1.5 Exponents and introduction to logarithms

An exponent is a compact way of writing repeated multiplication: $2^{10}$ is ten 2s multiplied together. Exponents describe anything that grows or shrinks by a constant factor, such as money earning compound interest, a population of cells or the decay of a radioactive substance. A logarithm answers the reverse question: given the result of such a multiplication, which exponent produced it? Logarithms turn multiplication into addition, compress very large ranges of numbers into small ones (as the pH scale does) and are the main tool for solving equations in which the unknown is an exponent.

Powers, bases and exponents

What a power means

For a positive integer $n$, the power $a^n$ means $n$ copies of $a$ multiplied together:

The number $a$ is the base and $n$ is the exponent. The whole expression $a^n$ is read "$a$ to the power $n$" or "$a$ to the $n$th". Other names for the exponent are the index (plural indices) and the power.

Small powers come up so often that it saves time to recognise them on sight. The table lists the ones used most in calculations without a calculator.

Negative bases and brackets

When the base is negative it must be written in brackets. The sign of the result depends on whether the exponent is even or odd: an even number of negative factors gives a positive product, and an odd number gives a negative one. So $(-2)^4 = 16$ but $(-2)^3 = -8$.

Without brackets, the exponent applies only to the number it is attached to. The expression $-2^4$ means $-(2^4) = -16$, because powers are evaluated before the minus sign in the order of operations.

The laws of exponents

The laws of exponents follow directly from the meaning of a power as repeated multiplication. Each one is a shortcut for counting factors, and seeing why each law holds makes them much easier to remember and to use correctly.

Multiplying and dividing powers with the same base

Multiplying $a^m$ by $a^n$ joins $m$ factors of $a$ to another $n$ factors of $a$, giving $m + n$ factors in total. For example, $2^3 \times 2^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2^7$.

Dividing $a^m$ by $a^n$ (with $m > n$) cancels $n$ of the factors in the numerator, leaving $m - n$. For example, $\frac{5^6}{5^2} = \frac{5 \times 5 \times 5 \times 5 \times 5 \times 5}{5 \times 5} = 5^4$.

Both laws need the same base. A product such as $2^3 \times 3^2$ cannot be combined into a single power; it is just $8 \times 9 = 72$.

Powers of powers, products and quotients

Raising a power to a power repeats a block of factors. $(a^m)^n$ is $n$ copies of $a^m$, and each copy contains $m$ factors of $a$, so there are $mn$ factors altogether: $(2^3)^2 = 2^3 \times 2^3 = 2^6$.

A product raised to a power can be split up, because the factors can be reordered: $(ab)^3 = ab \times ab \times ab = a^3 b^3$. The same reasoning works for a quotient.

The zero exponent

The division law can be used when the two exponents are equal. On one hand, $\frac{a^n}{a^n} = 1$, because any non-zero number divided by itself is 1. On the other hand, the division law gives $\frac{a^n}{a^n} = a^{n-n} = a^0$. For the laws to stay consistent, $a^0$ must equal 1.

So $7^0 = 1$, $(-3)^0 = 1$ and $\left(\frac{2}{5}\right)^0 = 1$. The expression $0^0$ is left undefined in this course.

Negative exponents

The division law also works when the exponent in the denominator is larger. Cancelling directly, $\frac{2^3}{2^5} = \frac{2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{2^2}$, while the law gives $2^{3-5} = 2^{-2}$. So $2^{-2}$ must mean $\frac{1}{2^2} = \frac{1}{4}$.

A negative exponent means "the reciprocal of", not "a negative number". The value of $2^{-2}$ is $\frac{1}{4}$, which is positive. For fractions, taking the reciprocal turns the fraction upside down:

so $\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

Summary of the laws

All the laws hold for any integers $m$ and $n$ (positive, zero or negative), with $a, b \neq 0$ wherever a division, a zero exponent or a negative exponent appears.

Simplifying algebraic expressions

The laws apply to letters exactly as they do to numbers. A good routine is to deal with the numerical coefficients first, then each letter in turn, and finally rewrite any negative exponents if the question asks for positive exponents only.

A coefficient in front of a power is not affected by the power's exponent. In $3x^{-2}$ the exponent $-2$ applies only to $x$, so $3x^{-2} = \frac{3}{x^2}$, whereas $(3x)^{-2} = \frac{1}{9x^2}$.

Writing numbers as powers of a common base

Many expressions become simple once every number is written as a power of the same prime. Powers of 2, 3 and 5 from the table at the start of the section are the ones that appear most often: $8 = 2^3$, $16 = 2^4$, $\frac{1}{4} = 2^{-2}$, $27 = 3^3$, $\frac{1}{9} = 3^{-2}$, $125 = 5^3$.

