International Baccalaureate · DP 1: Analysis and Approaches HL

Logarithms

⌚ ~202 min ▢ 23 blocks ✓ 94 exercises ➣ Prerequisites : Exponentials

I The Logarithm of a Number

I.1 Logarithms in Base 10

Recappreviously studied — click to reread

For a power of 10, the logarithm is simply the exponent read off the power. The table below shows this two-way reading:

$y$	$0.01$	$0.1$	$1$	$10$	$100$	$1000$
$y$ as a power of $10$	$10^{-2}$	$10^{-1}$	$10^{0}$	$10^{1}$	$10^{2}$	$10^{3}$
$\log y$	$-2$	$-1$	$0$	$1$	$2$	$3$

For example, $10^{-2} = 0.01$, so $\log 0.01 = -2$; and $10^{3} = 1000$, so $\log 1000 = 3$.
The two operations travel in opposite directions between a number $y$ and its exponent $x$: the logarithm extracts the exponent, and raising 10 to a power rebuilds the number.

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Definition — Logarithm

The (base 10) logarithm of a number $y$ (where $y > 0$) is the exponent to which 10 must be raised to obtain $y$. It is denoted $\log_{10} y$ and defined by the equivalence: $$\log_{10} y = x \quad \Leftrightarrow \quad 10^x = y$$ As on most calculators and in exams, $\log_{10} y$ is abbreviated $\log y$; so $\log y = \log_{10} y$. In other words, $\log y$ is the exponent $x$ such that $10^x = y$.

Why $\log 0$ and $\log(-5)$ are undefined

The logarithm $\log y$ is defined only for $y > 0$. Indeed, a power of 10 is always strictly positive: $10^x > 0$ for every real $x$. Since no power of 10 equals $0$ or a negative number, expressions such as $\log 0$ and $\log(-5)$ have no value.

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Proposition — Inverse Properties of Logarithms

For real numbers:

$10^{\log x} = x$ for any $x > 0$,
$\log (10^x) = x$ for any real $x$.

Example

Evaluate $\log 100$.

Answer

$\begin{aligned}[t]\log(100) &= \log\left(10^{2}\right) &&(100 = 10^2) \\&=2\\\end{aligned}$
Since $10^2 = 100$, the logarithm of $100$ (base 10) is $2$.

Skills to practice

Evaluating Logarithms without a Calculator

I.2 Logarithms in Base $a$

Recappreviously studied — click to reread

The logarithm in base $10$ of a number is the exponent to which $10$ must be raised in order to obtain that number.
We can use the same principle to define logarithms in other bases:

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Definition — Logarithm in Base a

For a positive number $a \neq 1$, the logarithm base $a$ of a number $y$ (where $y > 0$) is the exponent to which $a$ must be raised to obtain $y$. It is denoted as:$$\log_a y = x \quad \Leftrightarrow \quad a^x = y.$$

Proposition — Inverse Properties of Logarithms in Base a

Let $a>0$ with $a\neq 1$.

$a^{\log_a x} = x$ for any $x > 0$,
$\log_a (a^x) = x$ for any real $x$.

Example

Evaluate $\log_2 8$.

Answer

$\begin{aligned}[t]\log_2 8 &= \log_2 (2^3) &&(8 = 2^3)\\&=3\\\end{aligned}$

Method — Solving Exponential Equations with Base a

To solve $a^x = b$ (where $a > 0$, $a \neq 1$, $b > 0$):

Take the logarithm base $a$ of both sides: $\log_a (a^x) = \log_a b$
Simplify: $x = \log_a b$

Example

Solve $3^x = 81$.

Answer

$$\begin{aligned}3^x &= 81 \\ \log_3 (3^x) &= \log_3 (81) &&\text{(applying $\log_3$ to both sides)}\\ x &= \log_3 (3^4) &&(\log_3(3^x) = x,\ 81 = 3^4)\\ x &= 4\end{aligned}$$

Skills to practice

Evaluating Logarithms in Base $a$

I.3 The Natural Logarithm

The natural logarithm, often denoted $\ln x$, is the logarithm with base $e$ (where $e \approx 2.71828$ is the base of the natural exponential function). It is the inverse of the exponential function $e^x$ and plays a central role in calculus, growth models, and many scientific applications.

