SL 1.5 Apply exponent and logarithm laws | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Number and Algebra

IB Mathematics: Analysis and Approaches SL · Study topic SL 1.5

Powers and logarithms describe connected ideas: powers tell us the result of repeated multiplication, while logarithms tell us which exponent produces a given value. You may already know that multiplying powers with the same base adds their exponents. This lesson extends that familiar rule to other exponent laws and to logarithms. The main safeguard is to check restrictions: a real logarithm needs a positive argument, and its base must be positive and different from 1.

What you will learn

Apply exponent laws to simplify expressions, including those with negative and fractional exponents.
Apply logarithm laws while checking the conditions that make each logarithm valid.
Change between exponential and logarithmic forms and use these forms to solve simple equations.
Use a graphing calculator to check a result without replacing the mathematical reasoning.

1. Review: powers and their laws

For a non-zero number

a

, the expression

a^n

means that

a

is raised to the exponent

n

. For positive integer exponents, this is repeated multiplication. For example,

a^3=a\times a\times a

. The exponent laws let us rewrite expressions efficiently, but the bases must match where a law requires matching bases.

When multiplying powers with the same base, add the exponents. When dividing powers with the same non-zero base, subtract the exponents. A power raised to another power has its exponents multiplied. A power of a product can be applied to each factor. A negative exponent means reciprocal, and a zero exponent gives 1 for a non-zero base.

These rules also apply to fractional exponents when the expressions are defined as real numbers. In particular,

a^{1/n}

represents the positive

n

th root of positive

a

, and

a^{m/n}

represents a power of that root. To avoid ambiguity with real-valued fractional powers, use positive bases in this lesson.

a^m a^n=a^{m+n},\quad \frac{a^m}{a^n}=a^{m-n},\quad (a^m)^n=a^{mn},\quad a^{-n}=\frac{1}{a^n},\quad a^0=1

Add exponents when multiplying like bases; subtract them when dividing like bases.
A negative exponent does not make a value negative: it indicates a reciprocal.
For a real logarithm or the fractional powers used here, keep the relevant bases and arguments positive.

2. Logarithms: inverse notation and laws

A logarithm answers an exponent question. The statement

\log_a x=y

means that

a^y=x

. For this to define a real logarithm, the base must satisfy

a>0

and

a\ne1

, and the argument must satisfy

x>0

. For example,

\log_2 8=3

because

2^3=8

.

The logarithm laws follow from exponent laws. A product inside a logarithm becomes a sum, a quotient becomes a difference, and a power on the argument becomes a multiplier. These laws apply only when the logarithms involved are defined; in particular, the arguments of separate logarithms must be positive.

A logarithm does not distribute over addition. In general,

\log_a(x+y)

cannot be rewritten as

\log_a x+\log_a y

. Also, the base cannot be changed partway through an expression without a valid conversion. A useful conversion is

\log_a x=\frac{\log_b x}{\log_b a}

, where both logarithms on the right are defined. A calculator commonly provides base-10 logarithms and natural logarithms, so this conversion can evaluate other bases.

\log_a(xy)=\log_a x+\log_a y,\quad \log_a\!\left(\frac{x}{y}\right)=\log_a x-\log_a y,\quad \log_a(x^r)=r\log_a x

Convert between forms using $\log_a x=y$ if and only if $a^y=x$ .
Combine logarithms using product, quotient, and power laws only when their arguments are positive.
Do not split a logarithm over addition or subtraction.

3. Representations, equations, and technology

The exponential function

y=a^x

and the logarithmic function

y=\log_a x

describe inverse relationships when

a>0

and

a\ne1

. Their inputs and outputs switch: an exponential function accepts any real exponent and gives a positive output, while its logarithmic inverse accepts a positive input. For

a>1

, both graphs increase. This graphical picture supports the symbolic conversion between exponential and logarithmic forms.

Numerically, a logarithm can be interpreted as the exponent that produces a target value. Contextually, if a quantity is multiplied by the same factor for each equal time interval, an exponent can count those intervals; a logarithm can then identify how many intervals are needed to reach a target. No particular units are implied unless a problem supplies them.

For equations with the same positive base, rewrite both sides as powers of that base and compare exponents. If the bases do not match conveniently, isolate the exponential expression and take logarithms, checking that the isolated argument is positive. A graphing calculator can compare the two sides of an equation or evaluate a logarithm numerically. Use its display as a check: retain the exact form when available and state an appropriate rounding accuracy for decimal answers.

y=a^x\quad\Longleftrightarrow\quad x=\log_a y

The logarithm’s argument must be positive, including when solving an equation.
A calculator supports numerical checking; algebra explains why the result is valid.
Graphs help show that exponential and logarithmic functions reverse one another.

4. A reliable method

Before simplifying, identify the base and the operation: multiplication, division, a power of a power, or a logarithm of a product, quotient, or power. Apply only the matching law. After simplifying, check restrictions from the original expression, not only the rewritten one.

For an equation, state any domain restriction first, use exponent or logarithm laws to isolate the unknown, then substitute the result into the original equation. If using a calculator, keep sufficient digits during intermediate steps and round only at the end. This sequence makes a short solution both efficient and checkable.

