Transcript ELF.01.6 - Laws of Logarithms

T.1.2 - Laws of Logarithms
IB Math SL1 - Santowski
7/21/2015
IB Math SL1 - Santowski
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Lesson Objectives
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Understand the rationale behind the “laws of
logs”
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Apply the various laws of logarithms in
solving equations and simplifying expressions
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(A) Properties of Logarithms – Product
Law
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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
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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
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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
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loga(mn) = logam + logan
logam + logan = loga(mn)
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(A) Properties of Logarithms – Product
Law
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Express logam + logan as a single logarithm
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We will let logam = x and logan = y
So logam + logan becomes x + y
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But if logam = x, then ax = m and likewise ay = n
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Now take the product (m)(n) = (ax)(ay) = ax+y
Rewrite mn=ax+y in log form  loga(mn)=x + y
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But x + y = logam + logan
So thus loga(mn) = logam + logan
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(B) Properties of Logarithms – Quotient
Law
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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
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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
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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
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loga(m/n) = logam - logan
logam - logan = loga(m/n)
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(C) Properties of Logarithms- Logarithms
of Powers
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Now work with log5(625) = log5(54) = x :
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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
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So we can generalize as log5(54) = 4 [log5(5)]
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(C) Properties of Logarithms- Logarithms
of Powers
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So if log5(625) = log5(5)4 = 4 x log5(5)
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It would suggest a rule of logarithms 
logb(bx) = x
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Which we can generalize 
logb(ax) = xlogb(a)
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(D) Properties of Logarithms – Logs as
Exponents
log3 5
3
x
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Consider the example
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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!!!!
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We can demonstrate this algebraically as well
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(D) Properties of Logarithms – Logs as
Exponents
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Let’s take our exponential equation and write it in
logarithmic form
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So
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Since both sides of our equation have a log3 then x
= 5 as we had tried to reason out in the previous
slide
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So we can generalize that
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log3 5
3
x
becomes log3(x) = log3(5)
b
logb x
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x
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(E) Summary of Laws
Logs as
exponents
b
logb x
x
Product
Rule
loga(mn) = logam + logan
Quotient
Rule
loga(m/n) = loga(m) – loga(n)
Power Rule
Loga(mp) = (p) x (logam)
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(F) Examples
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(i) log354 + log3(3/2)
(ii) log2144 - log29
(iii) log30 + log(10/3)
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(iv) which has a greater value
 (a) log372 - log38 or (b) log500 + log2
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(v) express as a single value
 (a) 3log2x + 2log2y - 4log2a
 (b) log3(x+y) + log3(x-y) - (log3x + log3y)
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(vi) log2(4/3) - log2(24)
(vii) (log25 + log225.6) - (log216 + log39)
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(F) Examples
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Solve for x
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1
log x  log   2 log x
 x
1
3
log7 28  log7 4
3
3 log 5 50  log 2 2
3
log2 x  2 log2 7  log2 3
log2 x  log2 11  log2 99
1
log x  log13  log
91
3
1
3
log3 x  log3 16  log3 625
2
4
log5 x   2 log5 12  log25 3
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Simplify
1
2
1
2
log x  log y  0.5 log xy
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(F) Examples
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Given that log62 = 0.387, log63 = 0.613 &
log65 = 0.898, evaluate the following:
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(a) log618
(c) log6(3/2)
(e) log6(2.25)
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(b) log620
(d) log6300
(e) log61/√2
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(F) Examples

If logax = m and logay = n,
express the following in terms
of m and n:
x
loga  
 y
 
loga xy
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2 2
loga 5
x2
y
loga 2
1
x2 y
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Express as a single
logarithm:
1
1
1
(a) log5 x  log5 y  log5 z
2
3
4
1
(b) log4 x  3 log4 y   2log4 a  log4 b 
2
1
(c) 2 log2 x  3 log2 y   4 log2 z
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(F) Examples
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Solve and verify
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1
log2 x  log2 3  log2 3
3
1
log5 x  1  log5 x  5  log5
x3
1
log 250 log 2  3 log
x
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If a2 + b2 = 23ab,
prove that
 a  b  log a  logb
log

2
 5 
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(G) Internet Links
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Logarithm Rules Lesson from Purple Math
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College Algebra Tutorial on Logarithmic
Properties from West Texas AM
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(H) Homework
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Nelson textbook, p125, Q1-14
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HW
Ex 4C#1acgklo, 2adi, 3, 4ade;#7acde, 8, 9, 10
Ex 5C #1dgikl,2bdfh, 3ceg
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IB Packet #2, 3, 7, 8ab, 9a
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