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Definition — Natural Logarithm

The natural logarithm, denoted $\ln x$, is the logarithm base $e$, defined for $x>0$:$$\ln x = \log_e(x).$$

Proposition — Inverse Properties of Natural Logarithms

$e^{\ln x} = x$ for $x > 0$,
$\ln (e^x) = x$ for any real $x$.

Notation. A logarithm always has a base. Three forms are in use:

$\log_a x$ — the logarithm in base $a$ (the base is written explicitly);
$\log x = \log_{10} x$ — the base-10 logarithm; as on calculators and in exams, the base 10 is omitted;
$\ln x = \log_e x$ — the natural logarithm (base $e$).

So a bare $\log$ means $\log_{10}$, and $\ln$ means $\log_e$.

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Skills to practice

Evaluating Natural Logarithms

II Laws of Logarithms

II.1 The Laws of Logarithms

Recappreviously studied — click to reread

Proposition — Laws of Logarithms

For $x>0$, $y>0$, $a>0$ with $a\neq 1$, and for any real number $k$:

Product Rule: $\log_a (xy) = \log _a x + \log _a y$
Quotient Rule: $\log _a \left(\dfrac{x}{y}\right) = \log _a x - \log_a y$
Power Rule: $\log_a (x^k) = k \log_a x$

Proof

$$\begin{aligned}\log_a(xy) &= \log_a\bigl(a^{\log_a x} \times a^{\log_a y}\bigr) &&(x = a^{\log_a x})\\ &= \log_a\bigl(a^{\log_a x + \log_a y}\bigr) &&\text{(exponent product law)}\\ &= \log_a x + \log_a y &&(\log_a(a^u) = u)\\ \\ \log_a\left(\frac{x}{y}\right) &= \log_a\left(\frac{a^{\log_a x}}{a^{\log_a y}}\right) &&(x = a^{\log_a x})\\ &= \log_a\bigl(a^{\log_a x - \log_a y}\bigr) &&\text{(exponent quotient law)}\\ &= \log_a x - \log_a y &&(\log_a(a^u) = u)\\ \\ \log_a(x^k) &= \log_a\left(\bigl(a^{\log_a x}\bigr)^k\right) &&(x = a^{\log_a x})\\ &= \log_a\bigl(a^{k \log_a x}\bigr) &&\text{(power of a power)}\\ &= k \log_a x\\ \end{aligned}$$

Example

Write as a single logarithm: $\log (5)+\log (3)$.

Answer

$\begin{aligned}[t]\log (5)+\log (3) &= \log (5\times 3) &&\text{(product rule)} \\&= \log 15\\\end{aligned}$

Example

For $a>0$ and $b>0$, write $\log\left(\dfrac{a^2 b}{10}\right)$ in terms of $\log a$ and $\log b$.

Answer

$$\begin{aligned}[t] \log\left(\dfrac{a^2 b}{10}\right) &= \log(a^2 b) - \log 10 &&\text{(quotient rule)}\\ &= \log(a^2) + \log b - \log 10 &&\text{(product rule)}\\ &= 2\log a + \log b - \log 10 &&\text{(power rule)}\\ &= 2\log a + \log b - 1 \end{aligned}$$ The laws also work in this direction: they break a single logarithm into a combination of simpler logarithms.

Skills to practice

The Laws of Logarithms

II.2 Using the Laws with $\ln$ and in Base $a$

In particular, for the natural logarithm, for $x>0$ and $y>0$:

$\ln (xy) = \ln x + \ln y$
$\ln \left(\dfrac{x}{y}\right) = \ln x - \ln y$
$\ln (x^k) = k \ln x$

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Example

Evaluate $\ln \left(e^4\right)$.

Answer

$\begin{aligned}[t]\ln\left(e^4\right) &= 4 \ln e &&\text{(power rule)}\\&=4 \times 1 &&(\ln e = 1)\\&=4\\\end{aligned}$

Skills to practice

Writing as a Single Natural Logarithm
Writing as a Single Logarithm in Base $a$

II.3 The Change of Base Rule

Recappreviously studied — click to reread

Proposition — Change of Base Rule

For any $a>0$, $a \neq 1$, $b>0$, $b \neq 1$, and $x > 0$:$$\log_a x = \frac{\log_b x}{\log_b a}.$$

