M5 Indices and Logarithms (PDF)

M5: Indices and Logarithms Positive Integer Powers The most basic idea of a power comes from multiplying a value by itself a given number of times. Examples 1. Find 24 24 = 2 × 2 × 2 × 2 = 16 2 3 2. Find (− 5) 2 3 2 2...

M5: Indices and Logarithms Positive Integer Powers The most basic idea of a power comes from multiplying a value by itself a given number of times. Examples 1. Find 24 24 = 2 × 2 × 2 × 2 = 16 2 3 2. Find (− 5) 2 3 2 2 2 8 (− 5) = (− 5) × (− 5) × (− 5) = − 125 Questions 1. Find the following (try to do it without a calculator) a. 35 b. (−2)4 4 3 c. (3) 1 4 d. (2 2) 2. Find the value of 𝑛 in each of the following (again, try to do it without a calculator) a. 4𝑛 = 16 b. (−7)𝑛 = 2401 2 𝑛 125 c. (1 ) = 3 27 This idea leads naturally to the ‘Laws of Indices’: 𝑎 𝑛 × 𝑎 𝑚 = 𝑎 𝑛+𝑚 𝑎 𝑛 ÷ 𝑎 𝑚 = 𝑎 𝑛−𝑚 (𝑎 𝑛 )𝑚 = 𝑎 𝑛𝑚 By taking it that these laws should hold for all powers, not just positive integers, we can establish what it means to take a value to a zero power, or to negative and/or fractional powers. More General Powers Using the ‘Laws of Indices’ it is fairly straightforward to show that for all values of 𝑎 except zero (we will see why 𝑎 cannot be zero later): 𝑎0 = 1 And for negative powers, (again 𝑎 cannot be zero, as this time as it would result in division by zero): 1 1 1 𝑎 −1 = 𝑎 −2 = 𝑎 −𝑛 = 𝑎 𝑎2 𝑎𝑛 And for fractional powers, we have: 1⁄ 1⁄ 1⁄ 𝑎 2 = √𝑎 𝑎 3 = 3√𝑎 𝑎 𝑛 = 𝑛√𝑎 We can also combine results to get, for example: 1 𝑚⁄ 𝑛 𝑚 𝑎− 1⁄ 𝑛 = 𝑎 𝑛 = √𝑎 𝑚 = ( 𝑛√𝑎 ) 𝑛 √𝑎 Examples 2⁄ 1. Find 64 3 2⁄ 3 2 64 3 = (√64) = 42 = 16 1 2. Simplify 𝑎 2 × √𝑎 1 1⁄ 3⁄ 𝑎2 × = 𝑎2 × 𝑎− 2 =𝑎 2 √𝑎 Questions 1. Find a. 8−1 2⁄ b. 27 3 1⁄ c. 16− 2 3⁄ d. 25− 2 2. Simplify a. 𝑎 3 × (𝑎 2 )2 b. 𝑏 × √𝑏 2 3 c. ( 4√𝑐 ) × (√𝑐) 3 𝑑 2 × √𝑑 d. √𝑑 −1⁄2 𝑒×√𝑒 e. ( 3 6 ) ( √𝑒 ) Finding the Power Using the Laws of Indices and a bit of logic, we can find the value of the power in an equation if the numbers are ‘nice’. Examples 1 1. Find the value of 𝑛 for which 81𝑛 = 9. 1 To get from 81 to 9 , we have 1 taken a square root (power of 2) taken a reciprocal (power of −1) 1 We therefore have 𝑛 = − 2 2. Find the value of 𝑛 for which 8𝑛 = 4. To get from 8 to 4 (using 2 as a ‘stepping stone’), we have 1 taken a cube root (power of 3) taken a square (power of 2) 2 We therefore have 𝑛 = 3 Questions 1. Find the value of 𝑛, 1 a. 5𝑛 = 25 b. 4𝑛 = 8 1 c. 125𝑛 = 5 1 𝑛 d. (2) = 16 e. 9𝑛 = √3 2. Find the value of 𝑛 1 𝑛 2 a. (1 ) = 2 3 1 𝑛 1 b. (6 4) = 2 2 c. 2𝑛 = 16√2 5 d. 25𝑛 = √5 Logarithms When the numbers are not so ‘nice’, we can use logarithms to solve these Power equations. Most modern calculators have a log button on them - this is the most flexible way to find logarithms. For log 2 8 we say ‘log to the base 2 of 8’, and the answer is 3 because 23 = 8. A logarithm is basically asking which power the base needs to be taken to in order to get the other number. Examples 1. Find log 5 25. log 5 25 = 2 because 52 = 25 (you can check on your calculator) 1 2. Find log 8 2 1 1 1⁄ 1 log 8 2 = − 3 because 8− 3 = 2 (you can check on your calculator) Questions 1. Find the following ‘nice’ logarithms by logic, then use your calculator to check. a. log 4 64 1 b. log 5 25 c. log 9 3 d. log 2 2√2 2. Find the following logarithms on your calculator. Give your answers to 3 significant figures. a. log 5 10 b. log 7 55 1 c. log 3 2 d. log 100 20 Other Calculator Buttons There are two other buttons on your calculator that relate to logarithms. ‘log’ is shorthand for log to the base 10 and ‘ln’ means log to the base 𝑒, where 𝑒 = 2.7182818 … is an important number that we see a lot in Diploma Maths and beyond. ‘ln’ is called the ‘Natural Logarithm’. The base 10 ‘log’ button is there really as a continuation from the times before calculators when people used to use books of log tables in order to perform calculations. They were in base 10 due to their ease of use alongside the decimal system. Check out what it does by calculating log 10, log 100, log √10 and log 0.1. At this stage, there is no reason to use the base 𝑒 ‘ln’ button, but it will become indispensible (and therefore much more ‘natural’) during DP and University that you’ll wonder how you ever lived without it! Solving Equations with Logarithms Now that we have seen what logarithms do, we can use them to solve equations where the unknown is the power. Example 1. Solve 3𝑥 = 5. 3𝑥 = 5 𝑥 = log 3 5 = 1.46 (3sf) You can check that this has worked by typing 31.46… on your calculator. Questions 1. Solve, a. 2𝑥 = 10 b. 10𝑥 + 1 = 19 c. 2 × 5𝑥 = 18 d. 3𝑥+5 = 30 e. 710−3𝑥 = 500 f. (3𝑥 − 9)(3𝑥 − 11) = 0 (two answers) 2. Explain why typing log 2 (−10) into your calculator results in a ‘Math Error’. 3. Explain why typing log 3 0 into your calculator results in a ‘Math Error’. We can also see logarithms, and how they are connected to powers, through the ‘Log-Power Relationship’, log a 𝑏 = 𝑐 ⟺ 𝑎 𝑐 = 𝑏 It is, however, probably more easily remembered through the numerical example, log 2 8 = 3 ⟺ 23 = 8 Keeping this relationship in mind will help enormously when doing logarithms questions, especially as they become more complicated. The idea is to recall where the 2, 3 and 8 go in the above expression, and then follow the pattern with your particular case. E.g. to find log 25 5, 1 If 𝑛 = log 25 5, then 25𝑛 = 5, so 𝑛 = 2 1 log 25 5 = 2 Equations containing Logarithms We can use the Log-Power Relationship, or just logic for that matter, to turn Log equations into Power equations, and hence solve them. Examples 1. Solve log x 125 = 3. log x 125 = 3 𝑥 3 = 125 𝑥=5 2. Solve log 2 𝑥 = 5. log 2 𝑥 = 5 25 = 𝑥 𝑥 = 32 Questions 1. Find 𝑥, a. log x 125 = 3 b. log x 64 = 2 c. log x 0.1 = −1 1 d. log x 0.25 = − 2 2 e. log x 4 = 3 f. log x 50 = 3 (give answer to 3sf) g. log x 33 = −2 (give answer to 3sf) 2. Find 𝑥, a. log 3 𝑥 = 4 b. log 7 𝑥 = −2 c. log 2 (𝑥 + 5) = 3 d. log 4 (3𝑥 − 2) = 2 𝑥−2 e. log 3 ( ) = −1 𝑥+2 f. log 3 (𝑥 2 + 6𝑥) = 3 𝑥2 g. log 2 (𝑥+3) =2 Laws of Logarithms The ‘Laws of Indices’ (from Page 1) have their corresponding counterparts in logarithms. log x 𝑎 + log x 𝑏 = log x 𝑎𝑏 𝑎 log x 𝑎 − log x 𝑏 = log x 𝑏 𝑛log x 𝑎 = log x 𝑎 𝑛 These are the direct consequence of the ‘Log-Power Relationship’ and the ‘Laws of Indices’. Notice that the base (𝑥 in all the expressions above) has to be the same throughout. Examples 1. Write 2 log 7 5 + log 7 3 as a single logarithm. 2 log 7 5 + log 7 3 = log 7 52 + log 7 3 = log 7 25 × 3 = log 7 75 Questions 1. Write as a single logarithm, a. log 2 7 + log 2 3 b. log 3 6 − log 3 10 c. log 10 3 + log10 7 − log10 5 d. 3 log 9 2 + 2 log 9 5 e. 2 log 6 5 + log 6 7 − 3 log 6 3 2. Evaluate without using a calculator, a. log 6 4 + log 6 9 b. log 10 5000 − log10 5 c. 3 log 4 6 − log 4 27 d. log 7 4 − log 7 28 e. 2 log 3 6 + 3 log 3 18 − 5 log 3 2 3. Write the following in terms of log x 𝑎 and log x 𝑏, a. log x 𝑎 2 𝑏 𝑏 b. log x 𝑎 c. log x √𝑎𝑏 𝑎2 d. log x 3 √𝑏 e. log x 𝑎3 𝑏4 𝑥2 Change of Base Formula Sometimes it is useful to change to base in a logarithm, for example, to have all of the bases in an expression the same in order to use the Laws of Logarithms. For this we can use the ‘Change