1.7 Laws of exponents and logarithms

The laws of exponents in 1.5 were stated for integer exponents. Extending them to fractional exponents connects powers with roots, so that $\sqrt{x}$ can be written $x^{1/2}$ and handled with the same rules as any other power. Every law of exponents also has a matching law of logarithms, which turns products into sums and powers into multiples. Together, these laws make it possible to solve any equation in which the unknown is an exponent, such as finding how long an investment takes to double or when a decaying quantity falls below a given level.

Rational exponents

Roots as powers

The laws of exponents from 1.5 can be used to decide what a fractional exponent should mean. If the power-of-a-power law $(a^m)^n = a^{mn}$ is to keep working, then

So $a^{1/2}$ is a number whose square is $a$: it is the square root of $a$. In the same way, $\left(a^{1/3}\right)^3 = a$, so $a^{1/3}$ is the cube root of $a$. In general, $a^{1/n}$ is the nth root of $a$.

For example, $16^{1/2} = \sqrt{16} = 4$, $8^{1/3} = \sqrt[3]{8} = 2$ and $81^{1/4} = \sqrt[4]{81} = 3$, since $3^4 = 81$. When $n$ is odd, negative numbers also have an $n$th root: $(-8)^{1/3} = -2$, because $(-2)^3 = -8$. When $n$ is even, a negative number has no real $n$th root, because no real number raised to an even power is negative.

Exponents of the form m/n

A rational exponent $\frac{m}{n}$ combines a root and a power. By the power-of-a-power law, $a^{m/n}$ can be split in two ways:

The denominator $n$ is the root and the numerator $m$ is the power. Both orders give the same answer, but taking the root first keeps the numbers small: $27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$, while $\sqrt[3]{27^2} = \sqrt[3]{729}$ needs a less familiar cube root.

All the laws of exponents from 1.5 continue to hold for rational exponents when the base is positive. A negative rational exponent still means the reciprocal, so $a^{-m/n} = \frac{1}{a^{m/n}}$.

Simplifying expressions with roots

Expressions that mix roots and powers become easier to handle once every root is written as a power. Then the laws of exponents apply directly.

Writing roots as powers is also how expressions such as $\frac{1}{\sqrt{x}}$ are prepared for differentiation and integration in topic 5, where $\frac{1}{\sqrt{x}} = x^{-1/2}$ fits the rules for powers of $x$.

The laws of logarithms

Logarithms undo powers

By the definition in 1.5, $\log_a x$ is the exponent to which $a$ must be raised to give $x$, where $a > 0$, $a \neq 1$ and $x > 0$. Two results follow at once. Raising $a$ to that exponent gives back $x$, and the logarithm of a power of $a$ is just its exponent:

The second gives the special cases $\log_a 1 = \log_a a^0 = 0$ and $\log_a a = \log_a a^1 = 1$. These four facts are used constantly when simplifying, so they are worth knowing by heart. For example, $10^{\log 7} = 7$, $e^{\ln 5} = 5$ and $\log_3 3^{-4} = -4$.

The product, quotient and power laws

Since logarithms are exponents, each law of exponents gives a law of logarithms. Multiplying powers adds the exponents, so the logarithm of a product is the sum of the logarithms.

These three laws are in the formula booklet. Each one can be proved from the matching law of exponents.

Proof of the product law. Let $p = \log_a x$ and $q = \log_a y$. In exponential form, $x = a^p$ and $y = a^q$. Multiplying,

Rewriting this in logarithmic form gives $\log_a xy = p + q = \log_a x + \log_a y$.

Proof of the quotient law. With the same $p$ and $q$, dividing gives $\frac{x}{y} = \frac{a^p}{a^q} = a^{p-q}$, so $\log_a \frac{x}{y} = p - q = \log_a x - \log_a y$.

Proof of the power law. With $x = a^p$, raising to the power $m$ gives $x^m = (a^p)^m = a^{pm}$, so $\log_a x^m = pm = m\log_a x$.

