Logarithms * Rory Adams Free High School Science Texts Project Mark Horner Heather Williams. 1 Introduction

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1 OpenSta-CNX module: m Logarithms * Rory Adams Free High School Science Tets Project Mark Horner Heather Williams This work is produced y OpenSta-CNX and licensed under the Creative Commons Attriution License Introduction In mathematics many ideas are related. We saw that addition and sutraction are related and that multiplication and division are related. Similarly, eponentials and logarithms are related. Logarithms are commonly refered to as logs, are the "opposite" of eponentials, just as sutraction is the opposite of addition and division is the opposite of multiplication. Logs "undo" eponentials. Technically speaking, logs are the inverses of eponentials. The logarithm of a numer in the ase a is dened as the numer n such that a n =. So, if a n =, then: log a = n 1 aside: When we say inverse function we mean that the answer ecomes the question and the question ecomes the answer. For eample, in the equation a = the question is what is a raised to the power? The answer is. The inverse function would e log a = or y what power must we raise a to otain? The answer is. The mathematical symol for logarithm is log a and it is read log to the ase a of. For eample, log is log to the ase 10 of Logarithm Symols : Write the following out in words. The rst one is done for you. 1. log 2 4 is log to the ase 2 of 4 2. log log log 8. log y * Version 1.4: Jul 12, :30 am

2 OpenSta-CNX module: m Denition of Logarithms The logarithm of a numer is the value to which the ase must e raised to give that numer i.e. the eponent. From the rst eample of the activity log 2 4 means the power of 2 that will give 4. As 2 2 = 4, we see that The eponential-form is then 2 2 = 4 and the logarithmic-form is log 2 4 = 2. Denition 2: Logarithms If a n =, then: log a = n, where a > 0; a 1 and > 0. log 2 4 = Applying the denition : Find the value of: 1. log log log log Reasoning : 7 3 = 343 therefore, log = Logarithm Bases Logarithms, like eponentials, also have a ase and log 2 2 is not the same as log We generally use the common ase, 10, or the natural ase, e. The numer e is an irrational numer etween 2.71 and It comes up surprisingly often in Mathematics, ut for now suce it to say that it is one of the two common ases. 3.1 Natural Logarithm The natural logarithm symol ln is widely used in the sciences. The natural logarithm is to the ase e which is approimately e is like π and is another eample of an irrational numer. While the notation log 10 and log e may e used, log 10 is often written log in Science and log e is normally written as ln in oth Science and Mathematics. So, if you see the log symol without a ase, it means log 10. It is often necessary or convenient to convert a log from one ase to another. An engineer might need an approimate solution to a log in a ase for which he does not have a tale or calculator function, or it may e algeraically convenient to have two logs in the same ase. Logarithms can e changed from one ase to another, y using the change of ase formula: log a = log 4 log a where is any ase you nd convenient. Normally a and are known, therefore log a is normally a known, if irrational, numer. For eample, change log 2 12 in ase 10 is: log 2 12 = log 1012 log 10 2

3 OpenSta-CNX module: m Change of Base : Change the following to the indicated ase: 1. log 2 4 to ase 8 2. log to ase 2 3. log 16 4 to ase log 8 to ase y. log y to ase Khan academy video on logarithms - 1 This media oject is a Flash oject. Please view or download it at < Figure 4 Laws of Logarithms Just as for the eponents, logarithms have some laws which make working with them easier. These laws are ased on the eponential laws and are summarised rst and then eplained in detail. log a 1 = 0 log a a = 1 log a y = log a + log a y log a y = log a log a y log a = log a log a = log a 6 Logarithm Law 1: log a 1 = 0 For eample, Since a 0 = 1 Then, log a 1 = log a a 0 = 0 y definition of logarithm 7 and log 2 1 = 0 8 log 2 1 = 0 9

4 OpenSta-CNX module: m Logarithm Law 1: log a 1 = 0 : Simplify the following: 1. log log log log 1 + 2y. log y 1 6 Logarithm Law 2: log a a = 1 For eample, Since a 1 = a Then, log a a = log a a 1 = 1 y definition of logarithm 10 and log 2 2 = 1 11 log 2 2 = Logarithm Law 2: log a a = 1 : Simplify the following: 1. log log log log + 2y. log y y tip: Useful to know and rememer When the ase is 10, we do not need to state it. From the work done up to now, it is also useful to summarise the following facts: 1. log1 = 0 2. log10 = 1 3. log100 = 2 4. log1000 = 3

