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 PostsSFUCHAS selected applicants 2021/22 | St. Francis University College of Health and Allied Sciences selected applicants 2021/2022 February 4, 2024SFUCHAS Selected applicants 2023/2024 pdf: Have you applied for Diploma, certificate or bachelor degree admission at St. Francis University College of Health and Allied Sciences for 2023/2024 academic year and you have been anxiously waiting to see if you have been selected to join SFUCHAS and you don’t know how… Read More Matokeo ya darasa la nne 2023 – 2024 – Check online Standard Four National Assessment (SFNA) results February 1, 2024The Standard Four National Assessment (SFNA) Exam is an exam that takes place every year in October. 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