{"id":1922,"date":"2023-10-02T07:22:20","date_gmt":"2023-10-02T07:22:20","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=1922"},"modified":"2023-10-02T07:28:26","modified_gmt":"2023-10-02T07:28:26","slug":"week-3-ss1-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-ss1-first-term-further-mathematics-notes\/","title":{"rendered":"Week 3 &#8211; SS1 First Term Further Mathematics Notes"},"content":{"rendered":"<p>\u00a0<br \/>\n\u00a0<strong>WEEK THREE<br \/>\n<\/strong><strong>TOPIC: LOGARITHM &#8211; SOLVING PROBLEMS BASED ON LAWS OF LOGARITHM<\/strong><br \/>\n\t\t<strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Logarithm of numbers (Index &amp; Logarithmic Form)\n<\/li>\n<li>Laws of Logarithm\n<\/li>\n<li>Logarithmic Equation\n<\/li>\n<li>Change of Base\n<\/li>\n<li>Standard forms\n<\/li>\n<li>Logarithm of numbers greater than one\n<\/li>\n<li>Multiplication and divisions of numbers greater than one using logarithm\n<\/li>\n<li>Using logarithm to solve problems with roots and powers (no &gt; 1)\n<\/li>\n<li>Logarithm of numbers less than one.\n<\/li>\n<li>Multiplication and division of numbers less than one using logarithm\n<\/li>\n<li>Roots and powers of numbers less than one using logarithm\n<\/li>\n<\/ul>\n<p>\u00a0<strong>Logarithm of numbers (Index &amp; Logarithmic Form)<br \/>\n<\/strong>The logarithm to base <strong>a<\/strong> of a number <strong>P<\/strong>, is the index <strong>x<\/strong> to which <strong>a <\/strong>must be raised to be equal to <strong>P<\/strong>.<strong><br \/>\n\t\t\t<\/strong>Thus if P = a<sup>x<\/sup>, then x is the logarithm to the base <strong>a <\/strong>of <strong>P<\/strong>. We write this as x = log a P. The relationship log<sub>a<\/sub>P = x and<br \/>\na<sup>x <\/sup>=P are equivalent to each other.<br \/>\na<sup>x <\/sup>=P is called the index form and log<sub>a<\/sub>P = x is called the logarithm form<\/p>\n<p>\u00a0<strong>Conversion from Index to Logarithmic Form<br \/>\n<\/strong>Write each of the following in index form in their logarithmic form<br \/>\na)\u00a0\u00a0\u00a0\u00a02<sup>6<\/sup> = 64\u00a0\u00a0\u00a0\u00a0                      b)\u00a0\u00a0\u00a0\u00a025<sup>1\/2<\/sup> = 5 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0c)  4<sup>4<\/sup>= 1\/256<br \/>\nSolution<br \/>\na)\u00a0\u00a0\u00a0\u00a02<sup>6<\/sup> = 64<br \/>\n\u00a0\u00a0\u00a0\u00a0Log<sub>2<\/sub> 64 = 6<br \/>\nb)\u00a0\u00a0\u00a0\u00a025<sup>1\/2<\/sup> = 5<br \/>\n\u00a0\u00a0\u00a0\u00a0Log<sub>25<\/sub>5=1\/2<br \/>\nc)\u00a0\u00a0\u00a0\u00a04<sup>-4<\/sup>= 1\/256<br \/>\nLog<sub>4<\/sub>1\/256 = -4<\/p>\n<p>\u00a0<strong>Conversion from Logarithmic to Index form<br \/>\n<\/strong>a)\u00a0\u00a0\u00a0\u00a0Log<sub>2<\/sub>128 = 7\u00a0\u00a0\u00a0\u00a0         b)\u00a0\u00a0\u00a0\u00a0log<sub>10<\/sub> (0.01) = -2                  c)\u00a0\u00a0\u00a0\u00a0Log<sub>1.5<\/sub> 2.25 = 2<br \/>\nSolution<br \/>\na)\u00a0\u00a0\u00a0\u00a0Log<sub>2<\/sub>128 = 7<br \/>\n\u00a0\u00a0\u00a0\u00a02<sup>7<\/sup> = 128<br \/>\nb)\u00a0\u00a0\u00a0\u00a0Log<sub>10 <\/sub>(0.01) = -2<br \/>\n\u00a0\u00a0\u00a0\u00a010<sup>-2<\/sup>= 0.01<br \/>\nc)\u00a0\u00a0\u00a0\u00a0Log1.5 2.25 = 2<br \/>\n1.5<sup>2<\/sup> = 2.25<\/p>\n<p>\u00a0<strong>Laws of Logarithm<br \/>\n<\/strong>a)\u00a0\u00a0\u00a0\u00a0let P = b<sup>x<\/sup>, then log<sub>b<\/sub>P = x<br \/>\n\u00a0\u00a0\u00a0\u00a0Q = b<sup>y<\/sup>, then log<sub>b<\/sub>Q = y<br \/>\n\u00a0\u00a0\u00a0\u00a0PQ = b<sup>x<\/sup> X b<sup>y<\/sup> = b<sup>x+y<\/sup> (laws of indices)<br \/>\n\u00a0\u00a0\u00a0\u00a0Log<sub>b<\/sub> PQ = x + y<br \/>\n:.