Year 11 Methods — Logarithms (13E)

Today's lesson

So far you've solved $2^x=8$ by spotting that $8=2^3$. But what about $2^x=5$? The base $5$ isn't a "nice" power of $2$ — so we need a new tool. That tool is the logarithm. By the end of the lesson you should be able to:

Learning intentions

Translate between exponential form $a^x=y$ and logarithm form $\log_a(y)=x$
Evaluate logarithms by rewriting the argument as a power of the base (e.g. $\log_2(8)=3$)
Apply the three laws of logarithms — product, quotient, power — to simplify expressions
Use the change-of-base rule when the calculator only has $\log_{10}$ or $\ln$
Solve simple logarithmic equations like $\log_2(x)+\log_2(x+2)=3$ (and check the domain $x>0$)

Part 1 — What is a logarithm?

Definition. For $a>0$, $a\ne 1$ and $y>0$: $$ a^x=y \quad \Longleftrightarrow \quad \log_a(y) = x. $$ In plain English: "the logarithm of $y$, to the base $a$, is the exponent that turns $a$ into $y$."

📺 Walkthrough: the exponential form $2^3=8$ as the equivalent logarithm form $\log_2(8)=3$, colour-coded so you can see which number plays which role.

Exponential form
$2^3 = 8$
$5^4 = 625$
$7^{-2} = \tfrac{1}{49}$
$\left(\tfrac{1}{3}\right)^3 = \tfrac{1}{27}$

Logarithm form
$\log_2(8) = 3$
$\log_5(625) = 4$
$\log_7\!\left(\tfrac{1}{49}\right) = -2$
$\log_{1/3}\!\left(\tfrac{1}{27}\right) = 3$

EXAMPLE 1 — fill in the blanks

Complete each statement.
(a) $\log_2(16) = \underline{\;\;\;}$ (b) $\log_6(36) = \underline{\;\;\;}$ (c) $\log_{10}\!\left(\tfrac{1}{1000}\right) = \underline{\;\;\;}$
(d) $\log_5(\underline{\;\;\;}) = 3$ (e) $\log_2(\underline{\;\;\;}) = -1$ (f) $\log_{\underline{\;\;\;}}(81)=2$

(a) $2^?=16=2^4$, so $\boxed{\log_2(16)=4}$.
(b) $6^?=36=6^2$, so $\boxed{\log_6(36)=2}$.
(c) $10^?=\tfrac{1}{1000}=10^{-3}$, so $\boxed{\log_{10}\!\left(\tfrac{1}{1000}\right)=-3}$.
(d) $5^3=125$, so $\boxed{\log_5(125)=3}$.
(e) $2^{-1}=\tfrac{1}{2}$, so $\boxed{\log_2\!\left(\tfrac{1}{2}\right)=-1}$.
(f) $?^2=81$ ⇒ $?=9$, so $\boxed{\log_9(81)=2}$.

Part 2 — Evaluating logarithms

To evaluate $\log_a(N)$ without a calculator: ask yourself "what power of $a$ gives me $N$?" Rewrite $N$ as a power of $a$, then read off the exponent.

EXAMPLE 2 — straight powers

Evaluate (a) $\log_2(8)$ (b) $\log_3(9)$ (c) $\log_2\!\left(\tfrac{1}{8}\right)$ (d) $\log_3\!\left(\tfrac{1}{81}\right)$.

(a) $8=2^3$, so $\boxed{\log_2(8)=3}$.
(b) $9=3^2$, so $\boxed{\log_3(9)=2}$.
(c) $\tfrac{1}{8}=2^{-3}$, so $\boxed{\log_2\!\left(\tfrac{1}{8}\right)=-3}$.
(d) $\tfrac{1}{81}=3^{-4}$, so $\boxed{\log_3\!\left(\tfrac{1}{81}\right)=-4}$.

EXAMPLE 3 — fractional answers

Evaluate (a) $\log_{25}(5)$ (b) $\log_{16}(2)$ (c) $\log_4(8)$ (d) $\log_{27}(81)$.

