elf 01 6 laws of logarithms
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ELF.01.6 - Laws of Logarithms. MCB4U - Santowski. (A) Review. if f(x) = a x , find f -1 (x) so y = a x then x = a y and now isolate y in order to isolate the y term, the logarithmic notation or symbol or function was invented or created so we write x = a y as y = log a (x).

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a review
(A) Review
  • if f(x) = ax, find f -1(x) so y = ax then x = ay and now isolate y
  • in order to isolate the y term, the logarithmic notation or symbol or function was invented or created so we write x = ay as y = loga(x)
b properties of logarithms logarithms of powers
Consider this approach here in column 1

evaluate log5(625) = x

5x = 625

5x = 54

x = 4

Now try log5(625) = x tis way:

log5(54) = x

log5(5)4 = x

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

log5(5)4 = 4

4 x log5(5) = 4

4 x 1 = 4

(B) Properties of Logarithms- Logarithms of Powers
b properties of logarithms logarithms of powers1
(B) 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)
c properties of logarithms logs as exponents
(C) 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 x
  • So we are then asking you to place that same exponent (the x) on the same base of 3
  • Therefore taking the exponent that gave us 5 on the base of 3 (x ) onto a 3 again, must give us the same 5!!!!
  • We can demonstrate this algebraically as well
c properties of logarithms logs as exponents1
(C) 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
d properties of logarithms product law
(D) 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
e properties of logarithms quotient law
(E) Properties of Logarithms – Quotient Law
  • We could develop a similar “proof” to the quotient law as we did with the product law  except we ask you to express logam - logan as a single logarithm
  • As it turns out (or as is predictable), the simplification becomes as follows:
  • loga(m/n) = loga(m) – loga(n)
g examples
(G) 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)
g internet links
(G) Internet Links
  • Logarithm Rules Lesson from Purple Math
  • College Algebra Tutorial on Logarithmic Properties from West Texas AM
h homework
(H) Homework
  • Nelson textbook, p125, Q1-14
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