Form 2 Mathematics – LOGARITHMS msomimaktaba, November 13, 2018August 17, 2024 LOGARITHMSSTANDARD NOTATIONS Standard notation form is written in form of A x 10n whereby 1≤ A< 10 and n is any integersExampleWrite the following in standard form (i) 2380Solution: 2380 = 2.38 x 103(ii) 97Solution: 97 = 9.7 x 101(iii) 100000Solution: 100000 = 1 x 105(iv) 8 Solution: 8 = 8 x 100ExampleWrite the following in standard form(i) 0.00056 = 5.6 x 10-4(ii) 0.001 = 1 x 10-3(iii) 0.34 = 3.4 x 10 -1(iv) 2. 0001 = 2. 0001 x 100 EXERCISE 1:i). Write the following in standard form17000 = 1.7 x 104ii) 0.00998 = 9.98 x 10-3 iii). Write in standard form0.000625 = 6.25 x 10-48/300 correct to four significant figure8/300 = 0.02666Now 2.667 x 10-22.667 x 10-2iv) If a = br and a = 8.4 x 104 , b = 7.0 x 102 Find r.solution:a = 84 000b = 700Nowbr = a(700) (r) = 84000 r=120 r = 1.2 x 102DEFINITION OF LOGARITHMSConsider3 x 3 x 3 x 3 then3 x 3 x 3 x 3 = 34 = 81, the number 3 is the base ,and 4 is the exponent.Now we say;Logarithm of 81 to base 3 is equal to exponent 4log381 = 4 In short bn = alogba= nExample 1. Write the following in logarithmic formi) a5 = 10 loga10 = 5ii)10-3 = 0.001 10-3= 0.001 log100.001 = -3iii) 2-1 = ½ log21⁄2 = -1iv) 3 = 91/2 log39 = 1⁄2Example 2Write the following in exponential form (i) log3729= 6 36 = 729(ii) log31⁄3 = -1 3-1 = 1/3(iii) log100.01 = -2 10-2 = 0.01(iv)1⁄2 = log42 4(1/2) = 2Example 3If log100.01= y. Find ySolution:log100.01 = y10y = 0.0110y =1×10-210y=100×10-210y=10-2y= -2If log10x=-3 find xSolution: log10x = -310-3 = x x=0.001EXERCISE 1 1. Write in standard form i)405.06 ii) 0.912 Solution: i) 405.06 = 4.0506 x 102 ii) 0.912 = 9.12 x 10-12. Write in logarithimic form i)5-1 = 1⁄5 ii) 0.0001 =1 × 10-4 Solution: i)5-1 = 1⁄5 log5(1⁄5) = -1ii) 0.0001 = 10-4 log100.0001 = -43. Write in exponential form i) logax = n ii)-3 = log100.001 iii) log2(1⁄64) = -6Solution: i) logax = n an = xii)-3 =log100.001 10-3 = 0.001iii) log2(1⁄64) = -6 2-6 = 1⁄644. To solve for x i) log6x= 4 64 = x x = 1296ii)x = log36561 3x = 6561 x = 8iii)logx10= 1 x1 = 10 x = 10iv) log42 = x 4x = 2 22x = 21 2x= 1 x = 1⁄2BASE TEN LOGARITHM – Is an logarithm of a number to base 10. Also known as common logarithm example i) log105= log5 ii) log1075 = log75iii) log10p = log p SPECIAL CASES (1). logaa = x ax = a1 x = 1Generally logaa = 1Example i) log66 = 1 ii) log10 = 1(2) loga(an) = x ax = an x = nExample i) log4(45) = 5 ii) log10-3 = -3Example 1If log55 = log2m Find mSolution:log55 =log2mButlog55 = 11 = log2m21 = m1m = 2Example 2Given log525 + log4x = 6, Find xSolution:log525+log4x = 6log5(52)+log4x = 62log55+log4x = 62 +log4x = 6log4x = 4 x= 44x = 256EXERCISE 2.Evaluatei) log24096ii) log0.0001solutioni) log24096let x = log24096 2x = 40962x= 212x = 12∴log24096=12ii) log0.0001Solution:Let x = log0.0001 10x = 1/10000 10x = 1/(104) 10x = 10-4 x = -4∴log0001=12) If logk81 – log232= -1Solution: logk81 – 5log22 = -1 logk81 = -1 + 5 logk81= 4 k4 = 81 k4 = 34 k = 33. Given log6y = log7343. Find ySolution:log6y = 3log77log6y = 363 = y216 = yy = 2164) Solve for mi) log81 = m 8m = 1 since aº=1 then8m=80m=0ii) log5m + log327 = 8 log5m+ log333= 8 log5m +3 = 8 log5m= 5 m= 55m = 3125LAWS OF LOGARITHMSMULTIPLICATION LAWSuppose, logax = p and logay = q then logax= p….(i) logay= q….(ii) Write equation (i) and (ii) into exponential form. ap = x………(iii) aq = y……..(iv) Multiply equation (iii) and (iv)xy = ap x aq xy = a(p + q) …….(v)In equation (v) apply loga both sides loga (xy) = logaa(p + q) logaxy = (p + q) logaa logaxy= p + q But p = logax q = logay Examplei) log6(8 ×12) = log88 + log612 ii) log49 +log43 = log4(9 ×3)Example 1i)Find x , If log3x = log315 + log312Solution:log3x = log315 + log312 log3x = log3( 15 ×12) log3x = log3180 ∴x = 180Example 2Given log520 = log54 + log5x .Find xSolution:log520 = log54 + log5xlog520 = log5(4 × x) log520 = log54x ∴20 = 4xX = 5Example 3If log80.01= log8(m ×2). Find msolutionlog80.01 = log8( 2m)∴ 0.01 = 2mm = 0.01/2m = 0.005QUOTIENT LAWSuppose, logax= p and logay = q then logax = p……..