5 OpenSta-CNX module: m Logarithm Law 3: log a y = log a + log a y The derivation of this law is a it trickier than the rst two. Firstly, we need to relate and y to the ase a. So, assume that = a m and y = a n. Then from Equation 1, we have that: This means that we can write: log a = m and log a y = n 13 log a y = log a a m a n = log a a m+n Eponential laws = log a a log a +log a y = log a + log a y For eample, show that log = log10 + log100. Start with calculating the left hand side: 14 The right hand side: log = log 1000 = log 10 3 = 3 1 log10 + log100 = = 3 Both sides are equal. Therefore, log = log10 + log Logarithm Law 3: log a y = log a + log a y : Write as seperate logs: 1. log log log 16 y 4. log z 2y. log y 2 8 Logarithm Law 4: log a = log y a log a y The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an eercise. For eample, show that log 100 = log10 log100. Start with calculating the left hand side: log = log 1 10 = log 10 1 = 1 17

6 OpenSta-CNX module: m The right hand side: log10 log100 = 1 2 = 1 18 Both sides are equal. Therefore, log 100 = log10 log Logarithm Law 4: log a y = log a log a y : Write as seperate logs: 1. log log log 16 y 2 4. log z y. log y 2 9 Logarithm Law : log a = log a Once again, we need to relate to the ase a. So, we let = a m. Then, log a = log a a m = log a a m eponential laws But, m = log a Assumption that = a m log a = log a a log a = log a Definition of logarithm 19 For eample, we can show that log 2 3 = 3log 2. log 2 3 = log 2 = log 2 + log 2 + log 2 log a y = log a a m a n = 3log 2 20 Therefore, log 2 3 = 3log Logarithm Law : log a = log a : Simplify the following: 1. log log log 16 y 4. log z y. log y 2

7 OpenSta-CNX module: m Logarithm Law 6: log a = log a The derivation of this law is identical to the derivation of Logarithm Law and is left as an eercise. For eample, we can show that log 2 3 = log 2 Therefore, log 2 3 = log 2 3. log = log = 1 3 log 2 loga = log a = log Logarithm Law 6: log a = log a : Simplify the following: 1. log log log 16 y 4. log z y. log 2 y tip: The nal answer doesn't have to look simple. Khan academy video on logarithms - 2 This media oject is a Flash oject. Please view or download it at < Figure 21 Khan academy video on logarithms - 3 This media oject is a Flash oject. Please view or download it at < Figure 21 Eercise 1: Simplication of Logs Solution on p. 11. Simplify, without use of a calculator: 3log3 + log12 22

8 OpenSta-CNX module: m Eercise 2: Simplication of Logs Solution on p. 11. Simplify, without use of a calculator: log Eercise 3: Simplify to one log Solution on p. 11. Write 2log3 + log2 log as the logarithm of a single numer. tip: Eponent rule: a = a 11 Solving simple log equations In grade 10 you solved some eponential equations y trial and error, ecause you did not know the great power of logarithms yet. Now it is much easier to solve these equations y using logarithms. For eample to solve in 2 = 0 correct to two decimal places you simply apply the following reasoning. If the LHS = RHS then the logarithm of the LHS must e equal to the logarithm of the RHS. By applying Law, you will e ale to use your calculator to solve for. Eercise 4: Solving Log equations Solution on p. 11. Solve for : 2 = 0 correct to two decimal places. In general, the eponential equation should e simplied as much as possile. Then the aim is to make the unknown quantity i.e. the suject of the equation. For eample, the equation 2 +2 = 1 24 is solved y moving all terms with the unknown to one side of the equation and taking all constants to the other side of the equation Then, take the logarithm of each side = 1 log 2 = log log 2 = log = log 2 = 2log 2 Divide oth sides y log 2 = 2 Sustituting into the original equation, yields = 2 0 = 1 27

