5}{3}$ , an increasing function, as the two reflections predict.

c) The two points give $C b = 6$ and $C b^3 = 24$ . Divide the second equation by the first to eliminate $C$ : $b^2 = 4$ . The algebra offers $b = 2$ and $b = -2$ ; the base of an exponential function must be positive, so $b = -2$ is rejected, and $b = 2$ . Then $C = \frac{6}{2} = 3$ , so $f(x) = 3 \cdot 2^x$ , and $f(-1) = \frac{3}{2}$ . The figure agrees: the curve crosses the vertical axis at $f(0) = C = 3$ . Two common errors: joining the points by a line, which gives slope $9$ and $f(-1) = -12$ , a negative value that no exponential takes; and dividing the $y$ -values as if the $x$ -gap were $1$ , which gives $b = 4$ . The gap is $2$ , so the ratio $4$ is $b^2$ .

d) $(-2)^x$ is NOT an exponential function: $(-2)^{1/2} = \sqrt{-2}$ is not a real number, so it is not defined on an interval. $1^x = 1$ for every $x$ is a constant, excluded by the condition $b \ne 1$ (it would have no inverse, which is the point of the chapter). $\left(\frac{1}{2}\right)^x = 2^{-x}$ is exponential, with base in $(0, 1)$ , hence decreasing. $x^2$ is a POWER function: the variable is in the base, not in the exponent. $\pi^x$ is exponential with base $\pi > 1$ , increasing. $e^{-x} = \left(\frac{1}{e}\right)^x$ is exponential with base $\frac{1}{e} \in (0, 1)$ . The test is always the same: where is the variable, and is the base a fixed positive number different from $1$ ?

e) (i) FALSE: $2^x + 2^x = 2 \cdot 2^x = 2^{x+1}$ , while $4^x = 2^{2x}$ ; they agree only when $x + 1 = 2x$ , that is at $x = 1$ . At $x = 2$ : $8 \ne 16$ . There is no law for the SUM of two powers. (ii) FALSE: $(2^x)^3 = 2^{3x}$ , the exponents multiply; at $x = 2$ , $2^6 = 64$ while $2^{x^3} = 2^8 = 256$ . (iii) TRUE: $3^x \cdot 3^x = 3^{2x} = (3^2)^x = 9^x$ , or directly $(ab)^x = a^x b^x$ with $a = b = 3$ . (iv) FALSE for every $x$ : $e^{-x} = \frac{1}{e^x}$ is always positive, $-e^x$ always negative. A minus sign in the exponent means a RECIPROCAL, never a negative value; this is the error that makes students answer that $e^{-x}$ has negative values on its graph.

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Exercise 2: One-to-one functions and the inverse, with its domain

A function is one-to-one when it never takes the same value twice: $f(a) = f(b)$ forces $a = b$ . On a graph, no horizontal line meets the curve more than once. Only a one-to-one function has an inverse $f^{-1}$ , defined by $f^{-1}(y) = x \iff f(x) = y$ . The domain of $f^{-1}$ is the range of $f$ , and its range is the domain of $f$ . Method: write $y = f(x)$ , solve for $x$ , then rename.

The figure shows the graph of $f(x) = \sqrt{x + 3}$ on the interval $[-3, 6]$ , with four points marked.

a) Which of these are one-to-one? $f_1(x) = x^3 + 2$ , $f_2(x) = x^2 - 4x$ , $f_3(x) = |x - 1|$ , $f_4(x) = \frac{1}{x}$ . Prove it algebraically, or give two inputs with the same output.
b) Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$ , with its domain and range, and check it on one value.
c) The function $f_2$ of a) is not one-to-one. Restrict it to $[2, \infty)$ and find the inverse of the restriction. What is the inverse of the restriction to $(-\infty, 2]$ ?
d) For the function of the figure, read $f^{-1}(2)$ and $f^{-1}(0)$ , give the domain and range of $f^{-1}$ , and find a formula for $f^{-1}$ .
e) Sketch $f^{-1}$ by reflection, and compute $f^{-1}(f(x))$ and $f(f^{-1}(x))$ . On which intervals does each equal $x$ ?

Type your answers, the page tells you right or wrong 0/16

a)

f_1

one-to-one

f_2

one-to-one

f_3

one-to-one

f_4

one-to-one

b)

Domain of

f^{-1}

: excluded value

Range of

f^{-1}

: excluded value

f^{-1}(9)

c)

Domain of the inverse ,

Inverse of the restriction to

[2, \infty)

, at

5

Inverse of the restriction to

(-\infty, 2]

, at

5

d)

f^{-1}(2)

f^{-1}(0)

Domain of

f^{-1}

,

Range of

f^{-1}

,

e)

f^{-1}(f(x)) = x

on ,

f(f^{-1}(x)) = x

on ,

Show the solution

Answers

a) $f_1$ and $f_4$ are one-to-one; $f_2(0) = f_2(4) = 0$ and $f_3(0) = f_3(2) = 1$ are not.
b) $f^{-1}(x) = \frac{3x + 1}{x - 2}$ , domain $x \ne 2$ , range $y \ne 3$ ; $f(4) = 9$ and $f^{-1}(9) = 4$
c) $2 + \sqrt{x + 4}$ on $[-4, \infty)$ ; on $(-\infty, 2]$ the inverse is $2 - \sqrt{x + 4}$
d) $f^{-1}(2) = 1$ , $f^{-1}(0) = -3$ ; domain $[0, 3]$ , range $[-3, 6]$ ; $f^{-1}(x) = x^2 - 3$ for $0 \le x \le 3$
e) $f^{-1}(f(x)) = x$ on $[-3, 6]$ ; $f(f^{-1}(x)) = |x| = x$ on $[0, 3]$

a) $f_1$ : if $a^3 + 2 = b^3 + 2$ then $a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 0$ . The second factor equals $\left(a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}$ , which is $0$ only when $a = b = 0$ ; so in every case $a = b$ , and $f_1$ is one-to-one. $f_2(x) = x(x - 4)$ gives $f_2(0) = f_2(4) = 0$ : two inputs, one output, NOT one-to-one; the horizontal line $y = 0$ meets its parabola twice. $f_3(0) = f_3(2) = 1$ : not one-to-one. $f_4$ : $\frac{1}{a} = \frac{1}{b}$ gives $a = b$ directly, so $f_4$ is one-to-one although its graph is in two pieces. One counterexample settles a no; a yes needs the algebra for ALL inputs, and a few matching values prove nothing.

b) Write $y = \frac{2x + 1}{x - 3}$ and solve for $x$ : $y(x - 3) = 2x + 1$ , so $xy - 2x = 3y + 1$ , $x(y - 2) = 3y + 1$ and $x = \frac{3y + 1}{y - 2}$ , possible only when $y \ne 2$ . Renaming, $f^{-1}(x) = \frac{3x + 1}{x - 2}$ . Its domain is $x \ne 2$ , which is the range of $f$ (the value $2$ is never taken: $\frac{2x + 1}{x - 3} = 2$ would give $1 = -6$ ). Its range is $y \ne 3$ , the domain of $f$ . Check: $f(4) = \frac{9}{1} = 9$ and $f^{-1}(9) = \frac{28}{7} = 4$ . The step students skip is collecting the $x$ terms on ONE side before factoring; without it they end with $x$ on both sides.

