Lesson 4 ~35 min Logarithms · Path +85 XP

Using the Log Laws

Knowing the laws is not the same as using them. This lesson adds the four results worth knowing on sight, then puts everything to work turning long expressions into single numbers.

Today's hook: Evaluate $\log_2 80 - \log_2 5$ without a calculator. Neither logarithm is a whole number, yet the answer is. One law turns an impossible-looking question into a two-second one.

0/5QUESTS

Think First

warm-up

What is $\log_7 1$? And what is $\log_7 7$? Neither needs a calculator if you translate to index form first. Now try $\log_7 (7^5)$, and then the harder-looking $7^{\log_7 5}$. What is the pattern in all four?

Record your answer in your workbook.

The Big Idea

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Four results fall straight out of the definition and are worth knowing on sight, because almost every simplification ends by using one of them. Combined with the three laws, they let you collapse an expression rather than evaluate its parts.

$$\log_a 1 = 0, \quad \log_a a = 1, \quad \log_a(a^x) = x, \quad a^{\log_a x} = x$$

The last two say the same thing twice: a logarithm and an exponential of the same base undo each other. That is the definition of inverse functions, written out. The strategy that follows is always the same: use the laws to gather everything into ONE logarithm, then evaluate that single logarithm.

$\log_a a = 1, \quad \log_a(a^x) = x$

Gather, then evaluate

Combine into one logarithm first. Evaluating term by term usually cannot be done.

Coefficients go inside

$2\log_a 3$ is $\log_a 9$. Move the multiplier in before combining.

Look for the base

If the argument is a power of the base, you are already finished.

What You'll Master

objectives

Know

The results $\log_a 1 = 0$, $\log_a a = 1$, $\log_a(a^x) = x$ and $a^{\log_a x} = x$
That the three laws combine several logarithms of one base into a single logarithm

Understand

Why $\log_a(a^x) = x$ and $a^{\log_a x} = x$ are two statements of the same inverse relationship
Why gathering into one logarithm is almost always the productive first move

Can Do

Evaluate expressions such as $\log_2 80 - \log_2 5$ without a calculator
Simplify expressions combining coefficients, products and quotients of logarithms
Express a single logarithm in terms of given logarithms, such as writing $\log_a 24$ using $\log_a 2$ and $\log_a 3$

Words You Need

vocabulary

$\log_a 1 = 0$Because $a^0 = 1$ for every allowed base.

$\log_a a = 1$Because $a^1 = a$.

$\log_a(a^x) = x$The logarithm undoes the power.

$a^{\log_a x} = x$The power undoes the logarithm. Same fact, other order.

SimplifyTo rewrite as a single logarithm, or as a number, using the laws.

EvaluateTo produce an actual value rather than an equivalent expression.

The Four Standard Results

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Each one is a single translation away from the definition.

$\log_a 1 = 0$. Ask: what index turns $a$ into $1$? Since $a^0 = 1$ for every allowed base, the answer is always $0$. Every logarithm graph passes through $(1, 0)$ for this reason.

$\log_a a = 1$. What index turns $a$ into $a$? The index $1$.

$\log_a(a^x) = x$. What index turns $a$ into $a^x$? It is written in front of you.

$a^{\log_a x} = x$. Read the exponent as "the index that turns $a$ into $x$". Putting that index on $a$ must therefore give $x$.

HSC Tip

The last two are the pair examiners use to make an expression look worse than it is. $5^{\log_5 12}$ is just $12$.

Gather, Then Evaluate

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Faced with several logarithms of the same base, resist evaluating them one at a time. Most are not whole numbers, and the point of the laws is that you never need them individually.

Take $\log_2 80 - \log_2 5$. Neither term is a whole number. But the quotient law gathers them:

$$\log_2 80 - \log_2 5 = \log_2\left(\frac{80}{5}\right) = \log_2 16 = 4$$

The strategy in three words: gather, then evaluate. Gather everything into one logarithm using the laws, and only then ask what that single logarithm is worth.