A useful special case of the power law is $\log_a \frac{1}{x} = \log_a x^{-1} = -\log_a x$. The same laws hold for $\ln$ and for $\log$ (base 10), since these are logarithms with particular bases.

Combining and expanding logarithms

The laws work in both directions. Combining several logarithms into one is useful for evaluating expressions and for solving equations; expanding one logarithm into several is useful for simplifying and, later, for differentiation.

When combining, apply the power law first, so that every logarithm has coefficient 1, then use the product and quotient laws. When expanding, do the reverse: split products and quotients first, then bring down the powers.

A common question type gives the values of two logarithms as letters and asks for other logarithms in terms of them. The method is to write each new argument as a product, quotient or power of the given numbers.

Expressions with powers of e and 10

The rule $a^{\log_a x} = x$ often appears in combination with the other laws. To simplify $e^{2\ln 3}$, first use the power law to move the 2 inside: $2\ln 3 = \ln 9$, so $e^{2\ln 3} = e^{\ln 9} = 9$. Similarly, $e^{\ln x - \ln 2} = e^{\ln (x/2)} = \frac{x}{2}$ and $10^{1 + \log 5} = 10^1 \times 10^{\log 5} = 50$.

Change of base

The change of base formula

Calculators evaluate logarithms to base 10 and base $e$ directly. To evaluate a logarithm in any other base, such as $\log_2 10$, it is rewritten in terms of one of these.

Proof. Let $y = \log_a x$, so that $a^y = x$. Take logarithms to base $b$ of both sides and use the power law.

Since $y = \log_a x$, this is the formula. Division by $\log_b a$ is allowed because $a \neq 1$, so $\log_b a \neq 0$.

The change of base formula works with any new base $b$. Choosing $b = 10$ or $b = e$ gives a decimal value with a calculator; choosing a base that both numbers are powers of gives an exact value by hand.

Equations with logarithms in different bases

When an equation contains logarithms with different bases, change them all to one base before using the other laws. A base that the others are powers of is usually the best choice.

Solving exponential equations

An exponential equation has the unknown in an exponent. There are three main methods, and choosing the right one is the first step.

Equal bases

If $a^p = a^q$ with $a > 0$ and $a \neq 1$, then $p = q$, because each positive number has only one logarithm to base $a$. So when both sides can be written as powers of the same base, the exponents can be equated.

Taking logarithms of both sides

When the bases cannot be matched, take the logarithm of both sides. The power law then brings the unknown down from the exponent. Any base works; $\ln$ or $\log$ is usual because a calculator can evaluate them.

Solving for time in growth and decay

Exponential equations arise whenever a quantity is multiplied by the same factor in each time period, as in the geometric sequences of 1.3 and the financial models of 1.4. Logarithms find the number of periods. When the answer must be a whole number of periods, solve an inequality and round in the correct direction.

The doubling time in part (a) is close to $\frac{72}{6} = 12$. This is the rule of 72 used in finance: an amount growing at $r\%$ per year doubles in roughly $\frac{72}{r}$ years. The rule works because $t = \frac{\ln 2}{\ln(1 + r/100)}$ and, for small $r$, $\ln(1 + r/100)$ is close to $\frac{r}{100}$, giving $t \approx \frac{69.3}{r}$; 72 is used instead because it has many factors and is slightly more accurate for typical rates.

When the growth factor is less than 1, as in decay or depreciation, its logarithm is negative. Dividing an inequality by a negative number reverses its direction.

Equations that become quadratics

An equation containing both $a^{2x}$ and $a^x$ is a quadratic in disguise, because $a^{2x} = (a^x)^2$. Substituting $u = a^x$ turns it into an ordinary quadratic in $u$. After solving for $u$, each value is turned back into an equation for $x$. Since $a^x > 0$ for every $x$, any negative or zero value of $u$ gives no solution.

The graph of $y = 4^x - 3 \times 2^x + 2$ confirms the two solutions of part (a). It crosses the $x$-axis exactly twice, at $x = 0$ and $x = 1$, and dips below the axis between them.

