IB Applications and Interpretation SL · Basic algebra and arithmetic · Section 3 of 5

Integer exponents and logarithms

3 worked examples · By Álvaro López

Key idea

A power is a shorthand for repeated multiplication, and a logarithm does the reverse: it tells you which exponent is needed. Here you learn the rules for handling powers with integer exponents and how to use $lo g$ and $ln$ to solve for an unknown that sits in the exponent, something very common in growth, investment and decay problems in the IB exam.

Powers with integer exponents

To extend the idea to zero and negative exponents, the rules below must still hold. This gives $x^{0} = 1$ (if $x \neq = 0$ ), and a negative exponent means "take the reciprocal" of the power.

Property · Properties of powers

For bases $x, y \neq = 0$ and integer exponents $m, n$ :

Rule	Formula
Zero exponent	$x^{0} = 1$
Product of the same base	$x^{m} \cdot x^{n} = x^{m + n}$
Quotient of the same base	$\frac{x ^{m}}{x ^{n}} = x^{m - n}$
Power of a power	$(x^{m})^{n} = x^{mn}$
Power of a product	$(x y)^{n} = x^{n} y^{n}$
Power of a quotient	$(\frac{x}{y})^{n} = \frac{x ^{n}}{y ^{n}}$
Negative exponent	$x^{- n} = \frac{1}{x ^{n}}$
Fraction with negative exponent	$(\frac{x}{y})^{- n} = (\frac{y}{x})^{n}$

Why they are true. With positive exponents, $x^{m} \cdot x^{n}$ puts $m$ factors together with $n$ other factors, and there are $m + n$ in total. If in the quotient $x^{m} / x^{n}$ equal factors are cancelled, $m - n$ remain. When $m = n$ the quotient equals $1$ , which forces $x^{0} = 1$ . And if $m = 0$ , the quotient rule gives $x^{- n} = 1/ x^{n}$ .

An example that combines all the ideas: $\frac{2 ^{5} \cdot 2 ^{- 2}}{2 ^{4}} = \frac{2 ^{3}}{2 ^{4}} = 2^{- 1} = \frac{1}{2}$ .

Logarithms

The logarithm answers the inverse of the question a power answers: to which exponent must the base be raised to obtain a given number?

For example, $lo g_{3} 81 = 4$ because $3^{4} = 81$ ; $lo g_{5} \frac{1}{25} = - 2$ because $5^{- 2} = \frac{1}{25}$ ; and $lo g_{9} 9 = 1$ because $9^{1} = 9$ . From the definition these facts follow:

$lo g_{a} 1 = 0$ (because $a^{0} = 1$ ) and $lo g_{a} a = 1$ .
Logarithms can only be taken of positive numbers: no power of a positive base gives zero or a negative result.
$lo g_{a} (a^{k}) = k$ and $a^{l o g_{a} b} = b$ : raising to $a$ and taking $lo g_{a}$ undo each other.

With both definitions, $lo g (1 0^{k}) = k$ , $ln (e^{k}) = k$ , $1 0^{l o g x} = x$ and $e^{l n x} = x$ hold for all $x > 0$ . For example, $lo g 1000 = 3$ , $lo g \frac{1}{10} = - 1$ , $ln e^{4} = 4$ and $ln 1 = 0$ .

Equations with the unknown in the exponent

To solve for an unknown that is in an exponent, first isolate the power and then apply the definition of logarithm.

For example, for $4 e^{2 x} = 20$ divide by $4$ to get $e^{2 x} = 5$ . Then $2 x = ln 5$ and $x = \frac{ln 5}{2} \approx 0.805$ .

Worked examples

Example 1 Basic

Simplifying powers

Calculate, giving the result as an integer or as a fraction:

a) $5^{- 2}$

b) $\frac{6 ^{9}}{6 ^{7}}$

c) $(2^{3})^{2} \cdot 2^{- 4}$

d) $(\frac{2}{5})^{- 2}$

Solution

a) A negative exponent inverts the power: $5^{- 2} = \frac{1}{5 ^{2}} = \frac{1}{25}$ .

b) Same base, so the exponents are subtracted: $6^{9 - 7} = 6^{2} = 36$ .

c) First the power of a power, $(2^{3})^{2} = 2^{6}$ , and then the product of the same base: $2^{6} \cdot 2^{- 4} = 2^{6 - 4} = 2^{2} = 4$ .

d) The fraction is inverted and the exponent becomes positive: $(\frac{2}{5})^{- 2} = (\frac{5}{2})^{2} = \frac{25}{4}$ .

