1 DBM20023 ENGINEERING MATHEMATICS 2 SESI DIS2020 CHAPTER 1 INDICES & LOGARITHMS ISBN 978-967-16639-3-6 1.2 LAWS OF LOGARITHMS Logarithmic functions are the inverses of exponential functions, and any exponential function can be expressed in logarithmic form. Similarly, all logarithmic functions can be rewritten in exponential form. Laws of Logarithms log = 1 log 1 = 0 log = 1 log log = log + log log = log − log log = log log = log log On our calculators, "log" (without any base) is taken to mean "log base 10". So, for example "log 7" means "log107".
2 DBM20023 ENGINEERING MATHEMATICS 2 SESI DIS2020 CHAPTER 1 INDICES & LOGARITHMS ISBN 978-967-16639-3-6 1. Express 2 5 = 32 in logarithmic form. SOLUTION 2 5 = 32 log = … … … .. = = 2; = 5; = 32 log2 32 = 5 2. Convert log√2 8 = 6 to index form. SOLUTION log√2 8 = 6 log = … … … .. = = √2; = 6; = 8 2 8 6 = 3. Express the following expressions by using the laws of logarithms: a) log 2 3 b) log √( 3 ) SOLUTION a) log 2 3 = log 2 + log 3 = 2 log + 3log INPUT LEARNING 3
3 DBM20023 ENGINEERING MATHEMATICS 2 SESI DIS2020 CHAPTER 1 INDICES & LOGARITHMS ISBN 978-967-16639-3-6 b) log √( 3 ) = log( 3 ) 1 2 = 1 2 log + 3 2 log − 1 2 log 4. Express each of the following as a single logarithm. a) log 5 + 2 log 3 − log 7 b) 1 2 log + log 2 − 2 log SOLUTION a) log 5 + 2 log 3 − log 7 = 5 + 3 2 − log 7 = log( 5×9 7 ) = log( 45 7 ) b) 1 2 log + log 2 − 2 log = log 1/2 + log 2 − log() 2 2 2 2 2 1 log a b a a b = b a 2 1 = log 5. Find the value for log2 8 SOLUTION This expression can be solved by several methods. METHOD 1: By using comparison logarithm via exponent { = → = } log2 8 = =2, =8 = Subs into = 8 = 2 2 3 = 2 ∴ = 3
4 DBM20023 ENGINEERING MATHEMATICS 2 SESI DIS2020 CHAPTER 1 INDICES & LOGARITHMS ISBN 978-967-16639-3-6 METHOD 2: By using Laws of Logarithms 3 log2 8 = log2 2 x m x a m a = 3log2 2 log = log = 3 METHOD 3: By using scientific calculator or logarithmic based convertor a b b c c a log log log log 2 log 8 log2 8 = = 0.301 0.903 = = 3 6. Given that log3 2 = 0.631 and log3 5 = 1.465. Find the value of: a) log3 10 b) log3 2.5 c) log3 25 d) log2 3 SOLUTION a) log3 10 = log3(2 × 5) = log3 2 + log3 5 = 0.631+1.465 = 2.096 b) log3 2.5 = log3( 5 2 ) = log3 5 − log3 2 = 1.465 − 0.631 = 0.834 c) log3 25 = log3 5 2 = 2 log3 5 = 2(1.465) = 2.93 d) log2 3 log 2 log 3 3 3 = 0.631 1 = = 1.585
5 DBM20023 ENGINEERING MATHEMATICS 2 SESI DIS2020 CHAPTER 1 INDICES & LOGARITHMS ISBN 978-967-16639-3-6 7. Solve the following equations: a) 3 = 5 +2 b) log 8 + log 4 = 5 SOLUTION a) 3 = 5 +2 3 = 5 5 2 3 5 = 25 ( 3 5 ) = 25 **using log base 10 on both sides equations log10( 3 5 ) = log10 25 log10( 3 5 ) = log10 25 = log10 25 log10 3 5 = −6.301 b) log 8 + log 4 = 5 log (8 × 4) = 5 log 32 = 5 Convert to index form 5 = 32 5 = 2 5 = 2
6 DBM20023 ENGINEERING MATHEMATICS 2 SESI DIS2020 CHAPTER 1 INDICES & LOGARITHMS ISBN 978-967-16639-3-6 OUTPUT LEARNING 3 1. Given that log5 3 = 0.6826 and log5 6 = 1.1133 . Find the value of: a) log5 18 b) log5 2 c) log3 5 d) log5 1 2 e) log3 30 f) log√5 36 2. Solve each of the following equations a) 2 log3 = 8 b) 5 log2(−4) = 1 5 c) 3 log2 = 9 d) 2 log3(−2) = 1 2 3. Solve each of the following equations a) 2 = 5 b) 3 2 = 5 +2 c) 2 +1 = 5 −2 4. Solve each of the following equations a) log 3 = 3 2 − log 9 b) log3 3 = 2 + log3(1 − ) c) log2 = 4 + log2(3 − ) d) log2 ( + 2) + log4( + 2) = 3 2