c) Complete the square: $x^2 - 4x = (x - 2)^2 - 4$ . On $[2, \infty)$ the function increases from $-4$ , so its range is $[-4, \infty)$ . Solve $y = (x - 2)^2 - 4$ : $(x - 2)^2 = y + 4$ , so $x - 2 = \pm\sqrt{y + 4}$ . Here the domain decides the sign: $x \ge 2$ means $x - 2 \ge 0$ , so $x = 2 + \sqrt{y + 4}$ , and the inverse is $x \mapsto 2 + \sqrt{x + 4}$ on $[-4, \infty)$ . On $(-\infty, 2]$ , $x - 2 \le 0$ and the other sign is forced: $2 - \sqrt{x + 4}$ . Checks: $f_2(5) = 5$ and $2 + \sqrt 9 = 5$ ; $f_2(-1) = 5$ and $2 - \sqrt 9 = -1$ . Writing $2 \pm \sqrt{x + 4}$ is not an answer: it gives two outputs for one input, so it is not a function.

d) Read the figure backwards: $f^{-1}(2)$ is the input whose output is $2$ , and the marked point $(1, 2)$ gives $f^{-1}(2) = 1$ . Likewise $(-3, 0)$ gives $f^{-1}(0) = -3$ . The range of $f$ on $[-3, 6]$ is $[0, 3]$ , from $f(-3) = 0$ to $f(6) = 3$ , so $f^{-1}$ has domain $[0, 3]$ and range $[-3, 6]$ . Formula: $y = \sqrt{x + 3}$ gives $y^2 = x + 3$ with $y \ge 0$ , so $f^{-1}(x) = x^2 - 3$ for $0 \le x \le 3$ . The restriction is part of the answer: $x^2 - 3$ on all of $\mathbb{R}$ is a parabola that is not one-to-one, and it is not the inverse of anything.

e) Each point $(a, b)$ of $f$ gives the point $(b, a)$ of $f^{-1}$ : $(0, -3)$ , $(1, -2)$ , $(2, 1)$ , $(3, 6)$ , and the graph of $f^{-1}$ is the mirror image of $f$ in the line $y = x$ (figure of the solution, same scale on both axes, otherwise the mirror is distorted). For $x \in [-3, 6]$ : $f^{-1}(f(x)) = (\sqrt{x + 3})^2 - 3 = x$ . For $x \in [0, 3]$ : $f(f^{-1}(x)) = \sqrt{x^2 - 3 + 3} = \sqrt{x^2} = |x| = x$ because $x \ge 0$ . Outside $[0, 3]$ the formula would give $\sqrt{x^2} = |x|$ , which is $-x$ for negative $x$ : at $x = -2$ , $\sqrt{(-2)^2} = 2$ . The cancellation equations hold only on the right domains, and that is the thread of the whole chapter.

Exercise 3: Logarithms: exact values, the laws, and the domain they hide

$\log_b x$ is the inverse of $b^x$ : $\log_b x = y \iff b^y = x$ , defined only for $x > 0$ . The natural logarithm is $\ln x = \log_e x$ . The laws, valid for $x > 0$ and $y > 0$ ONLY: $\log_b(xy) = \log_b x + \log_b y$ , $\log_b \frac{x}{y} = \log_b x - \log_b y$ , $\log_b(x^r) = r \log_b x$ . Change of base: $\log_b x = \frac{\ln x}{\ln b}$ .

The figure shows the graph of $y = \ln(x^2 - 4x)$ with the vertical line $x = 4$ .

a) Find exactly: $\log_2 \frac{1}{32}$ , $\log_9 27$ , $\ln \frac{1}{\sqrt e}$ , $e^{3 \ln 2}$ , $\log_{10} 0.001$ .
b) Expand $\ln \frac{x^3 \sqrt{x + 1}}{e^2 (x - 2)}$ as a sum of simple terms, and say for which $x$ the expansion is valid.
c) Compute $\log_2 40 - \log_2 5$ and $\ln 12 - 2\ln 2 + \ln \frac{1}{3}$ , then write $3 \ln x - \frac{1}{2}\ln(x^2 + 1) + 1$ as a single logarithm.
d) A student rewrites $\ln(x^2 - 4x)$ as $\ln x + \ln(x - 4)$ . Find the domain of each expression, and explain which part of the figure the student's formula loses. Write an identity valid on the whole domain.
e) Compute $\log_3 5 \cdot \log_5 9$ and $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{31} 32$ . Between which two consecutive integers is $\log_2 7$ ?

Type your answers, the page tells you right or wrong 0/15

a)

\log_2 \frac{1}{32}

\log_9 27

\ln \frac{1}{\sqrt e}

e^{3 \ln 2}

\log_{10} 0.001

b)

Constant term

Valid for

x

in ,

c)

\log_2 40 - \log_2 5

\ln 12 - 2\ln 2 + \ln \frac{1}{3}

d)

Domain of

\ln(x^2 - 4x)

: first interval ,

Domain of

\ln(x^2 - 4x)

: second interval ,

Domain of

\ln x + \ln(x - 4)

,

e)

\log_3 5 \cdot \log_5 9

The long product

\log_2 7

lies in

(n, n + 1)

:

n =

Show the solution

Answers

a) $-5$ , $\frac{3}{2}$ , $-\frac{1}{2}$ , $8$ , $-3$
b) $3\ln x + \frac{1}{2}\ln(x + 1) - 2 - \ln(x - 2)$ , valid for $x > 2$
c) $3$ ; $0$ ; $\ln \frac{e x^3}{\sqrt{x^2 + 1}}$ for $x > 0$
d) $\ln(x^2 - 4x)$ : $x < 0$ or $x > 4$ ; $\ln x + \ln(x - 4)$ : $x > 4$ only, the left branch is lost. $\ln(x^2 - 4x) = \ln|x| + \ln|x - 4|$ .
e) $2$ ; $5$ ; $2 < \log_2 7 < 3$

a) Each value is a question about an exponent. $\log_2 \frac{1}{32}$ : $2^y = 2^{-5}$ , so $-5$ . $\log_9 27$ : $9^y = 27$ means $3^{2y} = 3^3$ , so $y = \frac{3}{2}$ . $\ln \frac{1}{\sqrt e} = \ln e^{-1/2} = -\frac{1}{2}$ . $e^{3\ln 2} = (e^{\ln 2})^3 = 2^3 = 8$ , by the cancellation $e^{\ln a} = a$ for $a > 0$ . $\log_{10} 0.001 = \log_{10} 10^{-3} = -3$ . A negative logarithm is normal: it says the number is between $0$ and $1$ . What does not exist is the logarithm OF a negative number.

b) For $x > 2$ every factor is positive, and the laws apply one by one: $\ln x^3 + \ln \sqrt{x + 1} - \ln e^2 - \ln(x - 2) = 3\ln x + \frac{1}{2}\ln(x + 1) - 2 - \ln(x - 2)$ . The condition matters, because the left side has a larger domain than the right. For $-1 < x < 0$ the numerator $x^3\sqrt{x + 1}$ and the denominator $e^2(x - 2)$ are both negative, so the quotient is positive: at $x = -\frac{1}{2}$ it equals $\frac{(-1/8)\sqrt{1/2}}{e^2(-5/2)} > 0$ , the left side exists, and $\ln x$ on the right does not. The expansion is therefore valid for $x > 2$ only, where every piece exists; announcing that condition is worth the mark here.

c) $\log_2 40 - \log_2 5 = \log_2 8 = 3$ . $\ln 12 - 2\ln 2 + \ln \frac{1}{3} = \ln \frac{12}{4 \cdot 3} = \ln 1 = 0$ . For the last one, turn every term into a logarithm first, including the constant: $1 = \ln e$ . Then $3 \ln x - \frac{1}{2}\ln(x^2 + 1) + \ln e = \ln \frac{e x^3}{\sqrt{x^2 + 1}}$ , valid for $x > 0$ (the original needs $\ln x$ ). Forgetting that $1 = \ln e$ and writing $\ln \frac{x^3}{\sqrt{x^2+1}} + 1$ is not wrong, but it is not a single logarithm, and the question loses its mark.