\u00a0\u00a0\u00a0\u00a0Log<sub>b<\/sub> PQ = log<sub>b<\/sub>P + Log<sub>b<\/sub>Q<\/p>\n<p>\u00a0b)\u00a0\u00a0\u00a0\u00a0P\u00f7Q = b<sup>x<\/sup>\u00f7by = b<sup>x+y<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a0Log<sub>b<\/sub>P\/Q = x \u2013y<br \/>\n:.\u00a0\u00a0\u00a0\u00a0Log<sub>b<\/sub>P\/Q = logbP \u2013 logbQ<\/p>\n<p>\u00a0c)\u00a0\u00a0\u00a0\u00a0P<sup>n<\/sup>= (b<sup>x<\/sup>)<sup>n<\/sup> = b<sup>xn<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a0Log<sub>b<\/sub>p<sup>n<\/sup> = nb<sup>x<br \/>\n<\/sup>:.\u00a0\u00a0\u00a0\u00a0LogP<sup>n<\/sup> = logbP<\/p>\n<p>\u00a0d)\u00a0\u00a0\u00a0\u00a0b = b<sup>1<\/sup><br \/>\n\t\t:. \u00a0\u00a0\u00a0\u00a0Log<sub>b<\/sub>b = 1<\/p>\n<p>\u00a0e)\u00a0\u00a0\u00a0\u00a01 = b<sup>0<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a0Logb1 = 0<\/p>\n<p>\u00a0<strong>Example<\/strong><br \/>\n\t\tSolve each of the following:<br \/>\na)\u00a0\u00a0\u00a0\u00a0Log<sub>3<\/sub>27 + 2log<sub>3<\/sub>9 \u2013 log<sub>3<\/sub>54<br \/>\nb)\u00a0\u00a0\u00a0\u00a0Log<sub>3<\/sub>13.5 \u2013 log<sub>3<\/sub>10.5<br \/>\nc)\u00a0\u00a0\u00a0\u00a0Log<sub>2<\/sub>8 + log<sub>2<\/sub>3<br \/>\nd)\u00a0\u00a0\u00a0\u00a0Given that log<sub>10<\/sub>2 = 0.3010 log<sub>10<\/sub>3 = 0.4771 and log<sub>10<\/sub>5 = 0.699 find the log<sub>10<\/sub>64 + log<sub>10<\/sub>27<br \/>\n<strong>Solution<br \/>\n<\/strong>a)\u00a0\u00a0\u00a0\u00a0Log<sub>3<\/sub>27 + 2 log<sub>3<\/sub>9 \u2013 log<sub>3<\/sub>54<br \/>\n= log<sub>3<\/sub> 27 + log<sub>3<\/sub> 9<sup>2<\/sup> \u2013log<sub>3<\/sub>54<br \/>\n= log<sub>3<\/sub> (27 x 9<sup>2<\/sup>\/54)<br \/>\n= log<sub>3<\/sub> (27<sup>1<\/sup> x 81\/54) = log<sub>3<\/sub> (81\/2)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n= log<sub>3<\/sub> 3<sup>4<\/sup>\/<sup><br \/>\n\t\t\t<\/sup>log32<br \/>\n= 4log<sub>3<\/sub> 3 \u2013 log<sub>3<\/sub> 2<br \/>\n= 4 x (1) \u2013 log<sub>3<\/sub> 2 = 4 \u2013 log<sub>3<\/sub> 2<br \/>\n= 4 &#8211; log<sub>3 <\/sub>2<\/p>\n<p>\u00a0b)\u00a0\u00a0\u00a0\u00a0log<sub>3 <\/sub>13.5 &#8211; log<sub>3 <\/sub>10.5<br \/>\n\u00a0\u00a0\u00a0\u00a0= log<sub>3 <\/sub>(13.5) &#8211; Log310.5 = log<sub>3 <\/sub>(135\/105)<br \/>\n\u00a0\u00a0\u00a0\u00a0= log<sub>3 <\/sub>(27\/21) = log<sub>3 <\/sub>27 &#8211; log<sub>3 <\/sub>21<br \/>\n\u00a0\u00a0\u00a0\u00a0= log<sub>3 <\/sub>3<sup>3<\/sup> &#8211; log<sub>3 <\/sub>(3 x 7)<br \/>\n= 3log<sub>3 <\/sub>3 &#8211; log<sub>3 <\/sub>3 -log<sub>3<\/sub>7<br \/>\n= 2 &#8211; Log<sub>3 <\/sub>7<\/p>\n<p>\u00a0c)\u00a0\u00a0\u00a0\u00a0Log<sub>2<\/sub>8 + Log<sub>3<\/sub>3<br \/>\n\u00a0\u00a0\u00a0\u00a0= log<sub>2<\/sub>2<sup>3<\/sup>+ log<sub>3<\/sub>3<br \/>\n\u00a0\u00a0\u00a0\u00a0= 3log<sub>2<\/sub>2 + log<sub>3<\/sub>3<br \/>\n             = 3 +1 = 4<\/p>\n<p>\u00a0d)\u00a0\u00a0\u00a0\u00a0log<sub>10<\/sub> 64 + log<sub>10<\/sub> 27<br \/>\n\u00a0\u00a0\u00a0\u00a0log<sub>10<\/sub> 2<sup>6<\/sup> + log<sub>10<\/sub>3<sup>3<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a06 log<sub>10<\/sub> 2 + 3 log<sub>10 <\/sub>3<br \/>\n\u00a0\u00a0\u00a0\u00a06 (0.3010) + 3(0.4771)<br \/>\n\u00a0\u00a0\u00a0\u00a01.806 + 1.4314 = 3.2373.<\/p>\n<p>\u00a0<strong>EVALUATION<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0Change the following index form into logarithmic form.