Proof

Let $y = \log_a x$.$$\begin{aligned}a^y &= a^{\log_a x} &&\text{(applying $a^{(\cdot)}$ to both sides)}\\ a^y &= x &&(a^{\log_a x} = x)\\ \log_b\left(a^y\right) &= \log_b x &&\text{(applying $\log_b$ to both sides)}\\ y \log_b a &= \log_b x &&\text{(power rule)}\\ y &= \frac{\log_b x}{\log_b a} &&\text{(dividing by $\log_b a$, not zero since $a \neq 1$)}\\ \log_a x &= \frac{\log_b x}{\log_b a} &&(y = \log_a x) \end{aligned}$$

Method — Using Change of Base

To compute $\log_a b$ (where $a > 0$, $a \neq 1$, $b > 0$) using base $10$ logarithms:$$\log_a b = \frac{\log b}{\log a}.$$ Using natural logarithms:$$\log_a b = \frac{\ln b}{\ln a}.$$

Example

Compute $\log_2 10$ using natural logarithms.

Answer

$$\begin{aligned}\log_2 10 &= \frac{\ln 10}{\ln 2} &&\text{(change of base)}\\ &\approx \frac{2.302585}{0.693147} &&\text{(using a calculator)}\\ &\approx 3.321928\end{aligned}$$

Skills to practice

Evaluating Logarithms by Changing the Base

III Solving Equations and Inequalities

III.1 Solving Exponential Equations with Logarithms

Recappreviously studied — click to reread

Some exponential equations are solved by writing both sides with the same base (e.g., $2^x = 8 \Rightarrow 2^x = 2^3$). However, most real-world problems involve values that are not perfect powers. To solve these, we use the power rule of logarithms to "bring down" the exponent.

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Method — Solving $a^x \equal b$

To solve an equation $a^x = b$ (where $a > 0$, $a \neq 1$, $b > 0$) in which the unknown is in the exponent:

Isolate the exponential term ($a^x$) if necessary.
Take the logarithm ($\log$ or $\ln$) of both sides.
Use the Power Rule: $\log(a^x) = x \log a$.
Solve for $x$ by dividing.

Example

Solve $2^x = 7$.

Answer

$$\begin{aligned}2^x &= 7 \\ \log(2^x) &= \log 7 &&\text{(taking the logarithm of both sides)}\\ x \log 2 &= \log 7 &&\text{(power rule)}\\ x &= \frac{\log 7}{\log 2} &&\text{(dividing)}\\ \end{aligned}$$

Skills to practice

Solving $a^x \equal b$ with a Logarithm
Solving Equations with $e^x$

III.2 Logarithmic Scales in Science

Recappreviously studied — click to reread

Logarithms are used to describe quantities that vary over many orders of magnitude. In the scientific formulas below, the base-10 logarithm is written in full as $\log_{10}$. Some important examples in science include:

pH scale in chemistry: $\text{pH} = -\log_{10} [H^+]$
(where $[H^+]$ is the numerical value of the hydrogen ion concentration in moles per litre)
Richter scale for earthquakes (simplified model): $\text{Magnitude} = \log_{10}\left(\dfrac{I}{I_0}\right)$
(where $I$ is the intensity of the earthquake, $I_0$ is a reference intensity)
Decibel (dB) scale for sound: $L = 10 \log_{10}\left(\dfrac{P}{P_0}\right)$
(where $P$ is the power/intensity measured, $P_0$ is a reference level)

On calculators, the $\log$ key gives $\log_{10}$.

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Example

The pH of a solution is $3.2$. Find the hydrogen ion concentration $[H^+]$.

Answer

We know: $$\begin{aligned} \text{pH} &= -\log_{10}[H^+] \\ 3.2 &= -\log_{10}[H^+] &&\text{(substituting the value)}\\ -3.2 &= \log_{10}[H^+] &&\text{(multiplying both sides by $-1$)}\\ 10^{-3.2} &= 10^{\log_{10}[H^+]} &&\text{(exponentiating both sides, base 10)}\\ [H^+] &= 10^{-3.2} \\ [H^+] &\approx 6.31 \times 10^{-4}\ \text{mol/L} &&\text{(using a calculator)} \end{aligned}$$

Skills to practice

Logarithmic Scales in Science

III.3 Equations in Quadratic Form

Example

Solve each equation, giving the exact solutions.