The same idea answers questions such as "find $n$ if $2^n = 64$". Since $64 = 2^6$, the exponent is $n = 6$. When the number is not an exact power of the base, as in $2^n = 50$, no integer works, and a new tool is needed to find the exponent. That tool is the logarithm.

Logarithms

The question a logarithm answers

An exponent and a logarithm describe the same relationship from two directions. The statement $2^3 = 8$ answers the question "what is 2 multiplied by itself 3 times?". The logarithm answers the reverse question "to what power must 2 be raised to give 8?". The answer, 3, is written $\log_2 8 = 3$ and read "log base 2 of 8 equals 3".

The double arrow $\iff$ means that each statement implies the other: they are two ways of writing the same fact. A logarithm is therefore always an exponent.

Evaluating logarithms without a calculator

To evaluate $\log_a b$ by hand, write $b$ as a power of $a$. The exponent of that power is the logarithm. It often helps to set the logarithm equal to $x$ and rewrite the statement in exponential form.

Parts (c) and (e) are special cases that hold for every valid base. Because $a^0 = 1$ and $a^1 = a$,

More generally, $\log_a a^x = x$: the logarithm to base $a$ undoes raising $a$ to a power. These facts are developed further, with the other laws of logarithms, in 1.7.

Logarithms that are not whole numbers

Most logarithms are not integers. Since $2^3 = 8$ and $2^4 = 16$, and 10 lies between 8 and 16, the number $\log_2 10$ lies between 3 and 4. A calculator gives $\log_2 10 = 3.3219\ldots$, which can be checked by evaluating $2^{3.3219} \approx 10.000$. Estimating between known powers like this is a useful check on any calculator answer.

Why the base and the argument must be positive

The number inside a logarithm is called its argument. For a positive base $a$, every power $a^x$ is positive, whatever the value of $x$. The graph of $y = 2^x$ shows this: the curve stays above the $x$-axis everywhere, approaching it on the left without ever reaching it.

So there is no exponent $x$ with $2^x = 0$ or $2^x = -4$, and $\log_2 0$ and $\log_2(-4)$ do not exist. The restrictions on the base have similar reasons. A base of 1 gives $1^x = 1$ for every $x$, so $\log_1 5$ could not have any value and $\log_1 1$ could have every value. A negative base produces powers that jump between positive and negative values and are not even defined for many fractional exponents, such as $(-2)^{1/2}$.

Common logarithms: base 10

Notation and powers of ten

Logarithms to base 10 are called common logarithms. They match the decimal number system, so they are the natural choice for measuring size on a scale of powers of ten. In IB notation, $\log x$ with no base written means $\log_{10} x$. Some calculators and books use $\lg x$ for the same thing.

Each time $x$ is multiplied by 10, $\log x$ increases by 1. Numbers between powers of ten have logarithms between the corresponding integers: $\log 50$ lies between $\log 10 = 1$ and $\log 100 = 2$, and a calculator gives $\log 50 = 1.69897\ldots$. The integer part of $\log x$ is one less than the number of digits before the decimal point of $x$ (for $x \ge 1$), which is why $\log x$ is a measure of order of magnitude.

Solving $10^x = b$

By the definition, $10^x = b$ has the solution $x = \log b$. A calculator gives a decimal value, and the solution can be checked by substituting it back.

A base 10 scale: pH

Quantities that range over many powers of ten are often measured on a logarithmic scale. In chemistry, the acidity of a solution depends on its concentration of hydrogen ions, $[\mathrm{H^+}]$, measured in moles per cubic decimetre ($\text{mol dm}^{-3}$). This concentration varies from about $1$ in strong acids to about $10^{-14}$ in strong alkalis, so it is described by its pH instead:

The minus sign makes the pH of most solutions positive, since $[\mathrm{H^+}] < 1$ gives a negative logarithm. Pure water at $25\,^\circ\text{C}$ has $[\mathrm{H^+}] = 10^{-7}$, so its pH is 7, which is called neutral. A lower pH means a higher concentration of hydrogen ions and a more acidic solution.

Because pH is a logarithm, a difference in pH corresponds to a ratio of concentrations. Two solutions whose pH values differ by $d$ have hydrogen ion concentrations in the ratio $10^d$. A solution of pH 3 is therefore $10^2 = 100$ times as concentrated in hydrogen ions as one of pH 5, not "a little more acidic".

Euler's number e and natural logarithms

Continuous compounding and the number e

Suppose 1 dollar is invested for one year at an interest rate of 100% per year. With the interest added once at the end of the year, the investment becomes $1 + 1 = 2$ dollars. If the interest is added twice, at 50% each half-year, it becomes $\left(1 + \frac{1}{2}\right)^2 = 2.25$ dollars, because the second half-year earns interest on the first half-year's interest. In general, compounding $n$ times a year at $\frac{100}{n}\%$ each time gives

dollars after one year. Compound interest itself is treated in 1.4; the question here is what happens as $n$ becomes very large.