<br \/>\n            (a)  6<sup>3<\/sup>= 216\u00a0\u00a0\u00a0\u00a0(b) 3<sup>3<\/sup> = 1\/27\u00a0\u00a0\u00a0\u00a0(c) 9<sup>2<\/sup> = 81<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Change the following logarithm form into index form.<br \/>\n            (a)  Log<sub>8<\/sub>8 = 1  \u00a0\u00a0\u00a0\u00a0(b) log <sub>\u00bd<\/sub>\u00bc = 2<br \/>\n3.\u00a0\u00a0\u00a0\u00a0Simplify the following<br \/>\n             a)  Log<sub>5<\/sub>12.5 + log52               b)\u00a0\u00a0\u00a0\u00a0\u00bd log<sub>4<\/sub>8 + log<sub>4<\/sub>32 \u2013 log<sub>4<\/sub>2      c) log<sub>3<\/sub>81<br \/>\n4. \u00a0\u00a0\u00a0\u00a0Given that log 2 = 0.3010, log3 0.4770, log5 = 0.6990,   find the value of log<sub><br \/>\n\t\t\t<\/sub>6.25 + log1.44 <\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>Logarithmic Equation<br \/>\n<\/strong>Solve the following equation:<br \/>\na)\u00a0\u00a0\u00a0\u00a0Log10 (x<sup>2<\/sup> \u2013 4x + 7) = 2<br \/>\n b)\u00a0\u00a0\u00a0\u00a0Log<sub>8<\/sub> (r<sup>2<\/sup> \u2013 8r + 18) = 1\/3<\/p>\n<p>\u00a0<strong>Solution<\/strong><br \/>\n\t\ta)\u00a0\u00a0\u00a0\u00a0Log<sub>10<\/sub> (x<sup>2<\/sup> \u2013 4x + 7) = 2<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 4x + 7 = 10<sup>2<\/sup> (index form)<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 4x + 7 = 100<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 4x + 7 \u2013 100 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 4x \u2013 93 = 0<br \/>\nUsing quadratic formula<br \/>\n           x\u00a0\u00a0\u00a0\u00a0= &#8211; b \u00b1\u221ab<sup>2<\/sup>\u2013 4ac<br \/>\n                                    2a\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0a = 1, b = &#8211; 4, c = &#8211; 93\u00a0\u00a0\u00a0\u00a0<br \/>\n         x = &#8211; (- 4) \u00b1 \u221a(- 4) <sup>2 <\/sup>\u2013 4 x 1 x (- 93)<br \/>\n\t\t                                2 x 1<br \/>\n\u00a0\u00a0\u00a0\u00a0= + 4 \u00b1 \u221a16 + 372<br \/>\n                             2<br \/>\n\u00a0\u00a0\u00a0\u00a0= + 4 \u00b1 \u221a388\/2<br \/>\n\u00a0\u00a0\u00a0\u00a0= x = 4 +\u221a 388\/2 or 4 &#8211; \u221a388\/2<br \/>\n\u00a0\u00a0\u00a0\u00a0 x =  11.84 or x = &#8211; 7.85<\/p>\n<p>\u00a02) \u00a0\u00a0\u00a0\u00a0Log<sub>8<\/sub> (x<sup>2<\/sup> \u2013 8x + 18) =1\/3<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 8x + 18 = 8<sup>1\/3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 8x + 18 = (2)<sup>3X1\/3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 8x + 18 =2<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 8x 18 \u2013 2 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 8x + 16 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0x<sup>2<\/sup> \u2013 4x \u2013 4x + 16 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0x(x &#8211; 4) -4 (x &#8211; 4) = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0(x &#8211; 4) (x &#8211; 4) = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0(x &#8211; 4) twice<br \/>\n\u00a0\u00a0\u00a0\u00a0 x = + 4 twice<\/p>\n<p>\u00a0<strong>Change of Base<br \/>\n<\/strong>Let log<sub>b<\/sub>P = x and this means P = b<sup>x<br \/>\n<\/sup>Log<sub>c<\/sub>P = log<sub>c<\/sub>b<sup>x<\/sup> = x log<sub>c<\/sub>b<br \/>\nIf x log<sub>c<\/sub>b = log<sub>c<\/sub>P<br \/>\n          x = log<sub>c<\/sub>P<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0     log<sub>c<\/sub> b<br \/>\n:.