$e^{2x} - 5e^x + 4 = 0$
$9^x - 3^{x+1} - 10 = 0$

Answer

Since $e^{2x} = \left(e^x\right)^2$, the equation is quadratic in $e^x$: substitute $u = e^x$, where $u \gt 0$. $$\begin{aligned}u^2 - 5u + 4 &= 0 &&\text{(substitution $u = e^x$)}\\ (u - 1)(u - 4) &= 0 &&\text{(factorisation)} \end{aligned}$$ So $u = 1$ or $u = 4$. Both values are positive, so each one gives a solution: $$\begin{aligned}e^x = 1 &\iff x = \ln 1 = 0 &&\text{(inverse property of $\ln$)}\\ e^x = 4 &\iff x = \ln 4 &&\text{(inverse property of $\ln$)} \end{aligned}$$ The solutions are $0$ and $\ln 4$. Check: $e^{2\ln 4} - 5e^{\ln 4} + 4 = 16 - 20 + 4 = 0$.
Since $9^x = \left(3^x\right)^2$ and $3^{x+1} = 3 \times 3^x$, the equation is quadratic in $3^x$: substitute $u = 3^x$, where $u \gt 0$. $$\begin{aligned}u^2 - 3u - 10 &= 0 &&\text{(substitution $u = 3^x$)}\\ (u - 5)(u + 2) &= 0 &&\text{(factorisation)} \end{aligned}$$ So $u = 5$ or $u = -2$. The value $u = -2$ is rejected: $3^x \gt 0$ for every real $x$. It remains $3^x = 5$: $$\begin{aligned}\ln\left(3^x\right) &= \ln 5 &&\text{(logarithm of both sides)}\\ x \ln 3 &= \ln 5 &&\text{(power rule)}\\ x &= \frac{\ln 5}{\ln 3} &&\text{(dividing by $\ln 3 \gt 0$)} \end{aligned}$$ The only solution is $\dfrac{\ln 5}{\ln 3}$, that is $\log_3 5$.

Skills to practice

Solving Equations in Quadratic Form with Logarithms

III.4 Exponential Inequalities

Proposition — Monotonicity of Logarithms

Let $a > 0$ with $a \neq 1$, and let $u$, $v$ be real numbers.

If $a > 1$, the function $\log_a$ is strictly increasing: the order is preserved, $$0 < u < v \quad \Leftrightarrow \quad \log_a u < \log_a v.$$
If $0 < a < 1$, the function $\log_a$ is strictly decreasing: the order is reversed, $$0 < u < v \quad \Leftrightarrow \quad \log_a u > \log_a v.$$

This proposition governs the solving of exponential inequalities such as $a^x \le b$ (with $b > 0$). Taking the logarithm of both sides and applying the power rule gives $x \log a \le \log b$ in every case, because $\log$ itself is increasing. The decisive step is the division by $\log a$: when $a > 1$, $\log a > 0$ and the inequality keeps its direction; when $0 < a < 1$, $\log a < 0$, and dividing both sides by a negative number reverses the inequality. The same reasoning applies with $\ln$ in place of $\log$.

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Method — Solving Exponential Inequalities

To solve an inequality such as $a^x \le b$ (where $a > 0$, $a \neq 1$, $b > 0$):

Isolate the exponential term ($a^x$) if necessary, then take the logarithm ($\log$ or $\ln$) of both sides: the direction is preserved, since $\log$ and $\ln$ are increasing.
Use the Power Rule: $\log(a^x) = x \log a$.
Divide both sides by $\log a$, checking its sign: if $a > 1$, $\log a > 0$ and the direction is kept; if $0 < a < 1$, $\log a < 0$ and the inequality reverses.

Example

Solve $2^x > 10$.

Answer

$$\begin{aligned}2^x &> 10 \\ \log(2^x) &> \log 10 &&\text{(taking the logarithm of both sides: $\log$ increasing)}\\ x \log 2 &> \log 10 &&\text{(power rule)}\\ x &> \frac{\log 10}{\log 2} &&\text{(dividing by $\log 2 > 0$: direction kept)}\\ x &> \frac{\log 10}{\log 2} \approx 3.32 \end{aligned}$$

Example

Solve $0.8^x \le 0.1$.