(a) $5 = 25^{1/2}$, so $\boxed{\log_{25}(5)=\tfrac{1}{2}}$.
(b) $2 = 16^{1/4}$, so $\boxed{\log_{16}(2)=\tfrac{1}{4}}$.
(c) Convert to base 2: $\log_4(8)=\log_4(4^{3/2})=\tfrac{3}{2}$. So $\boxed{\log_4(8)=\tfrac{3}{2}}$.
(d) Convert to base 3: $\log_{27}(81)=\log_{27}(27^{4/3})=\tfrac{4}{3}$. So $\boxed{\log_{27}(81)=\tfrac{4}{3}}$.

Now you try: $\log_{25}\!\left(\tfrac{1}{5}\right)$ and $\log_{81}\!\left(\tfrac{1}{3}\right)$. Answers: $-\tfrac{1}{2}$ and $-\tfrac{1}{4}$.

What can't a log do? $\log_a(N)$ is only defined for $N>0$. So $\log_2(-2)$ and $\log_2(0)$ are both undefined — no power of $2$ gives you a negative number or zero. Domain matters when you solve log equations later.

Practice 2.1 — evaluate.

$\log_2(32)$
$\log_3(27)$
$\log_5(125)$
$\log_{10}(0.01)$
$\log_4(2)$
$\log_{16}(4)$
$\log_a(a^2)$
$\log_a(1)$

a) $5$ b) $3$ c) $3$ d) $-2$ e) $\tfrac{1}{2}$ f) $\tfrac{1}{2}$ g) $2$ h) $0$ (any base — $a^0=1$).

Part 3 — The three laws of logarithms

The laws. For $a>0,\;a\ne 1$ and $m,n>0$:

Product: $\log_a(m)+\log_a(n) = \log_a(mn)$
Quotient: $\log_a(m)-\log_a(n) = \log_a\!\left(\tfrac{m}{n}\right)$
Power: $\log_a(m^p) = p\cdot\log_a(m)$

Two freebies: $\log_a(1)=0$ and $\log_a(a)=1$.

📺 Walkthrough: the three log laws side-by-side — add becomes multiply, subtract becomes divide, and the exponent jumps down out the front.

EXAMPLE 4 — sanity check on $\log_2(8)+\log_2(4)$

Verify the product rule by computing $\log_2(8)+\log_2(4)$ two different ways.

Direct: $\log_2(8)+\log_2(4)=3+2=5$.
Using the product rule: $\log_2(8\cdot 4)=\log_2(32)=\log_2(2^5)=5$. ✓

EXAMPLE 5 — simplify

Simplify (a) $\log_2(32)-\log_2(8)$ (b) $2\log_9(9)-\log_9\!\left(\tfrac{4}{9}\right)$ (c) $3\log_a(x)+\log_a(20)$.

(a) Quotient: $\log_2\!\left(\tfrac{32}{8}\right)=\log_2(4)=\boxed{2}$.
(b) $2\log_9(9)=2(1)=2$. Then $2-\log_9\!\left(\tfrac{4}{9}\right)=\log_9(81)-\log_9\!\left(\tfrac{4}{9}\right)=\log_9\!\left(\tfrac{81\cdot 9}{4}\right)=\boxed{\log_9\!\left(\tfrac{729}{4}\right)}$.
(c) Power first: $3\log_a(x)=\log_a(x^3)$. Then add: $\log_a(x^3)+\log_a(20)=\boxed{\log_a(20x^3)}$.

EXAMPLE 6 — combine into a single log

Write each as a single logarithm.
(a) $3\log_{10}(x^3y^6)-5\log_{10}(x)$ (b) $\tfrac{1}{2}\log_7(x^4y^6)-2\log_7(y)$.

(a) $\log_{10}(x^3y^6)^3-\log_{10}(x^5)=\log_{10}(x^9y^{18})-\log_{10}(x^5)=\log_{10}\!\left(\tfrac{x^9y^{18}}{x^5}\right)=\boxed{\log_{10}(x^4y^{18})}$.
(b) $\log_7\!\left((x^4y^6)^{1/2}\right)-\log_7(y^2)=\log_7(x^2y^3)-\log_7(y^2)=\log_7\!\left(\tfrac{x^2y^3}{y^2}\right)=\boxed{\log_7(x^2y)}$.