\u00a0\u00a0\u00a0\u00a0log<sub>c<\/sub>P = log<sub>c<\/sub>P<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  log<sub>c<\/sub>b<br \/>\n<strong>Example<\/strong> :<br \/>\nShows that log<sub>a<\/sub>b   x   log<sub>b<\/sub>a  = 1<br \/>\n\u00a0\u00a0\u00a0\u00a0     Log<sub>a<\/sub>b = log<sub>c<\/sub>b<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0       log<sub>c<\/sub>a<br \/>\n\u00a0\u00a0\u00a0\u00a0     Log<sub>b<\/sub>a = log<sub>c<\/sub>a<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0       log<sub>c<\/sub>b<br \/>\n:.\u00a0\u00a0\u00a0\u00a0log<sub>a<\/sub>b   x log<sub>b<\/sub>a  =  log<sub>c<\/sub>b  x log<sub>c<\/sub>a<br \/>\n\u00a0\u00a0\u00a0\u00a0log<sub>c<\/sub>a  +  log<sub>c<\/sub>b = 1<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>Evaluation<\/strong><br \/>\n\t\tSolve (i) Log<sub>3<\/sub> (x<sup>2<\/sup> + 7x + 21) = 2 (ii) Log<sub>10<\/sub> (x<sup>2<\/sup> \u2013 3x + 12) = 1<br \/>\n (iii) 5<sup>2x+1 <\/sup>&#8211; 26(5<sup>x<\/sup>) + 5 = 0 find the value of x<\/p>\n<p>\u00a0<strong>Logarithm of numbers greater than one<br \/>\n<\/strong>Numbers such as 1000 can be converted to its power of ten in the form 10<sup>n<\/sup> where n can be term as the number of times the decimal point is shifted to the front of the first significant figure i.e. 10000 = 10<sup>4<\/sup><br \/>\n\t\tNumber                                                 Power of 10<\/p>\n<ol>\n<li>10<sup>2<\/sup>\n\t\t\t<\/li>\n<li>10<sup>1<\/sup>\n\t\t\t<\/li>\n<li>10<sup>0<\/sup>\n\t\t\t<\/li>\n<li>10<sup>-3<\/sup>\n\t\t\t<\/li>\n<li>10<sup>-1<\/sup>\n\t\t\t<\/li>\n<\/ol>\n<p>Note: One tenth; one hundredth, etc are expressed as negative powers of 10 because the decimal point is shifted to the right while that of whole numbers are shifted to the left to be after the first significant figure.<br \/>\nA number in the form A x 10<sup>n<\/sup>, where A is a number between 1 and 10 i.e. 1 &lt; A &lt; 10 and n is an integer is said to be in <strong><em>standard form<\/em><\/strong>  e.g. 3.835 x 10<sup>3<\/sup> and 8.2 x 10<sup>-5<\/sup> are numbers in standard form.<\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>Examples<\/strong><br \/>\n\t\t Express the following in standard form<br \/>\n1)  7853   2)  382      3) 0.387   4)  0.00104<br \/>\n<strong>Solution<br \/>\n<\/strong>1)  7853 = 7.853 x 10<sup>3<\/sup><br \/>\n\t\t2)  382 = 3.82 x 10<sup>2<\/sup><br \/>\n\t\t3)  0.387 = 3.87 x 10<sup>-1<\/sup><br \/>\n\t\t4)  0.00104 = 1.04 x 10<sup>-3<\/sup><\/p>\n<p>\u00a0Base ten logarithm of a number is the power to which 10 is raised to give that number e.g.<br \/>\n628000 = 6.28 x10<sup>5<\/sup><br \/>\n\t\t628000 = 10<sup>0.7980<\/sup> x 10<sup>5<\/sup><br \/>\n\t\t             = 10<sup>0.7980<\/sup><br \/>\n\t\t\t<sup>+ 5<\/sup><br \/>\n\t\t             = 10<sup>5.7980<\/sup><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi1.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi2.png\" alt=\"\"\/>Log 628000 = 5.7980<strong><br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<br \/>\n\u00a0          Integer<strong><br \/>\n\t\t\t<\/strong>Fraction (mantissa)<br \/>\nIf a number is in its standard form, its power is its integer i.e. the integer of its logarithm e.g. log 7853 has integer 3 because 7853 = 7.853 x 10<sup>3<\/sup><br \/>\n\t\tExamples: Use tables (log) to find the complete logarithm of the following numbers.