Rules for logarithms We review the properties of logarithms from the - PowerPoint PPT Presentation

Rules for logarithms We review the properties of logarithms from the previous lecture. In that lecture, we developed the following identities. The “Product Property” is an identity involving the logarithm of product: Elementary Functions log b ( MN ) = log b M + log b N (1) Part 3, Exponential Functions & Logarithms Lecture 3.4a, Working With Logarithms The “Quotient Property” is an identity involving the log of quotient: Dr. Ken W. Smith log b ( M N ) = log b M − log b N (2) Sam Houston State University The “Exponent Property” allows us to rewrite the log of an expression 2013 with an exponent: log b ( M c ) = c · log b M (3) Smith (SHSU) Elementary Functions 2013 1 / 23 Smith (SHSU) Elementary Functions 2013 2 / 23 Exponential Functions Rules for logarithms There is the change of base equation: if c is a positive real number then log b x = log c x (4) log c b The first three equations here are properties of exponents translated into “logarithm language.” The fourth equation allows us to choose the base of our logarithm. There are two identities that express the inverse relationship between exponential and logarithmic functions: The last two equations in the list identify the logarithm as the inverse log b ( b x ) = x. function of the exponential function. (5) b log b x = x. (6) Smith (SHSU) Elementary Functions 2013 3 / 23 Smith (SHSU) Elementary Functions 2013 4 / 23

Practicing Logarithms Practicing Logarithms Let’s practice these properties of logarithms on some exercises. 2 By the quotient property Some worked problems. Expand the following expressions. e 3 1 ln ln (5( x + 1)) 3 x 2 +4 = ln((5( x + 1)) 3 ) − ln((5 x − 7) 2 ) . 2 ln (5( x +1)) 3 (5 x − 7) 2 (5 x − 7) 2 . 3 log 100 x 3 y 2 . z 4 4 log( x 3 y 2 z 5 ) 12 . By the exponent property Solutions. ln((5( x + 1)) 3 ) − ln((5 x − 7) 2 ) = 3 ln(5( x + 1)) − 2 ln(5 x − 7) . 1 By the quotient property e 3 x 2 + 4) = ln e 3 − ln( x 2 + 4) . ln( By the product property this is equal to 3(ln 5+ln( x +1)) − 2 ln(5 x − 7) = 3 ln 5 + 3 ln( x + 1) − 2 ln(5 x − 7) . By the first inverse property, since ln() stands for the logarithm base e , then ln e 3 = 3 so the answer is 3 − ln( x 2 + 4) . Smith (SHSU) Elementary Functions 2013 5 / 23 Smith (SHSU) Elementary Functions 2013 6 / 23 Practicing Logarithms Practicing Logarithms 3 By the quotient property, 4 By the exponent property, log 100 x 3 y 2 = log 100 x 3 y 2 − log z 4 . log( x 3 y 2 z 5 ) 12 = 12 log( x 3 y 2 z 5 ) z 4 By the product property By the product property, log 100 x 3 y 2 = log 100 + log x 3 + log y 2 . 12 log( x 3 y 2 z 5 ) = 12(log x 3 + log y 2 + log z 5 ) By the exponent property, we can rewrite all the exponents so that log 100+ log x 3 + log y 2 − log z 4 = log 100+ 3 log x + 2 log y − 4 log z. and then by the exponent property we have 12(3 log x + 2 log y + 5 log z ) . Since log() stands for the logarithm base ten then log 100 = 2 and so our final answer is (Also acceptable is 36 log x + 24 log y + 60 log z .) 2 + 3 log x + 2 log y − 4 log z . Smith (SHSU) Elementary Functions 2013 7 / 23 Smith (SHSU) Elementary Functions 2013 8 / 23

Simplification of logarithms Simplification of logarithms We can use our six logarithm identities to simplify expressions involving logs. Here are some worked examples. Use properties of logarithms to simplify the following expressions. √ 2 ) 3 log e ( e More worked problems. 4 10 log 10 (5) Use properties of logarithms to simplify the following expressions. 5 e − ln 3 , 1 log b ( b 3 b 5 ) 2 log b (( b 3 ) 5 ) Solutions. Solutions. √ √ 2 ) = 3 By the first inverse property log e ( e 2 . b 5 = b − 2 and then recognize that log b ( b − 2 ) = − 2 . 1 We could simplify b 3 4 By the second inverse property, 10 log 10 (5) = 5 . Or we could use the quotient property of logs and the first inverse 5 By the exponent property e − ln 3 = e ln(3 − 1 ) = e ln( 1 3 ) . By the first b 5 ) = log b b 3 − log b b 5 = 3 − 5 = − 2 . property to compute log b ( b 3 inverse property, e ln 1 3 = 1 3 . 2 By the exponent property log b (( b 3 ) 5 ) = 5 log b b 3 . By the first inverse property 5 log b b 3 = 5(3) = 15 . Smith (SHSU) Elementary Functions 2013 9 / 23 Smith (SHSU) Elementary Functions 2013 10 / 23 Simplification of logarithms Logarithms Sometimes a problem has an answer in a base which is intrinsic to the problem but it is not a base with which we can easily do computations. In that case we need to be prepared to change the base to one with which we can compute. Worked problems on changing the base of the logarithm. In the next presentation we continue to practice our logarithm properties. Use the “change of base” identity to write the following as fractions involving ln() . Use a calculator (or computer software program) to (END) approximate the answer. 1 log 2 (5) . 2 log 2 (125) . 3 log 16 (17) . 4 log(5) . 5 log 2 (1024) . Solutions. 1 log 2 (5) = ln 5 ln 2 ≈ 2 . 3219 . 2 log 2 (125) = ln 125 ln 2 ≈ 6 . 9658 . Smith (SHSU) Elementary Functions 2013 11 / 23 Smith (SHSU) Elementary Functions 2013 12 / 23 3 log 16 (17) = ln 17 ln 16 ≈ 1 . 0219 .

