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AS-Level Maths: Core 2 for Edexcel C2.6 Exponentials and logarithms This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 26 © Boardworks Ltd 2005 Exponential functions Contents Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions 2 of 26 © Boardworks Ltd 2005 Exponential functions So far in this course we have looked at many functions involving terms in xn. In an exponential function, however, the variable is in the index. For example: x y=2 x y=5 y = 0.1 x –x y=3 x+1 y=7 The general form of an exponential function to the base a is: y = ax where a > 0 and a ≠1. You have probably heard of exponential increase and decrease or exponential growth and decay. A quantity that changes exponentially either increases or decreases more and more rapidly as time goes on. 3 of 26 © Boardworks Ltd 2005 Graphs of exponential functions 4 of 26 © Boardworks Ltd 2005 Exponential functions When a > 1 the graph of y = ax has the following shape: When 0 < a < 1 the graph of y = ax has the following shape: y y 1 1 (1, a) x (1, a) x In both cases the graph passes through (0, 1) and (1, a). This is because: a0 = 1 and a1 = a for all a > 0. 5 of 26 © Boardworks Ltd 2005 Logarithms Contents Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions 6 of 26 © Boardworks Ltd 2005 Logarithms Find p if p3 = 343. We can solve this equation by finding the cube root of 343: p = 3 343 p =7 Now, consider the following equation: Find q if 3q = 343. We need to find the power of 3 that gives 343. One way to tackle this is by trial and improvement. Use the xy key on your calculator to find q to 2 decimal places. 7 of 26 © Boardworks Ltd 2005 Logarithms To avoid using trial and improvement we need to define the power y to which a given base a must be raised to equal a given number x. This is defined as: y = loga x “y is equal to the logarithm, to the base a, of x” The expressions y = loga x and ay = x are interchangeable. This can be written using the implication sign : y = loga x ay = x For example, 25 = 32 can be written in logarithmic form as: log2 32 = 5 8 of 26 © Boardworks Ltd 2005 Logarithms Taking a log and raising to a power are inverse operations. y = loga x ay = x We have that: aloga x = x So: y = loga ay Also: For example: log7 2 7 9 of 26 =2 and log3 36 = 6 © Boardworks Ltd 2005 Contents The laws of logarithms Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions 10 of 26 © Boardworks Ltd 2005 Some important results When studying indices we found the following important results: a1 = a This can be written in logarithmic form as: loga a = 1 a0 = 1 This can be written in logarithmic form as: loga 1 = 0 It is important to remember these results when manipulating logarithms. 11 of 26 © Boardworks Ltd 2005 The laws of logarithms The laws of logarithms follow from the laws of indices: The multiplication law Let: m = loga x and n = loga y So: x = am and y = an xy = am × an Using the multiplication law for indices: xy = am + n Writing this in log form gives: m + n = loga xy But m = loga x and n = loga y so: loga x + loga y = loga (xy) 12 of 26 © Boardworks Ltd 2005 The laws of logarithms The division law Let: m = loga x So: x = am and n = loga y and y = an x am = n y a Using the division law for indices: x = a mn y Writing this in log form gives: x m n loga y But m = loga x and n = loga y so: x loga x loga y loga y 13 of 26 © Boardworks Ltd 2005 The laws of logarithms The power law Let: m = loga x So: x = am xn =(am)n Using the power law for indices: xn =amn Writing this in log form gives: mn = loga xn But m = loga x so: n loga x = loga xn 14 of 26 © Boardworks Ltd 2005 The laws of logarithms These three laws can be used to combine several logarithms written to the same base. For example: Express 2loga 3 + loga 2 – 2loga 6 as a single logarithm. 