<br \/>\n(a)  80030\u00a0\u00a0\u00a0\u00a0   (b) 8      (c) 135.80<br \/>\n<strong><em>Solution:<br \/>\n<\/em><\/strong>(a)\u00a0\u00a0\u00a0\u00a080030\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a04.9033<br \/>\n(b)\u00a0\u00a0\u00a0\u00a08\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a00.9031<br \/>\n(c)\u00a0\u00a0\u00a0\u00a013580\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a02.1329<\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>Multiplication and Division of number greater than one using logarithm<br \/>\n<\/strong>To multiply and divide numbers using logarithms, first express the number as logarithm and then apply the addition and subtraction laws of indices to the logarithms. Add the logarithm when multiplying and subtract when dividing.<br \/>\nExamples: Evaluate using logarithm.<br \/>\n1.\u00a0\u00a0\u00a0\u00a04627 x 29.3<br \/>\n2.\u00a0\u00a0\u00a0\u00a08198 \u00f7 3.905<br \/>\n3.\u00a0\u00a0\u00a0\u00a048.63  x  8.53<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0      15.39<\/p>\n<p>\u00a0<strong><em>Solutions<br \/>\n<\/em><\/strong>1.\u00a0\u00a0\u00a0\u00a04627 x 29.3\u00a0\u00a0\u00a0\u00a0                      \u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi3.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<strong>\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0No\u00a0\u00a0\u00a0\u00a0  Log<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi4.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04627\u00a0\u00a0\u00a0\u00a0     3.6653<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a029.3\u00a0\u00a0\u00a0\u00a0  + 1.4669<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi5.png\" alt=\"\"\/>   Antilog \u2192   135600     5.1322<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi6.png\" alt=\"\"\/><strong>\u00a0\u00a0\u00a0\u00a0                                                              <\/strong><br \/>\n\t\t<strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>\\   4627 x  29.3 = <strong>135600<br \/>\n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0<br \/>\nTo find the Antilog of the log 5.1322 use the antilogarithm table:<br \/>\nCheck 13 under 2 diff 2 (add the value of the difference) the number is 0.1356. To place the decimal point at the appropriate place, add one to the integer of the log i.e. 5 + 1 = 6 then shift the decimal point of the antilog figure to the right (positive) in 6 places.<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi7.png\" alt=\"\"\/><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0       =    135600<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi8.png\" alt=\"\"\/>     2.\u00a0\u00a0\u00a0\u00a0       819.8 x 3.905\u00a0\u00a0\u00a0\u00a0                      \u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0<strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0No\u00a0\u00a0\u00a0\u00a0  Log<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0819.8\u00a0\u00a0\u00a0\u00a0  2.9137<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03.905\u00a0\u00a0\u00a0\u00a0  0.5916<br \/>\n           antilog \u2192    <\/p>\n<p>\t\t\t<strong>209.9<\/strong>         2.3221<br \/>\n<strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<\/strong><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>\\   819.8 \u00f7 3.905   = 209.9<strong><br \/>\n\t\t\t\t\t<\/strong>3.\u00a0\u00a0\u00a0\u00a048.63 x 8.53<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0      15.39<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi9.