t.1.2 - laws of logarithms
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T.1.2 - Laws of Logarithms. IB Math SL1 - Santowski. Lesson Objectives. Understand the rationale behind the “laws of logs” Apply the various laws of logarithms in solving equations and simplifying expressions. (A) Properties of Logarithms – Product Law. - PowerPoint PPT PresentationTRANSCRIPT
04/24/23 IB Math SL1 - Santowski 1
T.1.2 - Laws of Logarithms
IB Math SL1 - Santowski
04/24/23 IB Math SL1 - Santowski 2
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