Laws of logarithms ppt

laws of logarithmic function ppt and all formulas of logarithm
Prof.EvanBaros Profile Pic
Prof.EvanBaros,United Kingdom,Teacher
Published Date:26-07-2017
Your Website URL(Optional)
Comment
Laws of logarithms MATH 1510 Lili Shen Laws of Logarithms Proposition Let a 0 with a =6 1. Let A;B;C2R with A 0 and B 0. (1) log (AB) = log A+ log B. a a a A (2) log = log A log B. a a a B C (3) log (A ) = C log A. a aLaws of logarithms MATH 1510 Lili Shen Laws of Logarithms Example Evaluate each expression. (1) log 2+ log 32. 4 4 (2) log 80 log 5. 2 2 1 (3) log 8. 3Laws of logarithms MATH 1510 Lili Shen Laws of Logarithms Solution. (1) log 2+ log 32 = log (2 32) = log 64 = 3. 4 4 4 4 80 (2) log 80 log 5 = log = log 16 = 4. 2 2 2 2 5 1 1 1 3 (3) log 8 = log 8 = log 0:301. 3 2Expanding logarithmic expressions MATH 1510 Lili Shen Laws of Logarithms Example Expand each expression: (1) log (6x). 2 3 6 (2) log (x y ). 5 ab p (3) ln . 3 cExpanding logarithmic expressions MATH 1510 Lili Shen Laws of Logarithms Solution. (1) log (6x) = log 6+ log x. 2 2 2 3 6 3 6 (2) log (x y ) = log x + log y = 3 log x + 6 log y. 5 6 5 5 5 p ab 1 3 p (3) ln = ln(ab) ln c = lna+ lnb lnc. 3 3 cCombining logarithmic expressions MATH 1510 Lili Shen Laws of Logarithms Example Combine each expression into a single logarithm: 1 (1) 3 logx + log(x + 1). 2 1 2 (2) 3 lns+ lnt 4 ln(t + 1). 2Combining logarithmic expressions MATH 1510 Lili Shen Solution. (1) Laws of Logarithms 1 1 3 2 3 logx + log(x + 1) = logx + log(x + 1) 2 p 3 = log(x x + 1): (2) 1 1 2 3 2 4 2 3 lns+ lnt 4 ln(t + 1) = lns + lnt ln(t + 1) 2 p 3 s t = ln : 2 4 (t + 1)Combining logarithmic expressions MATH 1510 Lili Shen Laws of Logarithms Example Combine and simplify 1 1 5 6 2 3 log(x + 2) + log(x ) log(x x 6) : 5 3Combining logarithmic expressions MATH 1510 Lili Shen Solution. Laws of Logarithms 1 1 5 6 2 3 log(x + 2) + log(x ) log(x x 6) 5 3 2 2 = log(x + 2)+ log(x ) log(x x 6) 2 x (x + 2) = log 2 x x 6 2 x (x + 2) = log (x 3)(x + 2) 2 x = log : x 3The law of forgetting MATH 1510 Lili Shen Example Laws of Logarithms If a task is learned at a performance level P , then after a 0 time interval t the performance level P satisīŦes logP = logP c log(t + 1); 0 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 Solution. Lili Shen P 0 Laws of (1) logP = logP c log(t + 1) = log implies 0 Logarithms c (t + 1) P 0 P = : c (t + 1) (2) Here P = 90, c = 0:2, and t is measured in months. 0 90 In 2 months: P =  72. 0:2 3 90 In 1 year: P =  54. 0:2 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 log x a log x = . b log b aChange of base formula MATH 1510 Lili Shen Laws of Logarithms Corollary 1 (1) log b = . a log a b log b a (2) log b = . c a c log c log a b b (3) a = c .Change of base formula MATH 1510 Lili Shen The change of base formula is quite useful when evaluating Laws of Logarithms logarithms using calculators: Example Evaluate each logarithm: (1) log 5. 8 (2) log 20. 9 (3) log 3 log 64. 4 9 (4) log 3 log 4 log 5 log 2. 2 3 4 5 (5) log 27 log 32. 16 81Change of base formula MATH 1510 Lili Shen Solution. Laws of Logarithms log 5 (1) log 5 =  0:77398. 8 log 8 ln 20 (2) log 20 =  1:36342. 9 ln 9 log 3 log 8 1 3 (3) log 3 log 8 = = log 3 log 8 =  3 = . 2 9 9 2 log 2 log 9 2 2 (4) log 3 log 4 log 5 log 2 = log 3 log 4 log 5 log 2 = 1. 2 3 4 5 3 4 5 2 3 5 15 (5) log 27 log 32 = log 3 log 2 = . 16 81 2 3 4 4 16

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.