Appendix N: Derivation of the Logarithm Change of Base Formula
We take loga of each side of this equation which gives us loga by = loga x
MATHEMATICS 0110A CHANGE OF BASE Suppose that we have
So we get the following rule: Change of Base Formula: logb a = logc a logc b. Example 1. Express log3 10 using natural logarithms. log3 10 =.
Change of Base
Logarithms – University of Plymouth
16 jan. 2001 7. Quiz on Logarithms. 8. Change of Bases ... following important rules apply to logarithms. ... Proof that loga MN = loga M + loga N.
PlymouthUniversity MathsandStats logarithms
Secondary V Videos and Notes
Proof of the logarithm change of base rule https://youtu.be/1reblXFlM6I. Logarithm properties: review https://www.khanacademy.org/math/algebra2/.
Secondary V Videos and Notes
6.2 Properties of Logarithms
The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with bx logb(a) and use the Power Rule
S&Z . & .
Logarithms - changing the base
This leaflet gives this formula and shows how to use it. A formula for change of base. Suppose we want to calculate a logarithm to base 2. The formula states.
mc logs
Elementary Functions The logarithm as an inverse function
Each of these three properties is merely a restatement of a property of exponents. Smith (SHSU). Elementary Functions. 2013. 18 / 29. Changing the base.
. Logarithms (slides to )
Logarithms.pdf
16 nov. 2017 This law allows a logarithm with a given base to be changed to a new base ... The third law: (Power Rule) log log n a a. x n x. = Proof:.
Logarithms
ln b ∫ ln x dx.
The conventional proof of the integral of a logarithm utilises integration change of base rule and multiplication by a constant rule:.
an alternative proof of the integral of a logarithm
Introduction to Algorithms
I can prove this using the definition of big-Omega: This tells us that every positive power of the logarithm of n to the base b where b ¿ 1
cs lect fall notes
CS 5002: Discrete Structures Fall 2018
Lecture 9: November 8, 2018
1Instructors: Adrienne Slaughter, Tamara BonaciDisclaimer:These notes have not been subjected to the usual scrutiny reserved for formal publications.
They may be distributed outside this class only with the permission of the Instructor.Introduction to Algorithms
Readings for this week:
Rosen, Chapter 2.1, 2.2, 2.5, 2.6
Sets, Set Operations, Cardinality of Sets, Matrices9.1 OverviewObjective: Introduce algorithms.
1.Review logarithms
2.Asymptotic analysis
3.Dene Algorithm
4.Ho wto express or describ ean algorithm
5.Run time, space (Resource usage )
6.Determining Correctness
7.In troducerepresen tativeproblems
1. fo o9.2 Asymptotic Analysis
The goal with asymptotic analysis is to try to nd a bound, or an asymptote, of a function. This allows
us to come up with an \ordering" of functions, such that one function is denitely bigger than another,
in order to compare two functions. We do this by considering the value of the functions asngoes to innity, so for very large values ofn, as opposed to small values ofnthat may be easy to calculate. Once we have this ordering, we can introduce some terminology to describe the relationship of two functions. 9-19-2Lecture 9: November 8, 201823456781248163264128256512102420484096n!n
n2 nn2nlog(n)log(n)n
pn1Growth of Functions
From this chart, we see:
1lognpnnnlog(n)n22nn!nn(9.1)
ComplexityTerminology
(1)Constant (logn)Logarithmic (n)Linear (nlogn)Linearithmic nbPolynomial (bn) (whereb >1)Exponential (n!)Factorial9.2.1 Big-O: Upper BoundDenition 9.1 (Big-O: Upper Bound)f(n) =O(g(n))means there exists some constantcsuch
thatf(n)cg(n), for large enoughn(that is, asn! 1).We sayf(n) =O(g(n))
Example:I claim 3n2100n+ 6 =O(n2). I can prove this using the denition of big-O:Lecture 9: November 8, 20189-3
f(n) = 3n2100n+ 6 (9.2) g(n) =n2(9.3) )3n2100n+ 6cn2for somec(9.4)Ifc= 3 : 3n2100n+ 63n2(9.5)
To prove using Big-O:
Determinef(n) andg(n)
Write the equation based on the denition
Choose acsuch that the equation is true.
{If you can nd ad, thenf(n) =O(g(n)). If not, thenf(n)6=O(g(n)).These statements are all true:
3n2100n+ 6 =O(n2) (9.6)
3n2100n+ 6 =O(n3) (9.7)
3n2100n+ 66=O(n) (9.8)
Proving
9.7 f(n) = 3n2100n+ 6 (9.9) g(n) =n3(9.10) )3n2100n+ 6 =cn3(for somec) (9.11)Ifc= 1 : 3n2100n+ 6n3(whenn >3) (9.12)
We also know this to be true because order istransitive: iff(n) =O(g(n)), andg(n) =O(h(n)), then f(n) =O(h(n)). Sincen2=O(n3), then anyf(n) =O(n2) is alsoO(n3).Proving
9.8 f(n) = 3n2100n+ 6 (9.13) g(n) =n(9.14)For anyc:cn <3n2(whenn > c) (9.15)
9.2.2 Big-Omega: Lower BoundDenition 9.2 (Big-Omega: Lower Bound)f(n) =
(g(n))means there exists some constantc such thatf(n)cg(n), for large enoughn(that is, asn! 1).We sayf(n) =
(g(n))or \fofnis Big Omega ofgofn"9-4Lecture 9: November 8, 2018
Example:I claim 3n2100n+ 6 =
(n2). I can prove this using the denition of big-Omega: f(n) = 3n2100n+ 6 (9.16) g(n) =n2(9.17) )3n2100n+ 6cn2for somec(9.18)Ifc= 2 : 3n2100n+ 62n2(9.19)
We show Big-Omega the same way we show Big-O.