Solving logarithmic equations

Rewriting in exponential form

A logarithmic equation has the unknown inside a logarithm. The usual method is to use the laws of logarithms to collect everything into a single logarithm, $\log_a(\ldots) = c$, and then rewrite in exponential form: $(\ldots) = a^c$. If both sides are single logarithms with the same base, $\log_a P = \log_a Q$, then $P = Q$.

Checking the domain

Combining logarithms can produce values of $x$ that do not satisfy the original equation. For example, $\log x + \log(x - 3)$ is defined only when $x > 0$ and $x > 3$, but the combined form $\log x(x - 3)$ is also defined for negative $x$. Any value found by solving must therefore be substituted back into the original equation; a value that makes any argument zero or negative is rejected. Such a value is called an extraneous (or false) solution.

Logarithms on both sides and quadratic forms

When an equation has a multiple of a logarithm on one side, the power law moves the multiple inside first. Care is needed: $2\ln x$ is defined only for $x > 0$, even though $\ln x^2$ is defined for all $x \neq 0$, so the domain must be taken from the original form.

Exam-style questions on exponents and logarithms

Questions on these laws appear on both papers. Without a calculator, answers are exact and often in a given form such as $\frac{\ln a}{\ln b}$; with a GDC, the context is usually growth or decay, and the answer may be a whole number of time periods.

Further points that earn or lose marks:

• The laws of logarithms and the change of base formula are in the formula booklet; the conditions $a, x, y > 0$, $a \neq 1$ are not always printed beside them.

• An exact answer such as $\frac{\ln 5}{\ln 2} - 1$ is accepted in any equivalent form unless the question asks for a particular form.

• When a whole number of periods is asked for, give the integer and show the inequality that leads to it.

Self-check

• To do:

  I can evaluate powers with rational exponents, such as $16^{-3/4}$, without a calculator.

• To do:

  I can write roots as powers and simplify expressions such as $\frac{x\sqrt{x}}{\sqrt[3]{x}}$.

• To do:

  I can prove the product, quotient and power laws of logarithms from the laws of exponents.

• To do:

  I can combine several logarithms into one and expand one logarithm into several.

• To do:

  I can use the change of base formula to evaluate a logarithm and to solve equations with different bases.

• To do:

  I can solve exponential equations by equating exponents, by taking logarithms, or by substituting $u = a^x$.

• To do:

  I can reverse an inequality when dividing by the logarithm of a number between 0 and 1.

• To do:

  I can solve logarithmic equations and reject solutions that make an argument zero or negative.

$\mathrm{P}(\text{crack} \mid \text{positive}) = \dfrac{0.0276}{0.0858} = 0.322 \; (0.321678\ldots)$ A1

(b) The negative sign means reciprocal; then take the fourth root and cube.

(c) Turn the fraction upside down to make the exponent positive, then take the square root.

(d) Write the decimal as a fraction: $0.4 = \frac{2}{5}$. Then $32^{2/5} = \left(\sqrt[5]{32}\right)^2 = 2^2 = 4$.

So $k = \frac{7}{6}$.

(b) Raise each factor to the power $-\frac{1}{2}$.

(c) Raise both sides to the power $\frac{2}{3}$, the reciprocal of $\frac{3}{2}$, so that the exponent on the left becomes $\frac{3}{2} \times \frac{2}{3} = 1$.

Check: $9^{3/2} = \left(\sqrt{9}\right)^3 = 27$.

(b) Quotient law: $\log_2 40 - \log_2 5 = \log_2 \frac{40}{5} = \log_2 8 = 3$.

(c) Use the power law first, then the product law: $2\log 5 = \log 5^2 = \log 25$, so

(d) Quotient law: $\log_3 \frac{54}{2} = \log_3 27 = 3$.

(b) Write the constant $1$ as a logarithm to base 2: $1 = \log_2 2$. Then use the power law and combine.

