ELF.01.6 - Laws of ELF.01.6 - Laws of LogarithmsLogarithms

MCB4U - SantowskiMCB4U - Santowski

(A) Review(A) Review

► if f(x) = aif f(x) = axx, find f , find f -1-1(x) so y = a(x) so y = axx then x then x = a= ayy and now isolate y and now isolate y

► in order to isolate the y term, the in order to isolate the y term, the logarithmic notation or symbol or logarithmic notation or symbol or function was invented or created so function was invented or created so we write x = awe write x = ayy as y = log as y = logaa(x)(x)

(B) Properties of Logarithms- (B) Properties of Logarithms- Logarithms of PowersLogarithms of Powers

► Consider this approach here Consider this approach here in column 1in column 1

► evaluate logevaluate log55(625) = x (625) = x ► 55xx = 625 = 625► 55xx = 5 = 544

► x = 4x = 4

► Now try logNow try log55(625) = x tis (625) = x tis way:way:

► log5(5log5(544) = x) = x► loglog55(5)(5)44 = x = x

► Now I know from column 1 Now I know from column 1 that the answer is 4 … so …that the answer is 4 … so …

► loglog55(5)(5)44 = 4 = 4► 4 x log4 x log55(5) = 4(5) = 4► 4 x 1 = 44 x 1 = 4

(B) Properties of Logarithms- (B) Properties of Logarithms- Logarithms of PowersLogarithms of Powers

►So if logSo if log55(625) = log(625) = log55(5)(5)44 = 4 x log = 4 x log55(5)(5)

► It would suggest a rule of logarithms It would suggest a rule of logarithms ► loglogbb(b(bxx) = x) = x

►Which we can generalize Which we can generalize ► loglogbb(a(axx) = xlog) = xlogbb(a)(a)

(C) Properties of Logarithms – (C) Properties of Logarithms – Logs as ExponentsLogs as Exponents

► Consider the exampleConsider the example

► Recall that the expression logRecall that the expression log33(5) simply means (5) simply means “the exponent on 3 that gives 5” “the exponent on 3 that gives 5” let’s call that let’s call that xx

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

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

► We can demonstrate this algebraically as wellWe can demonstrate this algebraically as well

x5log33

(C) Properties of Logarithms – (C) Properties of Logarithms – Logs as ExponentsLogs as Exponents

► Let’s take our exponential equation and Let’s take our exponential equation and write it in logarithmic formwrite it in logarithmic form

► So becomes logSo becomes log33(x) = log(x) = log33(5)(5)

► Since both sides of our equation have a Since both sides of our equation have a loglog33 then x = 5 as we had tried to reason then x = 5 as we had tried to reason out in the previous slideout in the previous slide

► So we can generalize that So we can generalize that

x5log33

xb xb log

(D) Properties of Logarithms – (D) Properties of Logarithms – Product LawProduct Law

► Express logExpress logaam + logm + logaan as a single logarithmn as a single logarithm

► We will let logWe will let logaam = x and logm = x and logaan = yn = y► So logSo logaam + logm + logaan becomes x + yn becomes x + y

► But if logBut if logaam = x, then am = x, then axx = m and likewise a = m and likewise ayy = n = n

► Now take the product (m)(n) = (aNow take the product (m)(n) = (axx)(a)(ayy) = a) = ax+yx+y

► Rewrite mn=aRewrite mn=ax+yx+y in log form in log form log logaa(mn)=x + y(mn)=x + y

► But x + y = logBut x + y = logaam + logm + logaan n ► So thus logSo thus logaa(mn) = log(mn) = logaam + logm + logaan n

(E) Properties of Logarithms – (E) Properties of Logarithms – Quotient LawQuotient Law

►We could develop a similar “proof” to the We could develop a similar “proof” to the quotient law as we did with the product quotient law as we did with the product law law except we ask you to express except we ask you to express loglogaam - logm - logaan as a single logarithmn as a single logarithm

►As it turns out (or as is predictable), the As it turns out (or as is predictable), the simplification becomes as follows:simplification becomes as follows:

► loglogaa(m/n) = log(m/n) = logaa(m) – log(m) – logaa(n)(n)

(F) Summary of Laws(F) Summary of Laws

Logs as Logs as exponentsexponents

Product Product RuleRule

loglogaa(mn) = log(mn) = logaam + logm + logaan n

Quotient Quotient RuleRule

loglogaa(m/n) = log(m/n) = logaa(m) – log(m) – logaa(n)(n)

Power Power RuleRule

LogLogaa(m(mpp) = (p) x (log) = (p) x (logaam)m)

xb xb log

(G) Examples(G) Examples

► (i) log(i) log3354 + log54 + log33(3/2)(3/2)► (ii) log(ii) log22144 - log144 - log2299► (iii) log30 + log(10/3)(iii) log30 + log(10/3)

► (iv) which has a greater value (iv) which has a greater value (a) log(a) log3372 - log72 - log338 or (b) log500 + log28 or (b) log500 + log2

► (v) express as a single value(v) express as a single value (a) 3log(a) 3log22x + 2logx + 2log22y - 4logy - 4log22a a (b) log(b) log33(x+y) + log(x+y) + log33(x-y) - (log(x-y) - (log33x + logx + log33y)y)

► (vi) log(vi) log22(4/3) - log(4/3) - log22(24)(24)► (vii) (log(vii) (log225 + log5 + log2225.6) - (log25.6) - (log2216 + log16 + log339)9)

(G) Internet Links(G) Internet Links

►Logarithm Rules Lesson from Purple MLogarithm Rules Lesson from Purple Mathath

►College Algebra Tutorial on LogarithmiCollege Algebra Tutorial on Logarithmic Properties from West Texas AMc Properties from West Texas AM

(H) Homework(H) Homework

►Nelson textbook, p125, Q1-14Nelson textbook, p125, Q1-14