When a coefficient is in the way, the power law moves it inside first:

$$2\log_3 6 - \log_3 4 = \log_3(6^2) - \log_3 4 = \log_3\left(\frac{36}{4}\right) = \log_3 9 = 2$$

Expressing One Logarithm Using Others

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A common question gives you two logarithms and asks for a third. The method is to factorise the argument into the numbers you were given.

Suppose $\log_a 2 = p$ and $\log_a 3 = q$. To find $\log_a 24$, first factorise:

$$24 = 8 \times 3 = 2^3 \times 3$$

Now apply the product law, then the power law:

$$\log_a 24 = \log_a(2^3) + \log_a 3 = 3\log_a 2 + \log_a 3 = 3p + q$$

Fractions work the same way through the quotient law. With the same $p$ and $q$:

$$\log_a\left(\frac{9}{2}\right) = \log_a(3^2) - \log_a 2 = 2q - p$$

Working Backwards to a Single Logarithm

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The laws run in both directions. Read left to right they split a logarithm apart; read right to left they gather several into one.

$$\log_a x + 2\log_a y - 3\log_a z$$

Move every coefficient inside with the power law:

$$= \log_a x + \log_a(y^2) - \log_a(z^3)$$

Then combine, additions becoming multiplication and subtractions becoming division:

$$= \log_a\left(\frac{xy^2}{z^3}\right)$$

Everything that was added ends up on top, everything subtracted ends up underneath, and every coefficient becomes an index.

Common Pitfalls

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Leaving a coefficient outside when combining: writing $2\log_a 3 + \log_a 5$ as $\log_a(2 \times 3 \times 5)$.

Fix: the coefficient is an INDEX, not a factor. $2\log_a 3 = \log_a(3^2) = \log_a 9$, so the sum is $\log_a 45$, not $\log_a 30$.

Evaluating each logarithm separately, getting decimals, and losing an exact answer.

Fix: gather first. $\log_2 80 - \log_2 5$ is exactly $4$; computing $6.32 - 2.32$ gets there but invites rounding error and shows no method.

Reading $a^{\log_a x}$ as something complicated.

Fix: say it aloud. "$a$ raised to the index that turns $a$ into $x$" must be $x$. So $10^{\log_{10} 7} = 7$, immediately.

Watch Me Solve It · 3 examples

Watch Me Solve It · Evaluating with the quotient law

+15 XP per step

PROBLEM

Evaluate $\log_3 162 - \log_3 2$ without a calculator.

1

Recognise that neither term is exact

$\log_3 162 \text{ and } \log_3 2 \text{ are not whole numbers}$

So evaluating separately is a dead end; the laws are the route.
2

Gather with the quotient law

$\log_3 162 - \log_3 2 = \log_3\left(\frac{162}{2}\right)$

Same base, a subtraction, so the quotient law applies.
3

Simplify the argument

$= \log_3 81$
4

Evaluate the single logarithm

$81 = 3^4 \Rightarrow \log_3 81 = 4$

Write the argument as a power of the base and read off the index.

Answer$4$

Watch Me Solve It · Coefficients first, then gather

+15 XP per step

PROBLEM

Simplify $3\log_2 2 + \log_2 12 - \log_2 6$ to a single number.

1

Deal with the standard result

$\log_2 2 = 1 \Rightarrow 3\log_2 2 = 3$

$\log_a a = 1$, so that whole term is just $3$.
2

Gather the remaining two

$\log_2 12 - \log_2 6 = \log_2\left(\frac{12}{6}\right) = \log_2 2$

Quotient law, same base.
3

Evaluate what is left

$\log_2 2 = 1$
4

Add the parts

$3 + 1 = 4$

Answer$4$

Watch Me Solve It · Expressing in terms of given logarithms

+15 XP per step

PROBLEM

Given $\log_a 2 = p$ and $\log_a 5 = q$, express $\log_a\left(\dfrac{50}{a}\right)$ in terms of $p$, $q$ and numbers.