The values increase, but by less and less, and they settle towards a fixed number close to $2.71828$. More frequent compounding never makes the investment grow beyond this value. The limiting case, in which interest is added at every instant, is called continuous compounding, and the limiting value is the number $e$.

Euler's number is named after the Swiss mathematician Leonhard Euler. It is the base that makes calculus with exponential functions simplest: the function $e^x$ is its own derivative, a fact used throughout 5.6 and later. The idea of a value that a sequence approaches without reaching it is the idea of a limit, made precise in 5.1. On a calculator, $e^x$ is a separate key or function, and $e$ itself is $e^1$.

The same reasoning works for other interest rates. At an annual rate of 5% compounded $n$ times a year, one dollar grows to $\left(1 + \frac{0.05}{n}\right)^n$ after one year, and as $n$ increases this approaches $e^{0.05}$. Continuous compounding at a rate $r$ (as a decimal) multiplies an amount by $e^{r}$ each year, and by $e^{rt}$ after $t$ years.

The natural logarithm

A logarithm to base $e$ is called a natural logarithm and is written $\ln x$ (from the Latin logarithmus naturalis). Like any logarithm, it is an exponent: $\ln x$ is the power to which $e$ must be raised to give $x$.

The special cases of logarithms in any base apply to $\ln$ as well: $\ln 1 = 0$ because $e^0 = 1$, and $\ln e = 1$ because $e^1 = e$. More generally, $\ln e^x = x$ for every real $x$ and $e^{\ln x} = x$ for every $x > 0$, because taking $\ln$ undoes raising $e$ to a power, and raising $e$ to a power undoes taking $\ln$.

Solving simple exponential and logarithmic equations

Equations of the form $e^x = b$ and $10^x = b$

An equation such as $e^x = 3$ is solved by rewriting it in logarithmic form, $x = \ln 3$. This is the exact answer; a calculator gives the decimal value $1.0986\ldots$. When the exponential term is part of a larger expression, first rearrange to get the power on its own, then take the logarithm.

The method in three steps:

1. Rearrange so that the power ($e^{\ldots}$ or $10^{\ldots}$) is alone on one side.

2. Rewrite in logarithmic form with the matching base: $\ln$ for $e$, $\log$ for 10.

3. Solve the resulting linear equation for $x$, then evaluate if a decimal answer is wanted.

Equations with a logarithm

When the unknown is inside a logarithm, rewrite the equation in exponential form. A solution must give a positive argument, so it is worth checking each answer in the original equation.

Other bases

An equation such as $2^x = 10$ uses a base other than 10 or $e$. Its exact solution is $x = \log_2 10$, and many calculators can evaluate this directly. Alternatively, the solution can be found as the $x$-coordinate of the intersection of the graphs of $y = 2^x$ and $y = 10$ on a graphic display calculator (GDC), which gives $x = 3.32$ (3 s.f.). The algebraic method for any base, using the change of base formula $\log_2 10 = \frac{\ln 10}{\ln 2}$, is developed in 1.7.

Exam-style questions on exponents and logarithms

Questions on this content appear both without a calculator, where answers must be exact (such as $\ln 5$ or $\frac{1}{2} + \ln 3$), and with a GDC, often in a context such as pH or continuous growth. The two worked examples show both types with the setting out that earns full credit.

Further points that earn or lose marks on this content:

• When a question says "write down", the answer should follow from a known fact, such as $\log_3 27 = 3$ or $\ln 1 = 0$; no working is expected.

• An answer of $\ln(-2)$ or $\log 0$ is never valid. If an equation leads to one, say that there is no solution and give the reason.

• Final answers are exact or correct to 3 significant figures unless the question says otherwise.

Self-check

• To do:

  I can explain why $a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$ using the division law.

• To do:

  I can simplify expressions with integer exponents and write answers with positive exponents only.

• To do:

  I can convert between $a^x = b$ and $x = \log_a b$ in both directions.

• To do:

  I can evaluate logarithms such as $\log_2 32$, $\log_3 \frac{1}{9}$

• To do:

  I can explain why $\log_a b$ is undefined when $b \le 0$.

• To do:

  I can explain how $e$ arises from $\left(1 + \frac{1}{n}\right)^n$ and state its value to 5 decimal places.

• To do:

  I can solve equations such as $10^x = 50$, $3e^x + 2 = 20$ and $\ln(2x - 1) = -1$

• To do:

  I can calculate a pH from a hydrogen ion concentration, and the concentration from a pH.

1. (a)

   The regression line of $y$ on $t$ has equation $y = at + b$. Find the value of $a$ and the value of $b$.

   [2]2 marks

   Markscheme (a)

   $a = 5.20 \; (5.19642\ldots)$, $b = -4.56 \; (-4.55714\ldots)$ A1A1

2. (b)

   Write down the value of $r$, Pearson's product-moment correlation coefficient, for these data.