Part 4 — Change of base

Change-of-base rule. For any allowable bases $a,b$ and argument $x>0$: $$ \log_a(x) = \frac{\log_b(x)}{\log_b(a)}. $$ This is how your calculator handles, say, $\log_2(8)$ when it only has $\log_{10}$ (= LOG) and $\log_e$ (= LN): $$ \log_2(8) = \frac{\log_{10}(8)}{\log_{10}(2)} = \frac{0.9031\ldots}{0.3010\ldots} = 3. \;\checkmark $$

EXAMPLE 7 — change of base on the CAS

Use the change-of-base rule to evaluate $\log_2(8)$ via base $10$, and confirm with your TI-Nspire.

$\log_2(8)=\dfrac{\log_{10}(8)}{\log_{10}(2)}=\dfrac{3\log_{10}(2)}{\log_{10}(2)}=3.$
On the Nspire: type log(8,2) directly, or log(8)/log(2) — both return $3$.

Part 5 — Solving simple log equations

Combine multiple logs into one using the laws, then either equate the arguments or rewrite in exponential form. Always check the domain — the argument of every log must be positive.

EXAMPLE 8 — $\log_2(x)+\log_2(x+2)=3$

Solve $\log_2(x)+\log_2(x+2)=3$.

Combine with the product rule: $\log_2\!\big(x(x+2)\big)=3$.
Rewrite in exponential form: $x(x+2)=2^3=8$.
Expand and rearrange: $x^2+2x-8=0\Rightarrow (x+4)(x-2)=0$ ⇒ $x=-4$ or $x=2$.
Domain check. $\log_2(x)$ requires $x>0$, so reject $x=-4$.
Answer: $\boxed{x=2}$. Check: $\log_2(2)+\log_2(4)=1+2=3$ ✓.

EXAMPLE 9 — $\log_{10}(x+2)-\log_{10}(x)=2\log_{10}(4)$

Solve $\log_{10}(x+2)-\log_{10}(x)=2\log_{10}(4)$.

Tidy the RHS first: $2\log_{10}(4)=\log_{10}(16)$.
Tidy the LHS with the quotient rule: $\log_{10}\!\left(\tfrac{x+2}{x}\right)=\log_{10}(16)$.
Same base ⇒ equate arguments: $\dfrac{x+2}{x}=16\Rightarrow x+2=16x\Rightarrow 15x=2$.
$\boxed{x=\tfrac{2}{15}}$. (Domain: $x>0$ ✓.)

Practice 5.1 — solve for $x$.

$\log_2(x)+\log_2(3)=4$
$\log_3(x)-\log_3(2)=2$
$\log_5(x)+\log_5(x-4)=1$
$2\log_{10}(x)=\log_{10}(9)$

a) $\log_2(3x)=4\Rightarrow 3x=16\Rightarrow x=\tfrac{16}{3}$
b) $\log_3\!\left(\tfrac{x}{2}\right)=2\Rightarrow \tfrac{x}{2}=9 \Rightarrow x=18$
c) $\log_5(x(x-4))=1 \Rightarrow x^2-4x-5=0 \Rightarrow (x-5)(x+1)=0$. Domain $x>4$, so $x=5$.
d) $\log_{10}(x^2)=\log_{10}(9) \Rightarrow x^2=9 \Rightarrow x=3$ (reject $-3$ as $x>0$).

Part 6 — Quick quiz (5 min)

Working program — Cambridge Ex 13E

After the quiz, open Cambridge Chapter 13 and work through Exercise 13E:

Question	Foundation	Standard (set work)	Advanced
Q1 — exp/log form swap	1 (a,c,e)	1*	1*
Q2 — evaluate log	2 (a,c,e)	2*	2*
Q3 — fractional answers	3 (a,c)	3*	3*
Q4 — apply the three laws	4 (a)	4*	4*
Q5 — combine into a single log	—	5*	5*
Q6 — solve log equations	—	6*	6 + extension

Next lesson (13F): using logs to solve exponential equations like $2^x=5$.

Exit ticket — write in your book

Before you pack up, in your exercise book write one sentence each:

How do you read the statement $\log_5(125)=3$ in plain English?
State the product rule for logarithms in words.
Why must we always check the domain $x>0$ after solving a log equation?