Answer

$$\begin{aligned}0.8^x &\le 0.1 \\ \ln(0.8^x) &\le \ln 0.1 &&\text{(taking ln of both sides: $\ln$ increasing)}\\ x \ln 0.8 &\le \ln 0.1 &&\text{(power rule)}\\ x &\ge \frac{\ln 0.1}{\ln 0.8} &&\text{(dividing by $\ln 0.8 < 0$: inequality reverses)}\\ x &\ge \frac{\ln 0.1}{\ln 0.8} \approx 10.32 \end{aligned}$$

Method — Finding the Smallest Whole Number $n$

Many questions ask for a whole number of steps: after how many years, doses, copies…\ does a growing power $a^n$ ($a \gt 1$) exceed a value $b \gt 1$, or a decreasing power $a^n$ ($0 \lt a \lt 1$) fall below a value $b$ with $0 \lt b \lt 1$? In both cases $a^0 = 1$ does not satisfy the inequality yet, and such whole numbers exist. To find the smallest whole number $n$ such that $a^n \gt b$ (or $a^n \lt b$):

Solve the inequality as if $n$ were any real number, with the common logarithm; keep track of the sign of $\log a$ when dividing.
In both cases the solutions are the numbers greater than a bound $c \gt 0$: the answer is the smallest whole number strictly greater than $c$ (if $c$ is itself a whole number, the answer is $c + 1$).
Check with the calculator: the inequality is false for $n - 1$ and true for $n$.

Example

A machine loses $15\pourcent$ of its value every year. After how many whole years is its value less than half of its purchase price?

Answer

After $n$ years the value is multiplied by $0.85^n$; the question is the smallest whole number $n$ such that $0.85^n \lt 0.5$. $$\begin{aligned}0.85^n &\lt 0.5\\ \log(0.85^n) &\lt \log 0.5 &&\text{(logarithm of both sides: $\log$ increasing)}\\ n \log 0.85 &\lt \log 0.5 &&\text{(power rule)}\\ n &\gt \frac{\log 0.5}{\log 0.85} &&\text{(dividing by $\log 0.85 \lt 0$: the inequality reverses)}\\ n &\gt \frac{\log 0.5}{\log 0.85} \approx 4.27 \end{aligned}$$ The smallest whole number greater than $4.27$ is $5$. Check: $0.85^4 \approx 0.522 \gt 0.5$ and $0.85^5 \approx 0.444 \lt 0.5$. The value, computed at the end of each year, is below half of the price for the first time at the end of the fifth year: after $5$ whole years.

Skills to practice

Solving Exponential Inequalities
Finding the Smallest Whole Number $n$
Justifying the Direction of an Inequality

III.5 Logarithmic Equations

Method — Solving $\ln A \equal \ln B$ and $\ln A \le \ln B$

This is the method for a statement whose two sides each reduce to a single logarithm. It does not cover statements such as $(\ln x)^2 + \ln x = 0$, where $\ln x$ appears squared: those call for the substitution $X = \ln x$, which turns them into an ordinary equation in $X$.

Domain first. Every $\ln$ in the statement demands its argument strictly positive. One solves those conditions together and keeps their intersection: outside it, the statement has no meaning, and no solution can come from there.
One logarithm on each side. Using the algebraic rules, one rewrites the statement as $\ln A = \ln B$, or as $\ln A \le \ln B$.
Remove the logarithms. $\ln$ is strictly increasing on $]0\,;\,+\infty[$, so it preserves both equality and order. For $A > 0$ and $B > 0$ — which step 1 has secured: $$\ln A = \ln B \iff A = B \qquad \text{and} \qquad \ln A \le \ln B \iff A \le B.$$
Intersect with the domain. A value found at step 3 is a solution only if it lies in the domain of step 1.

Step 4 is not a formality, and it is worth being exact about why. Removing the logarithms is a true equivalence on the domain — nothing is lost and nothing is gained there. What widens the search is the next move: the equation obtained is then solved on all of $\mathbb{R}$, where it may well have roots the original statement never had. Those candidates are not errors of the method; they are values the domain must reject.

Example

Solve $\ln(x-1) + \ln(x+2) = \ln 4$.
Solve $\ln(2x-1) \le \ln(x+3)$.