These statements are all true:
3n2100n+ 6 =
(n2) (9.20)3n2100n+ 66=
(n3) (9.21)3n2100n+ 6 =
(n) (9.22)Proving
9.21CS 5002: Discrete Structures Fall 2018
Lecture 9: November 8, 2018
1Instructors: Adrienne Slaughter, Tamara BonaciDisclaimer:These notes have not been subjected to the usual scrutiny reserved for formal publications.
They may be distributed outside this class only with the permission of the Instructor.Introduction to Algorithms
Readings for this week:
Rosen, Chapter 2.1, 2.2, 2.5, 2.6
Sets, Set Operations, Cardinality of Sets, Matrices9.1 OverviewObjective: Introduce algorithms.
1.Review logarithms
2.Asymptotic analysis
3.Dene Algorithm
4.Ho wto express or describ ean algorithm
5.Run time, space (Resource usage )
6.Determining Correctness
7.In troducerepresen tativeproblems
1. fo o9.2 Asymptotic Analysis
The goal with asymptotic analysis is to try to nd a bound, or an asymptote, of a function. This allows
us to come up with an \ordering" of functions, such that one function is denitely bigger than another,
in order to compare two functions. We do this by considering the value of the functions asngoes to innity, so for very large values ofn, as opposed to small values ofnthat may be easy to calculate. Once we have this ordering, we can introduce some terminology to describe the relationship of two functions. 9-19-2Lecture 9: November 8, 201823456781248163264128256512102420484096n!n
n2 nn2nlog(n)log(n)n
pn1Growth of Functions
From this chart, we see:
1lognpnnnlog(n)n22nn!nn(9.1)
ComplexityTerminology
(1)Constant (logn)Logarithmic (n)Linear (nlogn)Linearithmic nbPolynomial (bn) (whereb >1)Exponential (n!)Factorial9.2.1 Big-O: Upper BoundDenition 9.1 (Big-O: Upper Bound)f(n) =O(g(n))means there exists some constantcsuch
thatf(n)cg(n), for large enoughn(that is, asn! 1).We sayf(n) =O(g(n))
Example:I claim 3n2100n+ 6 =O(n2). I can prove this using the denition of big-O:Lecture 9: November 8, 20189-3
f(n) = 3n2100n+ 6 (9.2) g(n) =n2(9.3) )3n2100n+ 6cn2for somec(9.4)Ifc= 3 : 3n2100n+ 63n2(9.5)
To prove using Big-O:
Determinef(n) andg(n)
Write the equation based on the denition
Choose acsuch that the equation is true.
{If you can nd ad, thenf(n) =O(g(n)). If not, thenf(n)6=O(g(n)).These statements are all true:
3n2100n+ 6 =O(n2) (9.6)
3n2100n+ 6 =O(n3) (9.7)
3n2100n+ 66=O(n) (9.8)
Proving
9.7 f(n) = 3n2100n+ 6 (9.9) g(n) =n3(9.10) )3n2100n+ 6 =cn3(for somec) (9.11)Ifc= 1 : 3n2100n+ 6n3(whenn >3) (9.12)
We also know this to be true because order istransitive: iff(n) =O(g(n)), andg(n) =O(h(n)), then f(n) =O(h(n)). Sincen2=O(n3), then anyf(n) =O(n2) is alsoO(n3).Proving
9.8 f(n) = 3n2100n+ 6 (9.13) g(n) =n(9.14)For anyc:cn <3n2(whenn > c) (9.15)
9.2.2 Big-Omega: Lower BoundDenition 9.2 (Big-Omega: Lower Bound)f(n) =
(g(n))means there exists some constantc such thatf(n)cg(n), for large enoughn(that is, asn! 1).We sayf(n) =
(g(n))or \fofnis Big Omega ofgofn"9-4Lecture 9: November 8, 2018
Example:I claim 3n2100n+ 6 =
(n2). I can prove this using the denition of big-Omega: f(n) = 3n2100n+ 6 (9.16) g(n) =n2(9.17) )3n2100n+ 6cn2for somec(9.18)Ifc= 2 : 3n2100n+ 62n2(9.19)
We show Big-Omega the same way we show Big-O.
These statements are all true:
3n2100n+ 6 =
(n2) (9.20)3n2100n+ 66=
(n3) (9.21)3n2100n+ 6 =
(n) (9.22)Proving
9.21- log change of base rule proof