Choose the law by the structure of the expression.
Check domains and verify equation solutions in the original statement.
Round only after the calculation is complete.

Worked example

Simplify a power expression

Simplify

\frac{(3x^2y^{-1})^2}{9xy^{-3}}

for

x>0

and

y>0

.

Apply the power to each factor
Square each factor in the numerator by multiplying each exponent by 2.
$(3x^2y^{-1})^2=9x^4y^{-2}$
Divide like bases
The coefficients cancel, and division of powers with the same base means subtracting exponents.
$\frac{9x^4y^{-2}}{9xy^{-3}}=x^{4-1}y^{-2-(-3)}$
Write with positive exponents
The resulting exponents are positive, so the simplified expression is already in the usual form.
$x^3y$

Answer:

x^3y

Check: For instance, with

x=2

and

y=3

, the original expression and

x^3y

both equal 24.

Worked example

Combine logarithms and respect the domain

For

x>2

, write

2\log_3(x-2)-\log_3(x+1)

as a single logarithm.

Use the power law
Move the coefficient 2 to the exponent of the first logarithm’s argument. The condition $x>2$ makes both arguments positive.
$2\log_3(x-2)=\log_3((x-2)^2)$
Use the quotient law
Subtracting logarithms with the same base gives the logarithm of the quotient of their arguments.
$\log_3((x-2)^2)-\log_3(x+1)=\log_3\!\left(\frac{(x-2)^2}{x+1}\right)$

Answer:

\log_3\!\left(\frac{(x-2)^2}{x+1}\right)

, for

x>2

.

Check: At

x=3

, the original expression is

2\log_3 1-\log_3 4=-\log_3 4

. The combined form is

\log_3(1/4)=-\log_3 4

.

Worked example

Solve an exponential equation

Solve

5^{2x-1}=17

. Give the answer to 3 significant figures.

Take logarithms
Since 17 is positive, take logarithms of both sides. The power law brings the exponent in front, allowing the equation to be rearranged for $x$ .
$(2x-1)\log 5=\log 17$
Isolate the unknown
Divide by $\log 5$ , then add 1 and divide by 2. A calculator can evaluate the exact logarithmic expression.
$x=\frac{1+\frac{\log 17}{\log 5}}{2}$
Evaluate and check
The calculator gives approximately 1.380, to 3 significant figures. Substitution gives an exponent of approximately 1.760, and $5^{1.760}$ is approximately 17.
$x\approx1.38$

Answer:

x=\frac{1+\frac{\log 17}{\log 5}}{2}\approx1.38

to 3 significant figures.

Check: On a graphing calculator, graph

y=5^{2x-1}

and

y=17

. Their intersection has an

x

-coordinate approximately 1.38, consistent with the algebra.

Common mistakes and how to avoid them

Writing

a^m+a^n=a^{m+n}

.

Correction: Adding powers is not an exponent law. The addition rule applies to multiplication:

a^m a^n=a^{m+n}

.

Treating

\log_a(x+y)

as

\log_a x+\log_a y

.

Correction: There is no logarithm law for a sum inside the argument. Keep the sum together unless another valid method applies.

Dropping domain restrictions after combining logarithms.

Correction: Check that every original logarithm argument is positive. A rewritten expression does not make an invalid original input valid.

Changing a negative exponent to a negative value.

Correction: A negative exponent means reciprocal:

a^{-n}=1/a^n

for

a\ne0

.

Lesson summary

Exponent laws simplify products, quotients, powers of powers, and negative exponents.
Logarithm laws convert products to sums, quotients to differences, and powers to multipliers.
A real logarithm requires a positive argument and a positive base different from 1.
Use calculator values and graphs to check analytical work, and state the required accuracy.

Check your understanding

Question 1

Simplify

\frac{p^5p^{-2}}{p}

for

p\ne0

.

$p^2$
$p^4$
$p^{-2}$
$p^8$

Show answer and explanation

p^2

The numerator is

p^{5+(-2)}=p^3

, and division by

p

gives

p^{3-1}=p^2

.

Question 2

Which expression equals

\log_2 12-\log_2 3

?

$\log_2 9$
$\log_2 4$
$\log_2 15$
$2\log_2 9$

Show answer and explanation

\log_2 4

The quotient law gives

\log_2(12/3)=\log_2 4=2

.

Question 3

What is the solution of

\log_4 x=3

?

$x=7$
$x=12$
$x=64$
$x=\frac{1}{64}$

Show answer and explanation

x=64

Convert to exponential form:

x=4^3=64

, which is positive and therefore valid as a logarithm argument.

Key terms

Exponent: The number indicating the power to which a base is raised.
Logarithm: The exponent required on a specified base to produce a given positive number.
Argument: The number or expression inside a logarithm.
Base: The fixed number raised to a power; for a real logarithm, it is positive and not equal to 1.

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SL 1.5 · Apply exponent and logarithm laws

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL · Study topic SL 1.5

What you will learn

1. Review: powers and their laws

2. Logarithms: inverse notation and laws

3. Representations, equations, and technology

4. A reliable method

Simplify a power expression

Combine logarithms and respect the domain

Solve an exponential equation

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review