d) $\ln(x^2 - 4x)$ needs $x(x - 4) > 0$ : both factors positive ( $x > 4$ ) or both negative ( $x < 0$ ). Its domain is $(-\infty, 0) \cup (4, \infty)$ : the two branches of the figure. The student's $\ln x + \ln(x - 4)$ needs $x > 0$ AND $x > 4$ , that is $x > 4$ only. The two expressions agree on $(4, \infty)$ , but the student's formula has thrown away the whole left branch: at $x = -1$ , $\ln(1 + 4) = \ln 5$ exists while $\ln(-1)$ does not. The law $\ln(ab) = \ln a + \ln b$ needs $a > 0$ AND $b > 0$ ; when only the product is positive, write $\ln(ab) = \ln|a| + \ln|b|$ , here $\ln|x| + \ln|x - 4|$ , valid on the whole domain.

e) By change of base, $\log_3 5 \cdot \log_5 9 = \frac{\ln 5}{\ln 3} \cdot \frac{\ln 9}{\ln 5} = \frac{\ln 9}{\ln 3} = \frac{2\ln 3}{\ln 3} = 2$ . The long product telescopes the same way: $\frac{\ln 3}{\ln 2} \cdot \frac{\ln 4}{\ln 3} \cdots \frac{\ln 32}{\ln 31} = \frac{\ln 32}{\ln 2} = \log_2 32 = 5$ . Finally, $\log_2$ is increasing and $4 < 7 < 8$ , so $\log_2 4 < \log_2 7 < \log_2 8$ , that is $2 < \log_2 7 < 3$ . The exact value is $\frac{\ln 7}{\ln 2}$ , and bracketing it by powers of the base is the no-calculator way to know its size.

Exercise 4: Exponential and logarithmic equations: solve, then check the domain

The method never changes: isolate the exponential or the logarithm, apply the inverse function to both sides ( $\ln$ undoes $e^x$ , $b^x$ undoes $\log_b$ ), solve, and CHECK every candidate in the ORIGINAL equation. The laws of logarithms and squaring can create candidates that the original equation refuses; they must be named and rejected.

No calculator: leave answers as exact expressions such as $\ln 3$ or $\frac{\ln 2}{\ln 5 - \ln 2}$ .

a) Solve $9^{x+1} = 27^{2x - 1}$ .
b) Solve $3e^{2x - 1} + 4 = 10$ .
c) Solve $e^{2x} - 2e^x - 3 = 0$ , then $4^x - 6 \cdot 2^x + 8 = 0$ .
d) Solve $\log_3 x + \log_3(x - 8) = 2$ , then $2\ln x = \ln(x + 6)$ .
e) Solve $5^x = 2^{x+1}$ , and show without a calculator that the solution lies between $0$ and $1$ .

Show the solution

Answers

a) $x = \frac{5}{4}$
b) $x = \frac{1 + \ln 2}{2}$
c) $x = \ln 3$ ( $e^x = -1$ rejected); $x = 1$ or $x = 2$
d) $x = 9$ ( $x = -1$ rejected); $x = 3$ ( $x = -2$ rejected)
e) $x = \frac{\ln 2}{\ln 5 - \ln 2} = \frac{\ln 2}{\ln(5/2)}$ , and $0 < \ln 2 < \ln \frac{5}{2}$ gives $0 < x < 1$

a) Both bases are powers of $3$ : $9^{x+1} = 3^{2x + 2}$ and $27^{2x - 1} = 3^{6x - 3}$ . Since $3^x$ is one-to-one, equal powers have equal exponents: $2x + 2 = 6x - 3$ , so $x = \frac{5}{4}$ . Check: $9^{9/4} = 3^{9/2}$ and $27^{3/2} = 3^{9/2}$ . The sentence that justifies the step is the one-to-one property; writing $x + 1 = 2x - 1$ (equating exponents of DIFFERENT bases) gives $x = 2$ , and $9^3 = 729 \ne 27^3$ .

b) Isolate the exponential first: $3e^{2x - 1} = 6$ , so $e^{2x - 1} = 2$ . Now apply $\ln$ : $2x - 1 = \ln 2$ and $x = \frac{1 + \ln 2}{2}$ . The frequent error is to take the logarithm too early, $\ln(3e^{2x-1} + 4) = \ln 10$ , and then to split the left side as $\ln 3 + (2x - 1) + \ln 4$ : that uses a law for $\ln$ of a SUM, which does not exist. The logarithm goes on only once the exponential stands alone.

c) Put $u = e^x$ , so $e^{2x} = u^2$ : $u^2 - 2u - 3 = (u - 3)(u + 1) = 0$ , $u = 3$ or $u = -1$ . Now go back to $x$ : $e^x = 3$ gives $x = \ln 3$ , and $e^x = -1$ has NO solution because $e^x > 0$ for every $x$ . One solution, $x = \ln 3$ . For the second, $u = 2^x$ gives $u^2 - 6u + 8 = (u - 2)(u - 4) = 0$ , so $2^x = 2$ or $2^x = 4$ : $x = 1$ or $x = 2$ , and both are kept, since both values of $u$ are positive. Rejecting is not a ritual: a candidate goes only when it violates a condition, and saying which condition is the mark.

d) First the domain: $\log_3 x$ needs $x > 0$ and $\log_3(x - 8)$ needs $x > 8$ , so $x > 8$ . Combine: $\log_3(x(x - 8)) = 2$ , so $x^2 - 8x = 9$ and $(x - 9)(x + 1) = 0$ . $x = 9$ is in the domain: $\log_3 9 + \log_3 1 = 2 + 0 = 2$ , correct. $x = -1$ is rejected, because $\log_3(-1)$ does not exist; it solves the COMBINED equation, $\log_3((-1)(-9)) = \log_3 9$ , which is exactly how combining created it. Second equation, domain $x > 0$ : $2\ln x = \ln x^2$ , so $x^2 = x + 6$ , $(x - 3)(x + 2) = 0$ . $x = 3$ works: $2\ln 3 = \ln 9$ . $x = -2$ satisfies $x^2 = x + 6$ but $2\ln(-2)$ is undefined: rejected. Two different equations, the same lesson: the domain is found BEFORE the algebra, and it decides at the end.

e) The bases differ and are not powers of a common base, so take $\ln$ of both sides (both are positive): $x \ln 5 = (x + 1)\ln 2$ . Collect the $x$ terms: $x(\ln 5 - \ln 2) = \ln 2$ , so $x = \frac{\ln 2}{\ln 5 - \ln 2} = \frac{\ln 2}{\ln(5/2)}$ , allowed since $\ln \frac{5}{2} \ne 0$ . Size without a calculator: $1 < 2 < \frac{5}{2}$ and $\ln$ is increasing, so $0 < \ln 2 < \ln \frac{5}{2}$ , and the quotient is between $0$ and $1$ . Another view: the equation is $\left(\frac{5}{2}\right)^x = 2$ , and $2$ lies between $\left(\frac{5}{2}\right)^0 = 1$ and $\left(\frac{5}{2}\right)^1 = \frac{5}{2}$ . With $\ln 2 \approx 0.69$ and $\ln \frac{5}{2} \approx 0.92$ , $x \approx 0.75$ , but the exact expression is the answer.

Exercise 5: Inverse trigonometric functions: exact values, restricted ranges and the triangle

Sine, cosine and tangent are not one-to-one, so each is restricted before being inverted. $\arcsin$ (Stewart writes $\sin^{-1}$ ) goes from $[-1, 1]$ to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; $\arccos$ goes from $[-1, 1]$ to $[0, \pi]$ ; $\arctan$ goes from $\mathbb{R}$ to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . An answer outside these ranges is wrong even if its sine, cosine or tangent is right.