Simplification of logarithms Elementary Functions A general exponential function has form y = ae bx where a and b are Part 3, Exponential Functions & Logarithms constants and the base of the exponential has been chosen to be e . Lecture 3.4b, Working With Logarithms, continued Occasionally we have an exponential function with a different base and need to change the function into this general form. Here are some Dr. Ken W. Smith examples. Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 13 / 23 Smith (SHSU) Elementary Functions 2013 14 / 23 Simplification of logarithms Understanding logarithms Suppose you do not have a calculator. You are asked to compute the Worked problems on general exponential form. logarithms, base 10 , of the first ten positive integers, 1,2,3, . . . , 10. That Write the following functions in the form y = ae bx . is, you are asked to fill out as much of the following table as possible. 1 y = 2 x log(1) 2 y = 2 2 x − 2 log(2) 3 y = 3 x log(3) 4 y = 10 x +1 log(4) log(5) Solutions. 1 Since 2 = e ln 2 then y = 2 x = ( e ln 2 ) x = e (ln 2) x . (Here a = 1 and log(6) log(7) b = ln 2 . ) 2 By basic properties of exponents, 2 2 x − 2 = 2 2 x log(8) 2 2 = ( 1 4 )2 2 x . Since 2 = e ln 2 then 2 2 x = ( e ln 2 ) 2 x = e (ln 2)(2 x ) = e (2 ln 2) x . So our answer log(9) log(10) 4 e 2(ln 2) x . (Here a = 1 1 is 4 and b = 2 ln 2 . ) Here are a series of questions designed to explore what we know about 3 Since 3 = e ln 3 then y = 3 x = ( e ln 3 ) x = e (ln 3) x . (Here a = 1 and logarithms. b = ln 3 . ) 1 Which of these ten logarithms can you compute immediately , without 4 By our basic properties of exponents, 10 x +1 = 10 x · 10 . Since any further information? Smith (SHSU) Elementary Functions 2013 15 / 23 Smith (SHSU) Elementary Functions 2013 16 / 23 10 = e ln 10 then y = 10 x +1 = 10( e ln 10 ) x = 10 e (ln 10) x . (Here a = 10

Understanding logarithms Understanding logarithms Solution. 1 Using properties of logs, without any further information, we know that log(1) = 0 and log(10) = 1 . So we can fill in the first and last line: 2 Suppose I tell you that log(2) = 0 . 30103 . Which of these ten log(1) 0 logarithms can you compute now? log(2) 3 Suppose I tell you that log(2) = 0 . 30103 and log(3) = 0 . 47712 . Now log(3) which of these ten logarithms can you compute? log(4) 4 Given the information for log(2) and log(3) , fill out as much of the log(5) table as possible. log(6) log(7) log(8) log(9) log(10) 1 Smith (SHSU) Elementary Functions 2013 17 / 23 Smith (SHSU) Elementary Functions 2013 18 / 23 Understanding logarithms Understanding logarithms So now we know: log(1) 0 log(2) 0 . 30103 2 Using log(2) = . 30103 , we have log(3) log(4) = log(2 2 ) = 2 log 2 = 2(0 . 30103) = 0 . 60206 and log(4) 0 . 60206 log(8) = log(2 3 ) = 3 log 2 = 3(0 . 30103) = . 90309 . log(5) 0 . 69897 log(6) We also have log(7) log(5) = log( 10 2 ) = log(10) − log(2) = 1 − . 30103 = . 69897 . log(8) 0 . 90309 log(9) log(10) 1 Smith (SHSU) Elementary Functions 2013 19 / 23 Smith (SHSU) Elementary Functions 2013 20 / 23

Understanding logarithms Understanding logarithms So now we know: log(1) 0 log(2) 0 . 30103 log(3) 0 . 47712 log(4) 0 . 60206 3 Using log(2) = . 30103 and log(3) = . 47712 we have log(5) 0 . 69897 log(6) = log(3 · 2) = log 2 + log 3 = . 30103 + . 47712 = . 77815 . log(6) 0 . 77815 log(7) and log(9) = log(3 2 ) = 2 log(3) = 2( . 47712) = . 95424 . log(8) 0 . 90309 log(9) 0 . 95424 log(10) 1 (We cannot obtain the log of 7 this way since we cannot write 7 as a product or quotient of powers of 2 and 3.) Challenge exercise: with this information on log 2 and log 3 how would you compute log 7 . 2 ? Smith (SHSU) Elementary Functions 2013 21 / 23 Smith (SHSU) Elementary Functions 2013 22 / 23 Logarithms In the next presentation we use our log properties to solve a variety of equations. (END) Smith (SHSU) Elementary Functions 2013 23 / 23