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<strong>\u00a0\u00a0\u00a0\u00a0No\u00a0\u00a0\u00a0\u00a0    Log<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a048.63\u00a0\u00a0\u00a0\u00a0    1.6869<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0              8.53\u00a0\u00a0\u00a0\u00a0  +0.9309<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    2.6178<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0        \u00f7 15.39        -1.1872<br \/>\nantilog \u2192        <strong>26.95<\/strong>         1.4306<br \/>\n<strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<\/strong><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>\\<br \/>\n\t\t\t\t48.63 \u00f7 8.53   = 26.96<strong><br \/>\n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0            15.39<br \/>\n<strong>Evaluation:<br \/>\n<\/strong>1.    Use table to find the complete logarithm of the following:<strong><br \/>\n\t\t\t<\/strong>(a)  183      (b) 89500     (c) 10.1300      (d) 7<\/p>\n<p>\u00a02    Use logarithm to calculate.     3612 x 750.9<br \/>\n\t\t                                                        113.2 x 9.98<\/p>\n<p>\u00a0<strong>Using logarithm to solve problems with powers and root (numbers greater than one).<br \/>\n<\/strong><strong>Examples:<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi10.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi11.png\" alt=\"\"\/>Evaluate<br \/>\n(a)\u00a0\u00a0\u00a0\u00a03.53<sup>3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b)    4    40000\u00a0\u00a0\u00a0\u00a0      (c)        94100 x 38.2<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0                   5.68<sup>3<\/sup> x 8.14          correct to 2s.f.<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi12.png\" alt=\"\"\/><strong>Solution<br \/>\n<\/strong><strong>No.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log_____<br \/>\n<\/strong>3.53<sup>3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.5478 x 3<br \/>\n44.00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01.6434<br \/>\n\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\\ 3.53<sup>3<\/sup> = 44.00<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi13.png\" alt=\"\"\/>(b)\u00a0\u00a0\u00a0\u00a04    4000<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi14.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi15.png\" alt=\"\"\/><strong>No.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log_____<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi16.png\" alt=\"\"\/><br \/>\n\t\t 4 4000      3.6021 \u00f7 4<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi17.png\" alt=\"\"\/><br \/>\n\t\t7.952\u00a0\u00a0\u00a0\u00a0       0.9005<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi18.png\" alt=\"\"\/><strong>\\    4 4000   =   7.952<br \/>\n\t\t\t\t\t<\/strong><br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi19.png\" alt=\"\"\/>(c)\u00a0\u00a0\u00a0\u00a0 94100  x 38.2<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0 5.683<sup>3<\/sup> x 8.14<\/p>\n<p>\u00a0Find the single logarithm representing the numerator and the single logarithm representing the denominator, subtract the logarithm then find the anti log.<br \/>\n(Numerator \u2013 Denominator).<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi20.