The figure shows the right triangle attached to $\theta = \arcsin x$ when $0 < x < 1$ : hypotenuse $1$ , side opposite $\theta$ equal to $x$ .

a) Find exactly: $\arcsin\left(-\frac{\sqrt 3}{2}\right)$ , $\arccos\left(-\frac{1}{2}\right)$ , $\arctan\left(-\frac{1}{\sqrt 3}\right)$ , $\arccos\left(-\frac{\sqrt 2}{2}\right)$ .
b) Find exactly, or say why it is undefined: $\arcsin\left(\sin \frac{2\pi}{3}\right)$ , $\arccos\left(\cos\left(-\frac{\pi}{4}\right)\right)$ , $\arctan\left(\tan \frac{5\pi}{6}\right)$ , $\sin(\arcsin 0.6)$ , $\cos(\arccos 2)$ .
c) Prove that $\cos(\arcsin x) = \sqrt{1 - x^2}$ for every $x \in [-1, 1]$ , including the negative values of $x$ that the triangle does not show.
d) Find $\sin\left(\arccos\left(-\frac{3}{5}\right)\right)$ and $\tan\left(\arccos\left(-\frac{3}{5}\right)\right)$ .
e) Find $\sin(2\arctan 3)$ and $\cos\left(2\arcsin \frac{1}{3}\right)$ .

Type your answers, the page tells you right or wrong 0/13

a)

\arcsin\left(-\frac{\sqrt 3}{2}\right)

(in multiples of

\pi

)

\arccos\left(-\frac{1}{2}\right)

(in multiples of

\pi

)

\arctan\left(-\frac{1}{\sqrt 3}\right)

(in multiples of

\pi

)

\arccos\left(-\frac{\sqrt 2}{2}\right)

(in multiples of

\pi

)

b)

\arcsin\left(\sin \frac{2\pi}{3}\right)

(in multiples of

\pi

)

\arccos\left(\cos\left(-\frac{\pi}{4}\right)\right)

(in multiples of

\pi

)

\arctan\left(\tan \frac{5\pi}{6}\right)

(in multiples of

\pi

)

\sin(\arcsin 0.6)

\cos(\arccos 2)

is

d)

\sin\left(\arccos\left(-\frac{3}{5}\right)\right)

\tan\left(\arccos\left(-\frac{3}{5}\right)\right)

e)

\sin(2\arctan 3)

\cos\left(2\arcsin \frac{1}{3}\right)

Show the solution

Answers

a) $-\frac{\pi}{3}$ , $\frac{2\pi}{3}$ , $-\frac{\pi}{6}$ , $\frac{3\pi}{4}$
b) $\frac{\pi}{3}$ , $\frac{\pi}{4}$ , $-\frac{\pi}{6}$ , $0.6$ ; $\cos(\arccos 2)$ undefined
c) $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ gives $\cos\theta \ge 0$ , so $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - x^2}$
d) $\frac{4}{5}$ and $-\frac{4}{3}$
e) $\frac{3}{5}$ and $\frac{7}{9}$

a) Each answer is the unique angle IN THE RANGE with the given value. $\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt 3}{2}$ and $-\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ : $\arcsin\left(-\frac{\sqrt 3}{2}\right) = -\frac{\pi}{3}$ (not $\frac{4\pi}{3}$ , which has the same sine). $\cos \frac{2\pi}{3} = -\frac{1}{2}$ with $\frac{2\pi}{3} \in [0, \pi]$ : $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ (not $-\frac{\pi}{3}$ , which is outside $[0, \pi]$ and has cosine $+\frac{1}{2}$ anyway). $\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt 3}$ : $\arctan\left(-\frac{1}{\sqrt 3}\right) = -\frac{\pi}{6}$ . $\arccos\left(-\frac{\sqrt 2}{2}\right) = \frac{3\pi}{4}$ . The pattern to remember: a negative input gives a negative angle for $\arcsin$ and $\arctan$ , but an angle between $\frac{\pi}{2}$ and $\pi$ for $\arccos$ .

b) $\sin \frac{2\pi}{3} = \frac{\sqrt 3}{2}$ , and the angle of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with that sine is $\frac{\pi}{3}$ : so $\arcsin\left(\sin \frac{2\pi}{3}\right) = \frac{\pi}{3}$ , NOT $\frac{2\pi}{3}$ , which is outside the range of $\arcsin$ . $\cos\left(-\frac{\pi}{4}\right) = \frac{\sqrt 2}{2}$ and $\arccos \frac{\sqrt 2}{2} = \frac{\pi}{4}$ . $\tan \frac{5\pi}{6} = -\frac{1}{\sqrt 3}$ , so $\arctan\left(\tan\frac{5\pi}{6}\right) = -\frac{\pi}{6}$ . In the other order the cancellation always works on the domain: $\sin(\arcsin 0.6) = 0.6$ because $0.6 \in [-1, 1]$ . But $\arccos 2$ does not exist, since no cosine equals $2$ , so $\cos(\arccos 2)$ is undefined, not $2$ . Summary: $\sin(\arcsin x) = x$ for $x \in [-1, 1]$ ; $\arcsin(\sin x) = x$ only for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

c) Let $\theta = \arcsin x$ , so $\sin\theta = x$ and $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . From $\sin^2\theta + \cos^2\theta = 1$ , $\cos\theta = \pm\sqrt{1 - x^2}$ . The range decides the sign: on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ the cosine is $\ge 0$ , so $\cos\theta = +\sqrt{1 - x^2}$ , for EVERY $x \in [-1, 1]$ . The triangle of the figure gives the same answer by Pythagoras, adjacent side $\sqrt{1^2 - x^2}$ and $\cos\theta = \frac{\sqrt{1 - x^2}}{1}$ , but a triangle only exists for $0 < x < 1$ ; the identity and the range cover the negative values and the endpoints too. Check at $x = -\frac{1}{2}$ : $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ , and $\cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt 3}{2} = \sqrt{1 - \frac{1}{4}}$ .

d) Let $\theta = \arccos\left(-\frac{3}{5}\right)$ : $\cos\theta = -\frac{3}{5}$ and $\theta \in [0, \pi]$ , in fact in $\left(\frac{\pi}{2}, \pi\right)$ since the cosine is negative. On $[0, \pi]$ the sine is $\ge 0$ , so $\sin\theta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ . Then $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{4/5}{-3/5} = -\frac{4}{3}$ . A $3$ - $4$ - $5$ triangle drawn without thinking gives $\frac{4}{3}$ : the triangle gives the SIZES, the range of the inverse function gives the SIGNS. An angle in the second quadrant has a negative tangent.

e) Let $\theta = \arctan 3$ , in $\left(0, \frac{\pi}{2}\right)$ since $3 > 0$ . A right triangle with opposite side $3$ and adjacent side $1$ has hypotenuse $\sqrt{10}$ , so $\sin\theta = \frac{3}{\sqrt{10}}$ and $\cos\theta = \frac{1}{\sqrt{10}}$ , both positive in the first quadrant. Then $\sin 2\theta = 2\sin\theta\cos\theta = \frac{6}{10} = \frac{3}{5}$ . For the second, $\phi = \arcsin\frac{1}{3}$ has $\sin\phi = \frac{1}{3}$ , and the double-angle identity in its sine-only form avoids the square root: $\cos 2\phi = 1 - 2\sin^2\phi = 1 - \frac{2}{9} = \frac{7}{9}$ . Choosing the form of $\cos 2\phi$ that uses what you already know is the time saver on an exam.