(b) $4.5 = \frac{9}{2} = \frac{3^2}{2}$, so $\log_a 4.5 = 2\log_a 3 - \log_a 2 = 2q - p$.

(c) Use the quotient law and $\log_a a^2 = 2$.

(b) Both 4 and 8 are powers of 2, so change to base 2.

Both 9 and 27 are powers of 3, so change to base 3: $\log_9 27 = \frac{\log_3 27}{\log_3 9} = \frac{3}{2}$.

(c) Change $\log_a b$ to base $b$, and use $\log_b b = 1$.

(b) Write both sides as powers of 3: $9^x = (3^2)^x = 3^{2x}$ and $27^{x-1} = (3^3)^{x-1} = 3^{3x-3}$.

(c) Write both sides as powers of 2: $\frac{1}{8} = 2^{-3}$ and $16 = 2^4$.

Numerically, $x = 2.3219\ldots - 1 = 1.3219\ldots \approx 1.32$. (Equivalently, $x = \log_2 5 - 1$.)

(b) Take natural logarithms of both sides.

Expand and collect the terms in $x$ on one side.

The last form uses $2\ln 3 - \ln 7 = \ln 9 - \ln 7 = \ln\frac{9}{7}$.

Then $2^x = 1$ gives $x = 0$, and $2^x = 2$ gives $x = 1$.

(b) Since $9^x = (3^x)^2$, let $u = 3^x$.

$3^x = 3$ gives $x = 1$. $3^x = -1$ has no solution, because $3^x > 0$ for all $x$. So the only solution is $x = 1$.

(c) Let $u = e^x$, so $e^{2x} = u^2$.

$e^x = 2$ gives $x = \ln 2$, and $e^x = 3$ gives $x = \ln 3$.

(b) Combine the logarithms with the product law, then rewrite in exponential form (base 10).

Check each value in the original equation. For $x = 5$: $\log 5 + \log 2 = \log 10 = 1$. For $x = -2$: $\log(-2)$ is undefined, so $x = -2$ is rejected. The only solution is $x = 5$.

(c) Combine the left side with the quotient law, then equate the arguments.

Check: $x = \frac{1}{2}$ makes both $x + 1$ and $x$ positive, so it is valid.

The original equation contains $\ln x$, which needs $x > 0$, so $x = -2$ is rejected. The solution is $x = 3$. Check: $2\ln 3 = \ln 9 = \ln(3 + 6)$.

(b) Collect the logarithms on one side and write $2$ as an exponent.

Check: $2x + 1 = \frac{27}{7} > 0$ and $x - 1 = \frac{3}{7} > 0$, so the solution is valid.

(c) This is a quadratic in $\log_2 x$. Let $u = \log_2 x$.

$\log_2 x = 1$ gives $x = 2$, and $\log_2 x = 2$ gives $x = 4$. Both are positive, so both are valid.

(b) Combine with the product law, then rewrite in exponential form.

So $x = 2$ or $x = -4$. Since $\log_2 x$ needs $x > 0$, reject $x = -4$. The solution is $x = 2$.

(c) Write $4^x = (2^x)^2$ and $2^{x+1} = 2 \times 2^x$, then let $u = 2^x$.

$2^x = 4$ gives $x = 2$. $2^x = -2$ has no solution, since $2^x > 0$. The solution is $x = 2$.

Take natural logarithms: $t = \frac{\ln 4}{\ln(1.08/0.95)} = 10.808\ldots \approx 10.8$ years. The same value comes from the intersection of the two graphs on the GDC.

(b) $A = 500 \times 1.08^{10.808\ldots} = 1148.8\ldots$, so both populations are about 1150 fish.

(c) Solve $500 \times 1.08^t > 1000$, so $1.08^t > 2$ and $t > \frac{\ln 2}{\ln 1.08} = 9.006\ldots$. Population $A$ first exceeds 1000 during the tenth year, so after 10 whole years (check: $A(9) = 999.5$ and $A(10) = 1079.5$).

[3]3 marks

[2]2 marks