of Base Formula’. log x 𝑏 log a 𝑏 = log x 𝑎 Here, you have completely free choice over what 𝑥 is, so it is a very flexible formula. Proof Let 𝑦 = log a 𝑏 Then, by the Log-Power Relationship, 𝑎 𝑦 = 𝑏 Taking log x of both sides, 𝑦 log x 𝑎 = log x 𝑏 log 𝑏 So, 𝑦 = logx 𝑎 and therefore, x log x 𝑏 log a 𝑏 = log x 𝑎 Example 1. Write log 3 𝑎 2 + 8 log 9 𝑎 in terms of log 3 𝑎. log3 𝑎 log 3 𝑎 2 + 8 log 9 𝑎 = 2 log 3 𝑎 + 8 log3 9 log3 𝑎 = 2 log 3 𝑎 + 8 2 = 2 log 3 𝑎 + 4 log 3 𝑎 = 6 log 3 𝑎 Questions 1. Write in terms of log 4 𝑎, a. log 16 𝑎 b. log 64 √𝑎 c. log 4 𝑎 3 − 4 log 16 𝑎 d. log 2 𝑎 e. log 2 8√𝑎 1 2. Show that log a 𝑏 = log𝑏 𝑎 3. Explain why trying to find log1 6 is equivalent to dividing by zero, and hence gives a ‘Math Error’. Further Equations containing Logarithms Now that we have some more advanced tools like the ‘Laws of Logarithms’, we are able to solve more complicated equations containing logarithms. Example 1. Solve log 2 𝑥 + log 2 (𝑥 + 2) = 3 log 2 𝑥 + log 2 (𝑥 + 2) = 3 log 2 𝑥(𝑥 + 2) = 3 𝑥(𝑥 + 2) = 8 𝑥 2 + 2𝑥 − 8 = 0 (𝑥 + 4)(𝑥 − 2) = 0 𝑥 = −4 or 𝑥 = 2 In fact, only 𝑥 = 2 is a solution as log 2 (−4) from the original equation is not a real number. It is therefore worth checking that the solutions you get at the end make sense in the original expression. You can sometimes get extra non-solutions appearing with these sorts of questions due to two minuses potentially cancelling one another out when you apply a Log Law. Questions 1. Solve the following equations, a. log 3 𝑥 + log 3 (𝑥 + 8) = 2 b. log 5 (𝑥 + 48) − log 5 𝑥 = 2 c. 2 log 2 𝑥 − log 2 (𝑥 + 6) = 3 2. Given that log 2 (𝑥 − 8) − log 4 𝑥 = 1 (𝑥−8)2 a. Show that log 4 ( )=1 𝑥 b. Hence find 𝑥. 3. Given that log 3 (2𝑥 + 3) − log 9 (𝑥 − 3) = 2 a. Show that 4𝑥 2 − 69𝑥 + 252 = 0 b. Hence find 𝑥. What is 𝟎𝟎 ? Typing 00 into your calculator results in a ‘Math Error’, but why is this? Surely, something as seemingly simple as this should have been resolved by now. Basically, there is a conflict between two fundamental results to do with powers, 0𝑛 = 0 for all values of 𝑛 > 0 𝑎0 = 1 for all values of 𝑎 ≠ 0 The first of these results would suggest that the most natural choice would be that 00 = 0, and the second that 00 = 1. Of course, it cannot be both! One way to try to resolve the problem is to consider the graph of 𝑦 = 𝑥 𝑥 for 𝑥 > 0 and consider what happens as 𝑥 gets closer to zero. A mathematical way of writing this is that we are considering lim 𝑥 𝑥 x→0+ The graph looks like this 4 y 3 2 1 x −4 −2 2 4 −1 −2 and suggests that 00 should be 1 as this is where the graph heads as we make 𝑥 increasingly close to zero, i.e. where it seems to hit the 𝑦-axis. −3 −4 However, we need to be careful as 𝑦 = 𝑥 𝑥 is not the only graph that we could use to try to get to grips with 00. It turns out though that almost all graphs that we can think of with the right properties, e.g. 𝑦 = 𝑥 2𝑥 , 𝑦 = (2𝑥 − 1)𝑥 , get closer to 1 as we make 𝑥 increasingly close to zero. This, of course, supports the case for 00 = 1. The problem is that we would need all such graphs to get closer to 1 in order to close the case and ‘define’ 00 to be 1. All it takes, however, is for just one appropriate graph not to get closer to 1 in order to prevent us from making such a ‘definition’. A graph that spoils everything is the rather trivial one: 𝑦 = 0𝑥. As we make 𝑥 get closer to zero, we approach what we might mean by 00 , but this time the graph just sits at 0 and therefore heads towards 0 on the 𝑦-axis rather than 1. This is why it is impossible to give a ‘consistent’ or ‘correct’ value for 00.