Exercise 6: An inverse from start to finish: a function trapped between 0 and 6

Let $f(x) = \frac{6}{1 + 2e^{-x}}$ , defined for every real $x$ . Its graph, in the figure, rises from near $0$ to near $6$ . This is the full exam routine on one function: one-to-one, range, inverse, domain of the inverse, and equations solved through the inverse.

Everything is done by algebra. Only the facts of the chapter are allowed: $e^{-x} > 0$ , $e^{-x}$ decreases, and $\ln$ and $e^x$ undo each other.

a) Show that $f$ is one-to-one and increasing, compute $f(0)$ , and prove that $0 < f(x) < 6$ for every $x$ .
b) Find $f^{-1}(x)$ and its domain, and explain why that domain proves that $f$ takes every value between $0$ and $6$ .
c) Check that $f^{-1}(f(0)) = 0$ , then compute $f^{-1}(3)$ and $f^{-1}(1)$ , and check $f^{-1}(3)$ in $f$ .
d) Solve $f(x) = 5$ and $f(x) = 7$ .
e) Find the domain of $h(x) = \ln(f(x) - 3)$ and of $g(x) = \sqrt{\ln(6 - f(x))}$ .

Show the solution

Answers

a) $f(a) = f(b) \Rightarrow e^{-a} = e^{-b} \Rightarrow a = b$ ; $f(0) = 2$ ; $1 + 2e^{-x} > 1$ gives $0 < f(x) < 6$
b) $f^{-1}(x) = \ln \frac{2x}{6 - x}$ , domain $(0, 6)$
c) $f^{-1}(2) = \ln 1 = 0$ ; $f^{-1}(3) = \ln 2$ , $f^{-1}(1) = \ln \frac{2}{5}$ ; $f(\ln 2) = 3$
d) $x = \ln 10$ ; no solution for $7$ (outside the range)
e) $h$ : $x > \ln 2$ ; $g$ : $x \le \ln 10$

a) If $f(a) = f(b)$ , the denominators are equal: $1 + 2e^{-a} = 1 + 2e^{-b}$ , so $e^{-a} = e^{-b}$ , and since the exponential is one-to-one, $-a = -b$ and $a = b$ . When $x$ increases, $e^{-x} = \frac{1}{e^x}$ decreases, so the positive denominator $1 + 2e^{-x}$ decreases and the quotient $f(x)$ increases. $f(0) = \frac{6}{1 + 2} = 2$ , the red point of the figure. Bounds: $e^{-x} > 0$ gives $1 + 2e^{-x} > 1$ , so $0 < \frac{6}{1 + 2e^{-x}} < 6$ . The dashed line $y = 6$ of the figure is a ceiling the graph approaches and never reaches, and it comes from this inequality, not from reading the picture.

b) Write $y = \frac{6}{1 + 2e^{-x}}$ and isolate the exponential one layer at a time: $1 + 2e^{-x} = \frac{6}{y}$ , so $2e^{-x} = \frac{6 - y}{y}$ and $e^{-x} = \frac{6 - y}{2y}$ . Now apply $\ln$ , legal only if the right side is positive: $-x = \ln\frac{6 - y}{2y}$ , so $x = \ln \frac{2y}{6 - y}$ . Renaming, $f^{-1}(x) = \ln \frac{2x}{6 - x}$ . Domain: $\frac{2x}{6 - x} > 0$ , and a sign table of the numerator ( $0$ at $x = 0$ ) and denominator ( $0$ at $x = 6$ ) shows the quotient is positive exactly on $(0, 6)$ . For EVERY $y$ in $(0, 6)$ the computation produced an $x$ with $f(x) = y$ , so $f$ takes every value of $(0, 6)$ : its range is exactly $(0, 6)$ , the domain of $f^{-1}$ . Finding the inverse is how the range is proved.

c) $f^{-1}(f(0)) = f^{-1}(2) = \ln\frac{4}{4} = \ln 1 = 0$ . Next, $f^{-1}(3) = \ln\frac{6}{3} = \ln 2$ , and the check in $f$ : $e^{-\ln 2} = \frac{1}{2}$ , so $f(\ln 2) = \frac{6}{1 + 1} = 3$ . Finally $f^{-1}(1) = \ln\frac{2}{5} = \ln 2 - \ln 5$ , a negative number, which is consistent: $f(0) = 2$ and $f$ is increasing, so the input giving the smaller output $1$ must be negative. Checking an inverse on one value takes twenty seconds and catches a swapped fraction every time.

d) $f(x) = 5$ : since $5$ is in the range $(0, 6)$ , the unique solution is $x = f^{-1}(5) = \ln\frac{10}{1} = \ln 10$ . $f(x) = 7$ : $7$ is outside the range, so there is NO solution. The formula agrees on its own: $f^{-1}(7) = \ln\frac{14}{-1}$ , the logarithm of a negative number, undefined. A student who writes $x = \ln(-14)$ , or worse $x = -\ln 14$ after moving the sign, has solved an equation that has no solution. The domain of $f^{-1}$ is the list of equations $f(x) = c$ that can be solved.

e) $h(x) = \ln(f(x) - 3)$ needs $f(x) > 3$ . Since $f$ is increasing and $f(\ln 2) = 3$ , this happens exactly when $x > \ln 2$ : the domain of $h$ is $(\ln 2, \infty)$ . $g(x) = \sqrt{\ln(6 - f(x))}$ needs two things: $6 - f(x) > 0$ , always true by a), and $\ln(6 - f(x)) \ge 0$ , that is $6 - f(x) \ge 1$ , that is $f(x) \le 5$ . As $f$ is increasing and $f(\ln 10) = 5$ , the domain of $g$ is $(-\infty, \ln 10]$ . Each condition on $f(x)$ became a condition on $x$ through $f^{-1}$ , and the direction of the inequality was kept only because $f$ is INCREASING; for a decreasing function it would flip.

Exercise 7: arcsin(sin x) is not x: the zigzag, and the solution the calculator never shows

The figure shows the graph of $y = \arcsin(\sin x)$ for $-7 \le x \le 7$ , and the dashed line $y = x$ . The two agree on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ only: outside, $\arcsin$ still has to return an angle of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and the graph zigzags.

The same fact explains why an inverse trigonometric function gives ONE solution of an equation like $\sin x = \frac{1}{3}$ , while the equation has two per period.

a) Explain why $\arcsin(\sin x)$ is defined for every real $x$ . Show that $\arcsin(\sin x) = \pi - x$ for $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ , and compute $\arcsin(\sin 3)$ and $\arcsin(\sin 5)$ exactly.
b) Find a formula for $\arccos(\cos x)$ on $[\pi, 2\pi]$ , then compute $\arccos(\cos 4)$ and $\arccos(\cos(-1))$ .
c) Solve $\sin x = \frac{1}{3}$ and $\cos x = -\frac{1}{4}$ on $[0, 2\pi]$ .
d) Solve $\tan x = 2$ on $(-\pi, \pi)$ , and locate each solution between two multiples of $\frac{\pi}{4}$ .
e) Prove that $\arcsin x + \arccos x = \frac{\pi}{2}$ for $x \in [-1, 1]$ . Then prove that $\arctan x + \arctan\frac{1}{x} = \frac{\pi}{2}$ for $x > 0$ , and find its value for $x < 0$ .