Example 2 Medium

Calculating and solving with logarithms

a) Without a calculator, calculate $lo g 10000$ .

b) Without a calculator, calculate $lo g \frac{1}{1000}$ .

c) Without a calculator, calculate $ln \frac{1}{e}$ .

d) Solve $1 0^{x} = 250$ and give $x$ to three significant figures.

Solution

a) $1 0^{4} = 10000$ , so $lo g 10000 = 4$ .

b) $\frac{1}{1000} = 1 0^{- 3}$ , so $lo g \frac{1}{1000} = - 3$ .

c) As $e = e^{1/2}$ , we have $\frac{1}{e} = e^{- 1/2}$ and therefore $ln \frac{1}{e} = - \frac{1}{2}$ .

d) By the definition of the common logarithm, $x = lo g 250$ . With the GDC, $lo g 250 = 2.39794 \dots$ , which to three significant figures is $x \approx 2.40$ . Check: $1 0^{2.4} \approx 251$ , a value very close to $250$ .

Example 3 Hard

An investment that grows over time

An investment of 2,500 euros is made in a fund whose value, in euros, after $t$ years is

$V (t) = 2500 e^{0.03 t}$

a) Calculate the value of the investment after $10$ years, rounded to the nearest euro.

b) Find how long it will take for the value to reach 5,000 euros. Give $t$ to three significant figures.

c) What is the first complete year in which the value already exceeds 5,000 euros?

Solution

a) Substitute $t = 10$ : $V (10) = 2500 e^{0.3}$ . With the GDC, $e^{0.3} = 1.34986 \dots$ , so $V (10) = 3374.65 \dots \approx 3375$ euros.

b) We set up $2500 e^{0.03 t} = 5000$ .

Isolate the power, dividing by $2500$ : $e^{0.03 t} = 2$ .
Apply the definition of the natural logarithm: $0.03 t = ln 2$ .
Solve: $t = \frac{ln 2}{0.03} = \frac{0.693147 \dots}{0.03} = 23.1049 \dots \approx 23.1$ years.

c) The value reaches 5,000 euros at $23.1$ years, so at $23$ years it has not yet exceeded it and at $24$ years it does. Check: $V (23) = 2500 e^{0.69} \approx 4984$ euros (less than $5000$ ) and $V (24) = 2500 e^{0.72} \approx 5136$ euros (more than $5000$ ). The first complete year is year $24$ .

In the exam: in context problems the set-up of the equation, the step with the logarithm and the result with the requested figures are marked separately. In c) rounding $23.1$ down to $23$ would be an error: the question asks for the value to exceed 5,000 euros, so you must move on to the following year.

Common mistakes

Summary

Power: $x^{n}$ is multiplying $x$ by itself $n$ times; $x^{0} = 1$ and $x^{- n} = \frac{1}{x ^{n}}$ .
With the same base: $x^{m} x^{n} = x^{m + n}$ , $\frac{x ^{m}}{x ^{n}} = x^{m - n}$ , $(x^{m})^{n} = x^{mn}$ ; also $(x y)^{n} = x^{n} y^{n}$ and $(\frac{x}{y})^{- n} = (\frac{y}{x})^{n}$ .
Logarithm: $lo g_{a} b = x ⟺ a^{x} = b$ , with $a > 0$ , $a \neq = 1$ and $b > 0$ ; it satisfies $lo g_{a} 1 = 0$ and $lo g_{a} a = 1$ .
$lo g x$ is the logarithm to base $10$ and $ln x$ the one to base $e \approx 2.71828$ ; $lo g 1 0^{k} = k$ and $ln e^{k} = k$ .
To solve $1 0^{u} = k$ or $e^{u} = k$ , isolate the power and take $u = lo g k$ or $u = ln k$ (with $k > 0$ ).
The GDC may be used in the exam, but write the set-up and round only at the end.

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