Answer

Domain. One needs $x - 1 > 0$ and $x + 2 > 0$, that is $x > 1$ and $x > -2$: the domain is $]1\,;\,+\infty[$. One logarithm each side, by the functional relation: $$\ln\bigl((x-1)(x+2)\bigr) = \ln 4.$$ Remove them, since $\ln$ is strictly increasing: $$\begin{aligned} (x-1)(x+2) &= 4 &&\color{gray}\text{(Strict monotonicity)} \\ x^2 + x - 2 &= 4 &&\color{gray}\text{(Expand)} \\ x^2 + x - 6 &= 0 &&\color{gray}\text{(Bring to zero)} \\ (x - 2)(x + 3) &= 0. &&\color{gray}\text{(Factorise: } (x-2)(x+3) = x^2 + x - 6\text{)} \end{aligned}$$ A product is zero when one of its factors is zero: $x = 2$ or $x = -3$. Intersect with the domain. $2 > 1$, so $2$ is kept. $-3$ is not in $]1\,;\,+\infty[$, and in the original statement $\ln(-4)$ does not exist. Note where $-3$ comes from: not from removing the logarithms, which was an equivalence on $]1\,;\,+\infty[$, but from solving $x^2 + x - 6 = 0$ on all of $\mathbb{R}$. Hence $S = \{2\}$. Check: $\ln 1 + \ln 4 = 0 + \ln 4 = \ln 4$. \checkmark
Domain. $2x - 1 > 0$ and $x + 3 > 0$ give $x > \frac{1}{2}$ and $x > -3$: the domain is $\left]\frac{1}{2}\,;\,+\infty\right[$. $\ln$ is strictly increasing, so it preserves the order: $$\ln(2x-1) \le \ln(x+3) \iff 2x - 1 \le x + 3 \iff x \le 4.$$ Intersect: $S = \left]\frac{1}{2}\,;\,4\right]$, half-open on the left because $\frac{1}{2}$ is excluded by the domain, closed on the right because the inequality is wide.

Skills to practice

Solving Logarithmic Equations

IV Logarithmic Functions

IV.1 The Logarithmic Function in Base $a$

Recappreviously studied — click to reread

Definition — Logarithmic Function in Base a

The logarithmic function in base a is $f(x) = \log_a(x)$, where $a$ is a positive constant and $a \neq 1$.

Domain: $(0, +\infty)$
Range: $(-\infty, +\infty)$

Proposition — Graph of the Logarithmic Function in Base a

The graph of $\textcolor{colordef}{y = \log_a(x)}$ is the reflection of $\textcolor{colorprop}{y = a^x}$ over the line $\textcolor{olive}{y = x}$. It has a vertical asymptote at $x=0$ and passes through the point $(1,0)$.

If $a > 1$, the function is increasing.
If $0 < a < 1$, the function is decreasing.

$\quad$

Example

Study the function $f(x) = \log(x - 1)$: domain, asymptote, intercept, and sketch.

Domain: the logarithm needs a positive input: $x - 1 > 0$, so the domain is $(1, +\infty)$.
Vertical asymptote: at the edge of the domain, $x = 1$ (the whole curve of $\log$ is shifted $1$ unit right, and its asymptote moves with it).
$x$-intercept: $f(x) = 0$ when $\log(x-1) = 0$, i.e. $x - 1 = 1$ (since $\log 1 = 0$): the point $(2, 0)$.
One exact point: $f(11) = \log(10) = 1$ (ten to the power $1$): the point $(11, 1)$.

Skills to practice

Finding Domains in Base $a$

IV.2 The Natural Logarithm Function

Definition — Natural Logarithm Function

The natural logarithm function is $x\mapsto \ln x$.

Domain: $(0, +\infty)$
Range: $(-\infty, +\infty)$

Proposition — Graph of the Natural Logarithm Function

The graph of $\textcolor{colordef}{y = \ln x}$ is the reflection of $\textcolor{colorprop}{y = e^x}$ over the line $\textcolor{olive}{y = x}$. It has a vertical asymptote at $x=0$, passes through (1,0), and is increasing.

Skills to practice

Finding Domains of Logarithmic Functions
Finding Inverses and Asymptotes
Graphing and Studying Logarithmic Functions

V Going further

Skills to practice

Proving Identities and Rejecting Solutions
Reasoning on Logarithmic Graphs
Solving Problems without Technology
Solving Problems with Technology
Quiz

\(y\)	\(0.01\)	\(0.1\)	\(1\)	\(10\)	\(100\)	\(1000\)
\(y\) as a power of \(10\)	\(10^{-2}\)	\(10^{-1}\)	\(10^{0}\)	\(10^{1}\)	\(10^{2}\)	\(10^{3}\)
\(\log y\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)