Type your answers, the page tells you right or wrong 0/11

a)

\arcsin(\sin 3)

\arcsin(\sin 5)

b)

\arccos(\cos 4)

\arccos(\cos(-1))

c)

\sin x = \frac{1}{3}

: smaller

x

Larger

x

\cos x = -\frac{1}{4}

: smaller

x

Larger

x

d)

Positive solution

Negative solution

e)

\arctan x + \arctan\frac{1}{x}

for

x < 0

(in multiples of

\pi

)

Show the solution

Answers

a) $\sin x \in [-1, 1]$ ; $\sin(\pi - x) = \sin x$ with $\pi - x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; $\pi - 3$ and $5 - 2\pi$
b) $2\pi - x$ ; $2\pi - 4$ and $1$
c) $\arcsin\frac{1}{3}$ and $\pi - \arcsin\frac{1}{3}$ ; $\arccos\left(-\frac{1}{4}\right)$ and $2\pi - \arccos\left(-\frac{1}{4}\right)$
d) $\arctan 2 \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ and $\arctan 2 - \pi \in \left(-\frac{3\pi}{4}, -\frac{\pi}{2}\right)$
e) $\arccos x = \frac{\pi}{2} - \arcsin x$ ; the arctan sum is $\frac{\pi}{2}$ for $x > 0$ and $-\frac{\pi}{2}$ for $x < 0$

a) $\sin x$ always lies in $[-1, 1]$ , the domain of $\arcsin$ , so the composite is defined for every $x$ . For $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ , the angle $\pi - x$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and has the same sine, $\sin(\pi - x) = \sin x$ . The unique angle of that interval with sine $\sin x$ is therefore $\pi - x$ : $\arcsin(\sin x) = \pi - x$ , the descending segment of the figure. Since $\frac{\pi}{2} \approx 1.57 \le 3 \le 4.71 \approx \frac{3\pi}{2}$ , $\arcsin(\sin 3) = \pi - 3$ , a small positive angle. For $5$ : it lies in $\left[\frac{3\pi}{2}, \frac{5\pi}{2}\right]$ , and $5 - 2\pi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has the same sine by periodicity, so $\arcsin(\sin 5) = 5 - 2\pi$ , a NEGATIVE angle. Answering $5$ is impossible, as $\arcsin$ never exceeds $\frac{\pi}{2}$ .

b) On $[\pi, 2\pi]$ , the angle $2\pi - x$ lies in $[0, \pi]$ , the range of $\arccos$ , and $\cos(2\pi - x) = \cos x$ . So $\arccos(\cos x) = 2\pi - x$ there. Since $\pi < 4 < 2\pi$ , $\arccos(\cos 4) = 2\pi - 4$ . For $-1$ : the cosine is even, $\cos(-1) = \cos 1$ , and $1 \in [0, \pi]$ , so $\arccos(\cos(-1)) = 1$ . The method is always the same: find the angle IN THE RANGE of the inverse that shares the value, using symmetry ( $\pi - x$ for sine, $-x$ for cosine) and periodicity ( $2\pi$ ).

c) $\arcsin\frac{1}{3}$ is one solution, in $\left(0, \frac{\pi}{2}\right)$ . The sine is also positive in the second quadrant, and $\sin(\pi - \alpha) = \sin \alpha$ gives the second one, $\pi - \arcsin\frac{1}{3}$ . Both are in $[0, 2\pi]$ , and there are no others. For $\cos x = -\frac{1}{4}$ : $\arccos\left(-\frac{1}{4}\right) \in \left(\frac{\pi}{2}, \pi\right)$ , and the cosine is even, so $-\arccos\left(-\frac{1}{4}\right)$ also works; it is negative, so add $2\pi$ to bring it into the interval: $2\pi - \arccos\left(-\frac{1}{4}\right)$ , in the third quadrant. A calculator key returns one value; an equation on $[0, 2\pi]$ asks for all of them, and a missing solution costs the same as a wrong one.

d) $\tan$ has period $\pi$ , so the solutions are $\arctan 2 + k\pi$ . In $(-\pi, \pi)$ : $\arctan 2$ and $\arctan 2 - \pi$ ( $\arctan 2 + \pi$ is already beyond $\pi$ ). Location: $\tan\frac{\pi}{4} = 1 < 2$ and $\tan$ increases on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so $\frac{\pi}{4} < \arctan 2 < \frac{\pi}{2}$ . Subtracting $\pi$ : $-\frac{3\pi}{4} < \arctan 2 - \pi < -\frac{\pi}{2}$ , in the third quadrant, where the tangent is positive as it should be. Adding $\pi$ for the tangent but $2\pi$ for sine and cosine is the difference in period, and mixing the two is a classic lost mark.

e) Let $\theta = \arcsin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Then $\frac{\pi}{2} - \theta \in [0, \pi]$ , the range of $\arccos$ , and $\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta = x$ . So $\arccos x = \frac{\pi}{2} - \theta$ , which is the identity. For $x > 0$ , let $\alpha = \arctan x \in \left(0, \frac{\pi}{2}\right)$ ; then $\frac{\pi}{2} - \alpha$ is also in $\left(0, \frac{\pi}{2}\right)$ and $\tan\left(\frac{\pi}{2} - \alpha\right) = \frac{\cos\alpha}{\sin\alpha} = \frac{1}{x}$ , so $\arctan\frac{1}{x} = \frac{\pi}{2} - \alpha$ . For $x < 0$ , $\arctan$ is odd: $\arctan x + \arctan\frac{1}{x} = -\left(\arctan(-x) + \arctan\frac{1}{-x}\right) = -\frac{\pi}{2}$ . Check at $x = -1$ : $-\frac{\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{2}$ . Stating the first identity for all $x \ne 0$ is the error; the range check in the proof is exactly what fails for negative $x$ .

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 140 assignment, and each is false. Say what is wrong, give a counterexample with exact values, and write the correct statement.

a) $\ln(x^2) = 2\ln x$ for every $x \ne 0$ .
b) The inverse of $f$ is $f^{-1}(x) = \frac{1}{f(x)}$ .
c) $\sin^{-1} x$ and $\frac{1}{\sin x}$ are two notations for the same function.
d) $\ln(a + b) = \ln a + \ln b$ for all $a, b > 0$ .
e) Since $x^2 = 9$ gives $x = \pm 3$ , the inverse of $f(x) = x^2$ is $f^{-1}(x) = \pm\sqrt x$ .

Show the solution

Answers

a) False at $x = -1$ : $\ln 1 = 0$ but $\ln(-1)$ is undefined. $\ln(x^2) = 2\ln|x|$ for $x \ne 0$ .
b) False: for $f(x) = 2x$ , $f^{-1}(4) = 2$ but $\frac{1}{f(4)} = \frac{1}{8}$ . $f^{-1}$ undoes $f$ by composition.
c) False: $\sin^{-1} 2$ is undefined while $\frac{1}{\sin 2}$ exists. $\frac{1}{\sin x} = \csc x$ .
d) False: $\ln 2 \ne \ln 1 + \ln 1 = 0$ . The law is $\ln(ab) = \ln a + \ln b$ .
e) False: $x^2$ is not one-to-one on $\mathbb{R}$ . Restricted to $[0, \infty)$ its inverse is $\sqrt x$ , to $(-\infty, 0]$ it is $-\sqrt x$ .

a) FALSE. At $x = -1$ : $\ln((-1)^2) = \ln 1 = 0$ , while $2\ln(-1)$ does not exist. The two sides have different domains, all $x \ne 0$ on the left and $x > 0$ on the right: the power law $\ln(a^r) = r\ln a$ requires $a > 0$ . Correct statement: $\ln(x^2) = 2\ln|x|$ for every $x \ne 0$ , and $\ln(x^2) = 2\ln x$ for $x > 0$ only. This one line is behind most of the lost solutions in log equations: replacing $\ln(x^2)$ by $2\ln x$ silently deletes the negative half of the domain.