(i) logay = q……..(ii)Write equation (i) and (ii) into exponent form ap = x……(iii) aq = y…..(iv)Divide equation (iii) and (iv) x/y = ap/aq x/y = a(p – q) ….. (v)In equation (v) apply log a both sides loga(x/y) = logaa(p-q) loga(x/y) = (p – q) logaa Butlogaa = 1 loga(x/y)= p – q ,where p= logax and q=logayi) log6( 8/12) = log68 – log612 ii) log49- log43 = log4(9/3)ExampleIf log220 = log2x – log28.Find xSolution:log220 = log2x – log28 log220 = log2(x/8) Now, 20 = x/8 X = 20 x 8 X = 160EXERCISE 3 1. Evaluatei) log63 + log62Solution:= log63 + log62= log6( 2 ×3)= log66 = 1ii) log 40 + log 5 + log40Solution:= log1040 + log105 + log1040 = log10( 40 ×5 ×40) =log108000iii) log1025 – log109 + log10360Solution:log1025 – log109 + log10360 log10( (25 ×360 )/9) =log101000 =log10103 =3log1010 =32. If log5ax = log5a9 + log5a12. Find xSolution:log5ax = log5a9 + log5a12 log5ax= log5a( 9 ×12) log5ax= log5a108 x = 1083. If log2a5 = log2ay + log2a0.001.Find YSolution:i) log2a5 = log2a(y×0.001)5 = 0.001yy= 5/0.001Y = 5000ii)Find y if log6100 = log65 + log680 – log6ySolution:log6100 = log6 (5 × 80)/ y100= 400/yy = 44. If log a = 0.9031, log b = 1.0792 and log c = 0.6990. Find log ac⁄bSolutionlog ac⁄b =log10 a + log10c – log10b =0.9031 +1.0792 -0.6990∴ log ac⁄b = 1.2833LOGARITHM OF POWERIf logax = p then X = ap Multiply by power in both sides xn = anpApply log a both sides logaxn = logaanp logaxn = npBut p = logax∴ logaxn= nlogax Example(1)Evaluatei) log2 (128)6 ii) log7 (343)8Solution i)log2 (128)6 = 6log2 27= (7 x 6) log22= 42 x 1= 42ii) Log7 (343)8 Solution: log7 3438 = 8log7 343 = 8log7 73 = (8 x 3) log7 7 = 24Example (2)If log5 625y = log3 7292 .Find y.Solution: log5 625y = log3 7292 log5 625y= 2log3 729 ylog5 54 = 2log3 36 (y x 4) log5 5 = (2 x 6) log3 3 4y log5 5 = 12 log3 3 4y = 12y=2/4y = 3LOGARITHM OF ROOTSExample (1)EXERCISE 4:1. Evaluatei) log 60 + log 40 – log 0.3 ii) log3 √(1⁄27)Solution: i) Log60 + log40 – log0.3 log10 60 + log10 40 – log10 0.3 log10 (60 x 40/0.3) = log10 (2400/0.3) = log10 8000 =3.9031 3. Given log2 x = 1 – log2 3. Find x Solution:log2 x = 1 – log2 3 log2 x= log2 2-log2 3 log2 x= log2 (2/3) x =2/34. Simplifyi) 2log5 + log36 – log9 ii) (log8-log4)/(log 4-log2)Solution:i) 2log5 + log36 – log9log52 + log36 – log9log1025 + log1036 – log109= log10 (25 x 36)/9= log10 (900/9)= log10100=log10 102=2 log10 10=2ii) (log8-log4)/(log 4-log2)Solution:(log8-log4)/(log 4-log2)= log10 (8/4) ÷log10 (4/2)= log102 ÷ log102 = 1 ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 2 Basic Mathematics Study Notes Msomi Maktaba All Notes FORM 2MATHEMATICSPost navigationPrevious postNext postRelated Posts Form Six Past Papers Commerce Form Six Past Papers – NECTA A Level October 16, 2019October 16, 2019Click The Links Below to download Form 6 Past Papers Commerce Year Questions/Answers 2018 Paper 1, Paper 2 2016 Paper 1, Paper 2 2015 Paper 1, Paper 2 2013 Paper 1, Paper 2 2012 Paper 1, Paper 2 2011 Paper 2 2009 Paper 1, Paper 2 A LEVEL NECTA… Read More Msomi Maktaba All Notes Matokeo Morocco vs Tanzania January 17, 2024 AFCON Leo February 2, 2024Matokeo ya Morocco vs Taifa Stars – African Cup of Nations – January 17, 2024: Live Score, Results, Squad, Form & Head to Head Welcome to our comprehensive blog post where we bring you live updates, squad details, and head-to-head information for the much-anticipated match, MATOKEO Tanzania vs Morocco Leo… Read More Msomi Maktaba All Notes Mauritius MES PSAC Results 2023/2024: Check Here February 3, 2024Mauritius MES PSAC results 2023/2024. 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