1

Split with the quotient law

$\log_a\left(\frac{50}{a}\right) = \log_a 50 - \log_a a$
2

Use the standard result

$\log_a a = 1$
3

Factorise the remaining argument

$50 = 2 \times 25 = 2 \times 5^2$

Break it into the numbers you were given values for.
4

Apply the product and power laws

$\log_a 50 = \log_a 2 + 2\log_a 5 = p + 2q$

$\text{so the answer is } p + 2q - 1$

Answer$p + 2q - 1$

Brain Trainer · 4 problems

Brain Trainer · Collapse each expression

4 problems

Four quick problems. Work each one, then reveal the answer.

1 Evaluate $\log_5 100 + \log_5 \dfrac{5}{4}$.

Product law: $\log_5\left(100 \times \dfrac{5}{4}\right) = \log_5 125$, and $125 = 5^3$.$3$
2 Simplify $7^{\log_7 11}$.

A power and a logarithm of the same base undo each other.$11$
3 Write $\log_a x - 3\log_a y$ as a single logarithm.

Move the coefficient inside, then use the quotient law.$\log_a\left(\dfrac{x}{y^3}\right)$
4 If $\log_a 3 = t$, find $\log_a 27$ in terms of $t$.

$27 = 3^3$, so the power law gives $3\log_a 3$.$3t$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

A standard result

+10 XP

For any allowed base $a$, the value of $\log_a 1$ is:

MC2

Gathering a difference

+10 XP

$\log_2 96 - \log_2 6$ equals:

MC3

Undoing a logarithm

+10 XP

$3^{\log_3 20}$ equals:

MC4

Combining into one logarithm

+10 XP

$2\log_a x + \log_a y$ written as a single logarithm is:

MC5

In terms of given values

+10 XP

If $\log_a 2 = p$ and $\log_a 7 = q$, then $\log_a 28$ equals:

Short Answer · 3 questions

Evaluate without a calculator

+15 XP

SHORT ANSWER

Evaluate each of the following exactly, showing the law you use at each step.
(a) $\log_4 32 + \log_4 2$
(b) $\log_6 108 - \log_6 3$
(c) $2\log_5 10 - \log_5 4$

Write your working in your book.

Single logarithm and back

+15 XP

SHORT ANSWER

(a) Write $3\log_a x - \dfrac{1}{2}\log_a y + \log_a z$ as a single logarithm.
(b) Write $\log_a\left(\dfrac{x^4 \sqrt{z}}{y}\right)$ as a sum and difference of logarithms of $x$, $y$ and $z$.

Write your working in your book.

Using given logarithms

+15 XP

SHORT ANSWER

You are given $\log_a 2 = 0.43$ and $\log_a 3 = 0.68$, correct to two decimal places. Without finding $a$, evaluate:
(a) $\log_a 6$
(b) $\log_a 8$
(c) $\log_a 1.5$
(d) $\log_a \sqrt{12}$

Write your working in your book.

Stretch Challenge · A chain that collapses

+25 XP

CHALLENGE

Consider $S = \log_2\left(\dfrac{2}{1}\right) + \log_2\left(\dfrac{3}{2}\right) + \log_2\left(\dfrac{4}{3}\right) + \cdots + \log_2\left(\dfrac{64}{63}\right)$.
(a) Use the product law to write $S$ as a single logarithm.
(b) Evaluate $S$.
(c) Explain what feature of the arguments made this collapse possible, and give one other chain of the same kind.

Quick Review

recap

Know on sight

$\log_a 1 = 0$, $\log_a a = 1$

Inverses

$\log_a(a^x) = x$ and $a^{\log_a x} = x$

Strategy

Gather into one logarithm, then evaluate

Coefficients

A multiplier becomes an index

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