b) FALSE. Take $f(x) = 2x$ : its inverse halves, $f^{-1}(x) = \frac{x}{2}$ , so $f^{-1}(4) = 2$ , while $\frac{1}{f(4)} = \frac{1}{8}$ . The exponent $-1$ in $f^{-1}$ means the inverse for COMPOSITION, $f^{-1}(f(x)) = x$ , not the reciprocal. Correct statement: $f^{-1}(y) = x$ exactly when $f(x) = y$ ; the reciprocal $\frac{1}{f(x)}$ is written $(f(x))^{-1}$ , with the parentheses.

c) FALSE. $\sin^{-1} x$ is $\arcsin x$ , defined on $[-1, 1]$ only, with values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . $\frac{1}{\sin x} = \csc x$ is defined wherever $\sin x \ne 0$ . At $x = 2$ : $\sin^{-1} 2$ does not exist, while $\frac{1}{\sin 2}$ does, because $\sin 2 \ne 0$ . At $x = \frac{1}{2}$ : $\sin^{-1}\frac{1}{2} = \frac{\pi}{6} < 1$ , while $\frac{1}{\sin(1/2)} > 1$ since $0 < \sin\frac{1}{2} < 1$ . The notation is a trap Stewart warns about: $\sin^{-1} x$ never means a reciprocal, whereas $\sin^2 x$ does mean $(\sin x)^2$ .

d) FALSE. With $a = b = 1$ : $\ln(1 + 1) = \ln 2 > 0$ , while $\ln 1 + \ln 1 = 0$ . There is NO law for the logarithm of a sum. Correct statement: $\ln(ab) = \ln a + \ln b$ for $a, b > 0$ , the logarithm turns PRODUCTS into sums, because $e^{u + v} = e^u e^v$ . The same error in disguise: averaging two pH values, or writing $\ln(x + 1) = \ln x + 1$ . Exercise 10 shows what it costs in a real computation.

e) FALSE. $f(x) = x^2$ is not one-to-one on $\mathbb{R}$ , since $f(-3) = f(3) = 9$ , so it has no inverse there. And $\pm\sqrt x$ is not a function at all: it gives two outputs for one input. Correct statement: the restriction of $x^2$ to $[0, \infty)$ is one-to-one with inverse $\sqrt x$ , and the restriction to $(-\infty, 0]$ has inverse $-\sqrt x$ . The equation $x^2 = 9$ has two solutions; a function has one output per input. Restricting a domain is the same gesture that builds $\arcsin$ from $\sin$ .

Exercise 9: Radiocarbon dating: the half-life function and its inverse

Carbon-14 has a half-life of $5730$ years: every $5730$ years, half of what is left disappears. In a sample that stopped exchanging carbon $t$ years ago, the fraction of the original carbon-14 still present is $F(t) = \left(\frac{1}{2}\right)^{t/5730}$ . A laboratory measures $F$ ; dating the sample means computing the INVERSE function, $t$ as a function of $F$ .

The figure shows $F$ , with $t$ in thousands of years, and the fractions left after one, two, three and four half-lives. No calculator: exact answers, with orders of magnitude from $\ln 2 \approx 0.69$ and $\ln 10 \approx 2.30$ .

a) Compute $F(11\,460)$ , $F(17\,190)$ and $F(2865)$ exactly. A student claims that after half a half-life, a quarter of the carbon-14 has disappeared. Correct this.
b) Explain why $F$ is one-to-one on $[0, \infty)$ and find its inverse $t = F^{-1}(y)$ , with its domain.
c) Date a bone in which $\frac{1}{16}$ of the carbon-14 is left, then a piece of charcoal in which $\frac{1}{5}$ is left: exact answer, then two consecutive multiples of $5730$ around it, then an order of magnitude.
d) Two samples have fractions $0.36$ and $0.09$ . Without computing either age, find their difference in age, and say which is older.
e) Instruments cannot measure a fraction below $\frac{1}{1000}$ reliably. Find the largest age the method can give, bracketed between two multiples of $5730$ .

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Answers

a) $\frac{1}{4}$ , $\frac{1}{8}$ , $\frac{\sqrt 2}{2}$ : about $29\%$ has disappeared, not $25\%$
b) $F^{-1}(y) = 5730 \log_2 \frac{1}{y} = \frac{5730 \ln(1/y)}{\ln 2}$ , domain $(0, 1]$
c) $22\,920$ years; $5730 \log_2 5$ , between $11\,460$ and $17\,190$ , about $13\,400$ years
d) $5730 \log_2 4 = 11\,460$ years; the sample at $0.09$ is older
e) $5730 \log_2 1000$ , between $51\,570$ and $57\,300$ years

a) $11\,460 = 2 \times 5730$ , so $F(11\,460) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ ; likewise $F(17\,190) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$ . For $2865 = \frac{5730}{2}$ : $F(2865) = \left(\frac{1}{2}\right)^{1/2} = \frac{1}{\sqrt 2} = \frac{\sqrt 2}{2}$ . Since $\sqrt 2 \approx 1.414$ , about $70.7\%$ is left and $29.3\%$ has disappeared. The student reasoned linearly: half a half-life, half of the half. But an exponential loses the same FRACTION in equal times, not the same amount, so the first half of the first half-life removes more than the second half: $1 - \frac{\sqrt 2}{2}$ then $\frac{\sqrt 2}{2} - \frac{1}{2}$ , and $1 - \frac{\sqrt 2}{2} \approx 0.29 > 0.21$ .

b) $F$ is an exponential function of $t$ with base $\left(\frac{1}{2}\right)^{1/5730}$ , a number between $0$ and $1$ , so it is decreasing, and a decreasing function never takes the same value twice: one-to-one. On $[0, \infty)$ its values fill $(0, 1]$ . Solve $y = \left(\frac{1}{2}\right)^{t/5730}$ for $t$ : $\frac{t}{5730} = \log_{1/2} y$ , and since $\log_{1/2} y = \frac{\ln y}{\ln(1/2)} = -\log_2 y = \log_2\frac{1}{y}$ , $t = F^{-1}(y) = 5730\log_2\frac{1}{y} = \frac{5730 \ln(1/y)}{\ln 2}$ , with domain $(0, 1]$ and range $[0, \infty)$ . Check: $F^{-1}\left(\frac{1}{2}\right) = 5730 \log_2 2 = 5730$ , one half-life.

c) Bone: $F^{-1}\left(\frac{1}{16}\right) = 5730\log_2 16 = 5730 \times 4 = 22\,920$ years. Charcoal: $F^{-1}\left(\frac{1}{5}\right) = 5730 \log_2 5 = \frac{5730\ln 5}{\ln 2}$ years, the exact answer. Bracket: $4 < 5 < 8$ and $\log_2$ is increasing, so $2 < \log_2 5 < 3$ , and the age is between $11\,460$ and $17\,190$ years, which the figure confirms: $\frac{1}{5}$ lies between the marked $\frac{1}{4}$ and $\frac{1}{8}$ . Order of magnitude: $\ln 5 = \ln 10 - \ln 2 \approx 2.30 - 0.69 = 1.61$ , so $\log_2 5 \approx \frac{1.61}{0.69} \approx 2.33$ and the age is about $13\,400$ years. The bracket is the part a marker trusts; the decimal is only a sanity check.

d) By the law of the quotient, $t_A - t_B = 5730\left(\log_2\frac{1}{F_A} - \log_2\frac{1}{F_B}\right) = 5730\log_2\frac{F_B}{F_A}$ . With $F_A = 0.09$ and $F_B = 0.36$ : $\frac{F_B}{F_A} = 4$ , so $t_A - t_B = 5730 \times 2 = 11\,460$ years. The sample with the SMALLER fraction, $0.09$ , is the older one, by exactly two half-lives. Only the ratio of the measurements matters, never their size: a logarithm turns a ratio of fractions into a difference of ages. Subtracting the fractions, $0.36 - 0.09 = 0.27$ , and converting that into years is the linear reflex again, and it has no meaning here.

e) $F(t) \ge \frac{1}{1000}$ means $t \le F^{-1}\left(\frac{1}{1000}\right) = 5730\log_2 1000$ , since $F^{-1}$ is decreasing. Bracket by powers of $2$ : $2^9 = 512 < 1000 < 1024 = 2^{10}$ , so $9 < \log_2 1000 < 10$ and the limit lies between $5730 \times 9 = 51\,570$ and $5730 \times 10 = 57\,300$ years. This is why radiocarbon dating stops at about fifty thousand years: a bone of $100\,000$ years keeps $\left(\frac{1}{2}\right)^{100000/5730} < \left(\frac{1}{2}\right)^{17} < \frac{1}{1000}$ of its carbon-14, since $\frac{100\,000}{5730} > 17$ , below what can be measured.

