04/24/23 IB Math SL1 - Santowski 1

T.1.2 - Laws of Logarithms

IB Math SL1 - Santowski

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Lesson Objectives

Understand the rationale behind the “laws of logs”

Apply the various laws of logarithms in solving equations and simplifying expressions

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(A) Properties of Logarithms – Product Law

Recall the laws for exponents product of powers (bx)(by) = b(x+y) so we ADD the exponents when we multiply powers

For example (23)(25) = 2(3+5)

So we have our POWERS 8 x 32 = 256

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(A) Properties of Logarithms – Product Law Now, let’s consider this from the INVERSE viewpoint We have the ADDITION of the exponents 3 + 5 = 8 But recall from our work with logarithms, that the

exponents are the OUTPUT of logarithmic functions So 3 + 5 = 8 becomes log28 + log232 = log2256 Now, HOW do we get the right side of our equation to

equal the left? Recall that 8 x 32 = 256 So log2(8 x 32) = log28 + log232 = log2256

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(A) Properties of Logarithms – Product Law

So we have our first law when adding two logarithms, we can simply write this as a single logarithm of the product of the 2 powers

loga(mn) = logam + logan logam + logan = loga(mn)

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(A) Properties of Logarithms – Product Law Express logam + logan as a single logarithm

We will let logam = x and logan = y So logam + logan becomes x + y

But if logam = x, then ax = m and likewise ay = n

Now take the product (m)(n) = (ax)(ay) = ax+y

Rewrite mn=ax+y in log form loga(mn)=x + y

But x + y = logam + logan So thus loga(mn) = logam + logan

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(B) Properties of Logarithms – Quotient Law

Recall the laws for exponents Quotient of powers (bx)/(by) = b(x-y) so we subtract the exponents when we multiply powers

For example (28)/(23) = 2(8-3)

So we have our POWERS 256 ÷ 8 = 32

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(B) Properties of Logarithms – Quotient Law Now, let’s consider this from the INVERSE viewpoint We have the SUBTRACTION of the exponents 8 - 3 = 5 But recall from our work with logarithms, that the

exponents are the OUTPUT of logarithmic functions So 8 - 3 = 5 becomes log2256 - log28 = log232 Now, HOW do we get the right side of our equation to

equal the left? Recall that 256/8 = 32 So log2(256/8) = log2256 - log28 = log232

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(B) Properties of Logarithms – Quotient Law

So we have our second law when subtracting two logarithms, we can simply write this as a single logarithm of the quotient of the 2 powers

loga(m/n) = logam - logan logam - logan = loga(m/n)

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(C) Properties of Logarithms- Logarithms of Powers

Now work with log5(625) = log5(54) = x :

we can rewrite as log5(5 x 5 x 5 x 5) = x we can rewrite as log5(5) + log5(5) + log5(5) + log5(5) = x We can rewrite as 4 [log5(5)] = 4 x 1 = 4

So we can generalize as log5(54) = 4 [log5(5)]

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(C) Properties of Logarithms- Logarithms of Powers So if log5(625) = log5(5)4 = 4 x log5(5)

It would suggest a rule of logarithms logb(bx) = x

Which we can generalize logb(ax) = xlogb(a)

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(D) Properties of Logarithms – Logs as Exponents Consider the example

Recall that the expression log3(5) simply means “the exponent on 3 that gives 5” let’s call that y

So we are then asking you to place that same exponent (the y) on the same base of 3

Therefore taking the exponent that gave us 5 on the base of 3 (y) onto a 3 again, must give us the same 5!!!!

We can demonstrate this algebraically as well

x5log33

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(D) Properties of Logarithms – Logs as Exponents Let’s take our exponential equation and write it in

logarithmic form

So becomes log3(x) = log3(5)

Since both sides of our equation have a log3 then x = 5 as we had tried to reason out in the previous slide

So we can generalize that

x5log33

xb xb log

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(E) Summary of Laws

Logs as exponents

Product Rule

loga(mn) = logam + logan

Quotient Rule

loga(m/n) = loga(m) – loga(n)

Power Rule Loga(mp) = (p) x (logam)

xb xb log

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(F) Examples (i) log354 + log3(3/2) (ii) log2144 - log29 (iii) log30 + log(10/3)

(iv) which has a greater value (a) log372 - log38 or (b) log500 + log2

(v) express as a single value (a) 3log2x + 2log2y - 4log2a (b) log3(x+y) + log3(x-y) - (log3x + log3y)

(vi) log2(4/3) - log2(24) (vii) (log25 + log225.6) - (log216 + log39)

(F) Examples

Solve for x Simplify

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911log13loglog

99log11loglog

3log7log2log

3

222

222

x

x

x

3log12log2log

625log4316log

21log

2555

333

x

xxyyx

xx

x

log5.0loglog

22log505log3

4log3128log

log21loglog

21

21

73

7

3

(F) Examples

Given that log62 = 0.387, log63 = 0.613 & log65 = 0.898, evaluate the following:

(a) log618 (b) log620 (c) log6(3/2) (d) log6300 (e) log6(2.25) (e) log61/√2

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(F) Examples

If logax = m and logay = n, express the following in terms of m and n:

Express as a single logarithm:

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yx

yx

xy

yx

a

a

a

a

2

5

2

22

1log

log

log

log

2

zyx

bayx

zyx

222

4444

555

log4log3log231 (c)

loglog2log3log21 (b)

log41log

31log

21 (a)

(F) Examples

Solve and verify If a2 + b2 = 23ab, prove that

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x

xxx

x

1log32log250log

31log5log1log

3log3log31log

555

222

2loglog

5log baba

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(G) Internet Links

Logarithm Rules Lesson from Purple Math

College Algebra Tutorial on Logarithmic Properties from West Texas AM

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(H) Homework

Nelson textbook, p125, Q1-14

HW Ex 4C#1acgklo, 2adi, 3, 4ade;#7acde, 8, 9, 10 Ex 5C #1dgikl,2bdfh, 3ceg

IB Packet #2, 3, 7, 8ab, 9a