2loga 3 + loga 2 2loga 6 = loga 32 + loga 2 loga 62 = loga 9 + loga 2 loga 36 9×2 = loga 36 = loga 15 of 26 1 2 © Boardworks Ltd 2005 The laws of logarithms The laws of logarithms can also be used to break down a single logarithm. For example: 2 Express log10 1 2 ab in terms of log10 a, log10 b and log10 c. 4 c 2 1 2 1 ab 2 2 log10 4 = log10 a b log10 c 4 c 1 2 2 = log10 a + log10 b log10 c4 = 2log10 a + 21 log10 b 4log10 c Logarithms to the base 10 are usually written as log or lg. We can therefore write this expression as: 2log a + 21 log b 4log c 16 of 26 © Boardworks Ltd 2005 Logarithms to the base 10 and to the base e Although the base of a logarithm can be any positive number, there are only two bases that are commonly used. These are: Logarithms to the base 10 Logarithms to the base e Logarithms to the base 10 are useful because our number system is based on powers of 10. They can be found by using the log key on a calculator. Logarithms to the base e are called Napierian or natural logarithms and have many applications in maths and science. They can be found by using the 17 of 26 ln key on a calculator. © Boardworks Ltd 2005 Changing the base of a logarithm Suppose we wish to calculate the value of log5 8. We can’t calculate this directly using a calculator because it only find logs to the base 10 or the base e. We can change the base of the logarithm as follows: Let x = log5 8 5x = 8 So: Taking the log to the base 10 of both sides: log 5x = log 8 So: 18 of 26 x log 5 = log 8 log 8 x= log 5 log 8 log5 8 = = 1.29 (to 3 s.f.) log 5 © Boardworks Ltd 2005 Changing the base of a logarithm If we had used log to the base e instead we would have had: ln 8 log5 8 = = 1.29 (to 3 s.f.) ln 5 In general, to find loga b: Let x = loga b, so we can write ax = b Taking the log to the base c of both sides gives: logc ax = logc b xlogc a = logc b logc b x= logc a So: 19 of 26 logc b loga b = logc a © Boardworks Ltd 2005 Contents Solving equations using logarithms Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions 20 of 26 © Boardworks Ltd 2005 Solving equations involving logarithms We can use the laws of logarithms to solve equations. For example: Solve log5 x + 2 = log5 10. To solve this equation we have to write the constant value 2 in logarithmic form: 2 = 2 log5 5 because log5 5 = 1 = log5 52 = log5 25 The equation can now be written as: log5 x + log5 25 = log5 10 log5 25x = log5 10 25x = 10 x = 0.4 21 of 26 © Boardworks Ltd 2005 Solving equations of the form ax = b We can use logarithms to solve equations of the form ax = b. For example: Find x to 3 significant figures if 52x = 30. We can solve this by taking logs of both sides: log 52x = log 30 Using a calculator: 2x log 5 = log 30 log 30 2x = log 5 log 30 x= 2log 5 x = 1.06 (to 3 s.f.) 22 of 26 © Boardworks Ltd 2005 Solving equations of the form ax = b Find x to 3 significant figures if 43x+1 = 7x+2. Taking logs of both sides: log 43 x 1 = log 7 x 2 (3 x +1)log 4 = ( x + 2)log 7 3 x log 4 +log 4 = x log 7 + 2log 7 x(3log 4 log 7) = 2log 7 log 4 2log 7 log 4 x= 3log 4 log 7 x = 1.13 (to 3 s.f.) 23 of 26 © Boardworks Ltd 2005 Solving equations of the form ax = b Solve 32x –5(3x) + 4 = 0 to 3 significant figures. If we let y = 3x we can write the equation as: y2 5 y + 4 = 0 ( y 1)( y 4) = 0 y =1 or y = 4 So: 3x =1 or 3x = 4 If 3x = 1 then x = 0. Now, solving 3x = 4 by taking logs of both sides: log 3 x = log 4 xlog 3 = log 4 log 4 x= log 3 x = 1.26 (to 3 s.f.) 24 of 26 © Boardworks Ltd 2005 Contents Examination-style questions Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions 25 of 26 © Boardworks Ltd 2005 Examination-style question Julia starts a new job on a salary of £15 000 per annum. She is promised that her salary will increase by 4.5% at the end of each year. If she stays in the same job how long will it be before she earns more than double her starting salary? 15 000 × 1.045n = 30 000 1.045n = 2 log 1.045n = log 2 n log 1.045 = log 2 log 2 n= 15.7 log 1.045 Julia’s starting salary will have doubled after 16 years. 26 of 26 © Boardworks Ltd 2005