MATH 1510
Lili Shen
Laws of
Logarithms
Fundamentals of Mathematics
(MATH 1510)
Instructor: Lili Shen
Email: [email protected]
Department of Mathematics and Statistics
York University
January 18-22, 2016
Outline
MATH 1510
Lili Shen
Laws of
Logarithms
1
Laws of Logarithms
Laws of logarithms
MATH 1510
Lili Shen
Laws of
Logarithms
Proposition
Let a > 0 with a 6= 1. Let A, B, C ∈ R with A > 0 and B > 0.
(1) loga (AB) = loga A + loga B.
A
(2) loga = loga A − loga B.
B
(3) loga (AC ) = C loga A.
Laws of logarithms
MATH 1510
Lili Shen
Laws of
Logarithms
Example
Evaluate each expression.
(1) log4 2 + log4 32.
(2) log2 80 − log2 5.
1
(3) − log 8.
3
Laws of logarithms
MATH 1510
Lili Shen
Laws of
Logarithms
Solution.
(1) log4 2 + log4 32 = log4 (2 · 32) = log4 64 = 3.
80
= log2 16 = 4.
(2) log2 80 − log2 5 = log2
5
1
1
1
(3) − log 8 = log 8− 3 = log ≈ −0.301.
3
2
Expanding logarithmic expressions
MATH 1510
Lili Shen
Laws of
Logarithms
Example
Expand each expression:
(1) log2 (6x).
(2) log5 (x 3 y 6 ).
ab
(3) ln √
.
3
c
Expanding logarithmic expressions
MATH 1510
Lili Shen
Laws of
Logarithms
Solution.
(1) log2 (6x) = log2 6 + log2 x.
(2) log5 (x 3 y 6 ) = log6 x 3 + log5 y 6 = 3 log5 x + 6 log5 y.
√
ab
1
(3) ln √
= ln(ab) − ln 3 c = ln a + ln b − ln c.
3
3
c
Combining logarithmic expressions
MATH 1510
Lili Shen
Laws of
Logarithms
Example
Combine each expression into a single logarithm:
1
(1) 3 log x + log(x + 1).
2
1
(2) 3 ln s + ln t − 4 ln(t 2 + 1).
2
Combining logarithmic expressions
MATH 1510
Lili Shen
Laws of
Logarithms
Solution.
(1)
3 log x +
1
1
log(x + 1) = log x 3 + log(x + 1) 2
2
√
= log(x 3 x + 1).
(2)
3 ln s +
1
1
ln t − 4 ln(t 2 + 1) = ln s3 + ln t 2 − ln(t 2 + 1)4
2
√
s3 t
.
= ln 2
(t + 1)4
Combining logarithmic expressions
MATH 1510
Lili Shen
Laws of
Logarithms
Example
Combine and simplify
1
1
log(x + 2)5 + [log(x 6 ) − log(x 2 − x − 6)3 ].
5
3
Combining logarithmic expressions
MATH 1510
Lili Shen
Laws of
Logarithms
Solution.
1
1
log(x + 2)5 + [log(x 6 ) − log(x 2 − x − 6)3 ]
5
3
= log(x + 2) + log(x 2 ) − log(x 2 − x − 6)
x 2 (x + 2)
x2 − x − 6
x 2 (x + 2)
= log
(x − 3)(x + 2)
x2
= log
.
x −3
= log
The law of forgetting
MATH 1510
Lili Shen
Example
Laws of
Logarithms
If a task is learned at a performance level P0 , then after a
time interval t the performance level P satisfies
log P = log P0 − c log(t + 1),
where c is a constant that depends on the type of task and t
is measured in months.
(1) Solve for P.
(2) If your score on a history test is 90, what score would
you expect to get on a similar test after two months?
After a year? (Assume that c = 0.2.)
The law of forgetting
MATH 1510
Lili Shen
Laws of
Logarithms
Solution.
(1) log P = log P0 − c log(t + 1) = log
P=
P0
implies
(t + 1)c
P0
.
(t + 1)c
(2) Here P0 = 90, c = 0.2, and t is measured in months.
90
In 2 months: P = 0.2 ≈ 72.
3
90
In 1 year: P = 0.2 ≈ 54.
13
Your expected scores after 2 months and after 1 year
are 72 and 54, respectively.
Change of base formula
MATH 1510
Lili Shen
Laws of
Logarithms
Proposition
logb x =
loga x
.
loga b
Change of base formula
MATH 1510
Lili Shen
Laws of
Logarithms
Corollary
1
.
logb a
loga b
(2) logac b =
.
c
(3) alogb c = c logb a .
(1) loga b =
Change of base formula
MATH 1510
Lili Shen
Laws of
Logarithms
The change of base formula is quite useful when evaluating
logarithms using calculators:
Example
Evaluate each logarithm:
(1) log8 5.
(2) log9 20.
(3) log4 3 · log9 64.
(4) log2 3 · log3 4 · log4 5 · log5 2.
(5) log16 27 · log81 32.
Change of base formula
MATH 1510
Lili Shen
Laws of
Logarithms
Solution.
log 5
≈ 0.77398.
log 8
ln 20
≈ 1.36342.
(2) log9 20 =
ln 9
log 3 · log 8
1
3
(3) log2 3 · log9 8 =
= log9 3 · log2 8 = · 3 = .
log 2 · log 9
2
2
(4) log2 3 · log3 4 · log4 5 · log5 2 = log3 3 · log4 4 · log5 5 · log2 2 = 1.
3
5
15
(5) log16 27 · log81 32 = log2 3 · log3 2 =
.
4
4
16
(1) log8 5 =