9 OpenSta-CNX module: m Similarly, = 3 4 is solved as follows: = = = 3 4 take the logarithm of oth sides log = log log 3 = 4log 3 divide oth sides y log = 4 4 = 2 28 Sustituting into the original equation, yields = = = 3 22 = Eercise : Eponential Equation Solution on p. 11. Solve for in = Eercises Solve for : 1. log 3 = log27 = = Logarithmic applications in the Real World Logarithms are part of a numer of formulae used in the Physical Sciences. There are formulae that deal with earthquakes, with sound, and ph-levels to mention a few. To work out time periods is growth or decay, logs are used to solve the particular equation. Eercise 6: Using the growth formula Solution on p. 12. A city grows % every 2 years. How long will it take for the city to triple its size? Eercise 7: Logs in Compound Interest Solution on p. 12. I have R to invest. I need the money to grow to at least R If it is invested at a compound interest rate of 13% per annum, for how long in full years does my investment need to grow? 12.1 Eercises 1. The population of a certain acteria is epected to grow eponentially at a rate of 1 % every hour. If the initial population is 000, how long will it take for the population to reach ? 2. Plus Bank is oering a savings account with an interest rate if 10 % per annum compounded monthly. You can aord to save R 300 per month. How long will it take you to save R ? Give your answer in years and months

10 OpenSta-CNX module: m End of Chapter Eercises 1. Show that log a = log y a log a y Show that log a = log a Without using a calculator show that: 4. Given that n = and n = log 2 y a. Write y in terms of n. Epress log 8 4y in terms of n c. Epress 0 n+1 in terms of and y. Simplify, without the use of a calculator: a log log 3 9 log 2 c. + log 3 9 2, Simplify to a single numer, without use of a calculator: a. log 12 + log32 log8 log8. log3 log0, 3 7. Given: log 3 6 = a and log 6 = a. Epress log 3 2 in terms of a.. Hence, or otherwise, nd log 3 10 in terms of a and. log log 32 + log = log Given: pq k = qp 1 Prove: k = 1 2log q p 9. Evaluate without using a calculator: log log log If log = 0, 7, determine, without using a calculator: a. log ,4 11. Given: M = log log 2 3 a. Determine the values of for which M is dened.. Solve for if M = Solve: 3 log = 10 2 Answers may e left in surd form, if necessary. 13. Find the value of log without the use of a calculator. 14. Simplify By using a calculator: log log Write log400 in terms of a and if 2 = 10 a and 9 = Calculate: Solve the following equation for without the use of a calculator and using the fact that 10 3, 16 : 2log + 1 = 6 log Solve the following equation for : 6 6 = 66 Give answer correct to 2 decimal places.

11 OpenSta-CNX module: m Solutions to Eercises in this Module Solution to Eercise p. 7 Step can e written as 3. Step 2. 3log3 + log12 = 3log3 + log 3 = 3log3 + 3log log a = log a 34 Step 3. We cannot simplify any further. The nal answer is: Solution to Eercise p. 7 Step 1. 8 can e written as can e written as 2. Step 2. Step 3. We can use: 3log3 + 3log log2 32 = log log a = log a 37 Step 4. Step. We can now use log a a = 1 Step 6. Step 7. The nal answer is: log 2 2 = log log 2 2 = = 4 + = log2 32 = 9 40 Solution to Eercise p. 8 Step 1. 2log3 + log2 log = log3 2 + log2 log Step 2. = log Step 3. = log3, 6 Solution to Eercise p. 8 Step 1. log2 = log0 Step 2. log2 = log0 Step 3. = log0 log2 = 1, Step 4. = 1, 22 Solution to Eercise p. 9 Step 1. There are two possile ases: and 7. is an eponent of. Step 2. In order to eliminate 7, divide oth sides of the equation y 7 to give: 3+3 = 41 Step 3. log 3+3 = log 42

12 OpenSta-CNX module: m Step log = log divide oth sides of the equation y log = 1 3 = 2 = Step = = 7 1 = 3 44 Solution to Eercise p. 9 Step 1. A = P 1 + i n Assume P =, then A = 3. therefore the n is halved for this purpose. Step 2. 3 = 1, 0 n 2 For this eample n represents a period of 2 years, log3 = n 2 log1, 0 using law n = 2log3 log1, 0 n = 4, 034 Step 3. So it will take approimately 4 years for the population to triple in size. Solution to Eercise p. 9 Step 1. Step 2. 4 A = P 1 + i n < , 13 n 1, 13 n > 2 nlog 1, 13 > log 2, n > log 2, log 1, 13 n > 7, Step 3. In this case we round up, ecause 7 years will not yet deliver the required R The investment need to stay in the ank for at least 8 years. 47