   [1]1 mark

   Markscheme (b)

   $r = 0.954 \; (0.953849\ldots)$ A1

3. (c)

   A scatter diagram suggests that the relationship between $t$ and $y$ is not linear. Let $z = \ln y$.

   The regression line of $z$ on $t$ has equation $z = ct + d$

4. (d)

   Find the value of Pearson's product-moment correlation coefficient for $t$ and $z$. Hence state which of the two regression lines gives the better model for the data.

   [2]2 marks

   Markscheme (d)

   $r = 0.999 \; (0.999274\ldots)$ A1

   the line of $z$

5. (e)

   Show that the equation $z = ct + d$ can be written in the form $y = Ae^{kt}$, and find the value of $A$

6. (f)

   Use the model $y = Ae^{kt}$ to estimate the number of downloads in week 9.

   [2]2 marks

   Markscheme (f)

   $y = 3.17601\ldots \times e^{0.349165\ldots \times 9}$

7. (g)

   Comment on the reliability of your estimate in part (f).

   [1]1 mark

   Markscheme (g)

   $t = 9$ is outside the range of the data, so this is an extrapolation and may be unreliable R1

8. (h)

   According to the model $y = Ae^{kt}$, find the first week in which the number of downloads is more than 100 thousand.

   [3]3 marks

   Markscheme (h)

   $3.17601\ldots e^{0.349165\ldots t} > 100$

(b) Turn the fraction upside down and make the exponent positive: $\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$.

(c) $5^0 = 1$ and $5^{-1} = \frac{1}{5}$, so the sum is $1 + \frac{1}{5} = \frac{6}{5}$.

(d) Combine the exponents in the numerator first, then divide.

Divide the coefficients, then subtract the exponents of each letter.

Rewrite $y^{-7}$ as $\frac{1}{y^7}$ so that every exponent is positive.

(b) Reverse the process: the base of the logarithm is raised to the value of the logarithm.

The second statement says that the square root of 4 is 2. Fractional exponents such as $\frac{1}{2}$ are explained fully in 1.7.

(b) Let $x = \log_3 \frac{1}{9}$. Then $3^x = \frac{1}{9} = 3^{-2}$, so $x = -2$.

(c) Let $x = \log_7 1$. Then $7^x = 1 = 7^0$, so $x = 0$.

(d) Let $x = \log_{10} 0.001$. Then $10^x = 0.001 = 10^{-3}$, so $x = -3$.

(e) Let $x = \log_5 5$. Then $5^x = 5 = 5^1$, so $x = 1$.

The exact answer is $x = 1 + \log 0.4$. The value of $\log 0.4$ is negative because $0.4 < 1$.

(c) Rewrite in exponential form: $x = 10^{2.5} = 316.227\ldots \approx 316$.

This lies between 3 and 4, as expected, since $10^{-4} < 3.2 \times 10^{-4} < 10^{-3}$.

(b) Use the exponential form:

(c) The ratio of the concentrations is

The hydrogen ion concentration of A is about 31.6 times that of B. The answer depends only on the difference in pH, $1.5$.

(c) $\ln 7$ is the power of $e$ that gives 7, so raising $e$ to that power gives $e^{\ln 7} = 7$.

(d) $\sqrt{e} = e^{1/2}$, so $\ln \sqrt{e} = \frac{1}{2}$.

(c) Isolate the power first.

(d) Since $e^x > 0$ for every real $x$, the equation $e^x = -2$ has no solution. Writing $x = \ln(-2)$ is not an answer, because $\ln$ of a negative number is undefined.

Check: $2x - 1 = e^{-1} > 0$, so the logarithm is defined.

(c) Divide by 5 first: $\log(x + 3) = 2$. Then $x + 3 = 10^2 = 100$, so $x = 97$.

One mark for $8x^6$ (each part of the bracket raised to the power 3), one for the final answer.

(b) "Write down" means no working is expected: $2^5 = 32$, so $\log_2 32 = 5$.

(c) Write the argument as a power of 10: $\sqrt{1000} = (10^3)^{1/2} = 10^{3/2}$. Hence

(d) Rewrite in logarithmic form, then solve the linear equation.

So $a = 1$, $b = 5$, $c = 2$.

(c) "Hence" means use part (b).

This equals $10^{1.5}$, as a difference of $1.5$ in pH predicts.

[2]2 marks

[3]3 marks

$73.6$ thousand $\; (73.5609\ldots)$ A1

Note: Accept $73.5$ from 3 s.f. values.

$t > 9.87939\ldots$ (A1)

week 10 A1

Note: Accept an equation, or a table of values showing $73.6$ thousand in week 9 and $104$ thousand in week 10, for the (M1)(A1).

[2]2 marks

[1]1 mark