Exercise 10: A final exam question: the pH scale, a logarithm that cannot be averaged

The acidity of a solution is measured by its pH, $p = -\log_{10} c$ , where $c$ is the concentration of hydrogen ions in moles per litre. A logarithmic scale compresses concentrations from $1$ down to $10^{-14}$ into numbers from $0$ to $14$ , and it is exactly that compression that makes pH easy to read and easy to misuse.

No calculator. Use $\log_{10} 2 \approx 0.30$ only for orders of magnitude; the answers are exact.

a) Find the pH of solutions with $c = 10^{-3}$ , $c = 4 \times 10^{-6}$ and $c = 1$ .
b) Write $c$ as a function of $p$ , the inverse function, and give the concentration of blood, whose pH is $7.4$ , exactly and to one significant figure.
c) Show that multiplying $c$ by $10$ lowers the pH by exactly $1$ . How many times more concentrated in hydrogen ions is a juice of pH $2$ than one of pH $4$ ? What does doubling $c$ do to the pH?
d) Equal volumes of a solution of pH $2$ and a solution of pH $4$ are mixed, so the concentration of the mixture is the average of the two concentrations. Find the pH of the mixture exactly, show that it lies between $2$ and $2 + \log_{10} 2$ , and explain why the answer $3$ is wrong.
e) Normal rain has pH $5.6$ and an acid rain pH $4.3$ . How many times more concentrated in hydrogen ions is the acid rain? Exact answer, then an order of magnitude.

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Answers

a) $3$ ; $6 - 2\log_{10} 2 \approx 5.4$ ; $0$
b) $c = 10^{-p}$ ; blood: $10^{-7.4} \approx 4 \times 10^{-8}$ mol/L
c) $-\log_{10}(10c) = p - 1$ ; $100$ times; the pH drops by $\log_{10} 2 \approx 0.30$
d) $p = \log_{10}\frac{20000}{101} = 4 + \log_{10} 2 - \log_{10} 101$ , between $2$ and $2 + \log_{10} 2 \approx 2.30$ ; $3$ averages logarithms
e) $10^{1.3} = 10 \cdot 10^{0.3} \approx 20$ times

a) $p = -\log_{10} 10^{-3} = 3$ . For $c = 4 \times 10^{-6}$ , the product law splits the logarithm: $\log_{10}(4 \times 10^{-6}) = \log_{10} 4 + \log_{10} 10^{-6} = 2\log_{10} 2 - 6$ , so $p = 6 - 2\log_{10} 2$ , about $6 - 0.60 = 5.4$ . For $c = 1$ , $p = -\log_{10} 1 = 0$ . Order check: $4 \times 10^{-6}$ lies between $10^{-6}$ and $10^{-5}$ , so its pH must lie between $5$ and $6$ , and $5.4$ does. The minus sign in the definition is why a LARGER concentration gives a SMALLER pH.

b) Solve $p = -\log_{10} c$ for $c$ : $\log_{10} c = -p$ , so $c = 10^{-p}$ , the inverse function, defined for every real $p$ (in practice $0 \le p \le 14$ ). Blood: $c = 10^{-7.4}$ mol/L exactly. To size it, write $-7.4 = 0.6 - 8$ : $10^{-7.4} = 10^{0.6} \times 10^{-8}$ , and $10^{0.6} = (10^{0.3})^2 \approx 2^2 = 4$ because $\log_{10} 2 \approx 0.30$ means $10^{0.3} \approx 2$ . So $c \approx 4 \times 10^{-8}$ mol/L. The trap is to write $10^{-7.4} = 10^{-7} \times 10^{-0.4}$ and then guess $10^{-0.4}$ ; splitting the exponent into a positive decimal and an integer is the clean way.

c) $-\log_{10}(10c) = -(\log_{10} 10 + \log_{10} c) = -1 - \log_{10} c = p - 1$ : a tenfold concentration lowers the pH by exactly $1$ , whatever $c$ is. Hence pH $2$ against pH $4$ is a factor $10^{-2}/10^{-4} = 10^2 = 100$ in concentration, not a factor $2$ and not a difference of $2$ units of acidity. Doubling $c$ gives $-\log_{10}(2c) = p - \log_{10} 2$ , a drop of about $0.30$ . Equal STEPS in pH are equal RATIOS in concentration: that is the whole meaning of a logarithmic scale.

d) The concentrations are $10^{-2}$ and $10^{-4}$ , and the mixture has $c = \frac{10^{-2} + 10^{-4}}{2} = \frac{100 + 1}{2 \times 10^4} = \frac{101}{20\,000}$ . Its pH is $p = -\log_{10}\frac{101}{20\,000} = \log_{10}\frac{20\,000}{101} = \log_{10} 2 + 4 - \log_{10} 101$ , the exact answer. Bracket: $100 < 101 < 200$ gives $2 < \log_{10} 101 < 2 + \log_{10} 2$ , so $2 < p < 2 + \log_{10} 2 \approx 2.30$ . The answer $3$ averages the pH values, that is, averages the LOGARITHMS; but a logarithm of an average is not an average of logarithms, the same error as $\ln(a + b) = \ln a + \ln b$ . The pH $3$ corresponds to $c = 10^{-3}$ , about five times too small: the stronger acid carries almost all the hydrogen ions of the mixture, and the pH stays close to $2$ .

e) The ratio is $\frac{10^{-4.3}}{10^{-5.6}} = 10^{5.6 - 4.3} = 10^{1.3}$ , the exact answer. Then $10^{1.3} = 10 \times 10^{0.3} \approx 10 \times 2 = 20$ : the acid rain carries about twenty times more hydrogen ions, although its pH is only $1.3$ units lower. Reading $1.3$ units as a $1.3$ times or $23\%$ increase is the linear reflex that a logarithmic scale is designed to defeat. The pattern of the three last questions is the thread of the chapter: to compare or combine quantities given on a log scale, come back through the INVERSE function, $c = 10^{-p}$ , do the arithmetic there, then take the logarithm again.

Corrected exercises: exponential, inverse, logarithmic and inverse trigonometric functions (MATH 140)

Course recap

Part A: the basics (/50)

Exercise 1: Exponential functions: the laws of exponents and a curve read from two points

Exercise 2: One-to-one functions and the inverse, with its domain

Exercise 3: Logarithms: exact values, the laws, and the domain they hide

Exercise 4: Exponential and logarithmic equations: solve, then check the domain

Exercise 5: Inverse trigonometric functions: exact values, restricted ranges and the triangle

Part B: problems and reasoning (/50)

Exercise 6: An inverse from start to finish: a function trapped between 0 and 6

Exercise 7: arcsin(sin x) is not x: the zigzag, and the solution the calculator never shows

Exercise 8: Five statements to correct

Exercise 9: Radiocarbon dating: the half-life function and its inverse

Exercise 10: A final exam question: the pH scale, a logarithm that cannot be averaged

See also

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