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<strong>No\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi21.png\" alt=\"\"\/><br \/>\n\t\t       94100\u00a0\u00a0\u00a0\u00a0                 4.9736 \u00f7 2   = 2.4868<br \/>\n          38.2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0              1.5821<br \/>\n\t\tNumerator\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  <strong>4.0689<\/strong>       \u2192 4.0689<br \/>\n5.68<sup>3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   0.7543 x 3   = 2.2629\u00a0\u00a0\u00a0\u00a0<br \/>\n8.14\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.9106<br \/>\n\t\t<strong>Denominator\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0              3.1735        \u2192 3.1735<br \/>\n\t\t\t<\/strong><strong>7.859\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   0.8954<br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi22.png\" alt=\"\"\/>\\<br \/>\n\t\t\t\t94100  x  38.2   =  7.859\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0        5.68<sup>3<\/sup> x 8.14<br \/>\n<strong>      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0~<br \/>\n\t\t\t\t\t    7.9 (2.sf)<br \/>\n<\/strong><strong>Evaluation:<\/strong> \u00a0\u00a0\u00a0\u00a0<br \/>\nEvaluate using logarithm.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi23.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a095.3 x    318.4<br \/>\n\u00a0\u00a0\u00a0\u00a0 1.29<sup>5<\/sup> x 2.03<strong><br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<strong>Logarithm of number less than one<br \/>\n<\/strong>To find the logarithm of number less than one, use negative power of 10 e.g.<br \/>\n\u00a0\u00a0\u00a0\u00a00.037\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a03.7 x 10<sup>-2<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010 <sup>0.5682 x 10-2<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010 <sup>0.5682 + (-2)<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010<sup>-2 5682<\/sup><br \/>\n\t\tLog 0.037 =  2 . 5682<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi24.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi25.png\" alt=\"\"\/>2     .\u00a0\u00a0\u00a0\u00a0   5682<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi26.png\" alt=\"\"\/><br \/>\n\t\tInteger\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0decimal fraction (mantissa)<\/p>\n<p>\u00a0<strong>Example:<br \/>\n<\/strong> Find the complete log of the following.<br \/>\n(a)\u00a0\u00a0\u00a0\u00a00.004863     (b) 0.853    (c) 0.293<\/p>\n<p>\u00a0<strong>Solution<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi27.png\" alt=\"\"\/>Log 0.004863\u00a0\u00a0\u00a0\u00a0                =\u00a0\u00a0\u00a0\u00a03.6369<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi28.png\" alt=\"\"\/>Log 0.0853\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   =\u00a0\u00a0\u00a0\u00a02.9309<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi29.png\" alt=\"\"\/>Log 0.293\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   =\u00a0\u00a0\u00a0\u00a01.4669<\/p>\n<p>\u00a0<strong> Evaluation<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0Find the logarithm of the following:<strong><br \/>\n\t\t\t\t<\/strong>(a)    0.064       (b)   0.002     (c) 0.802<br \/>\n<strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong><br \/>\n\t\t<strong>Using logarithm to evaluate problems of Multiplication, Division, Powers and roots with numbers less than One<br \/>\n<\/strong><br \/>\n\u00a0<strong>Examples:<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a00.6735 x 0.928               2.    0.005692 \u00b8 0.0943        3.\u00a0\u00a0\u00a0\u00a00.6104<sup>3<br \/>\n<\/sup><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi30.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi31.png\" alt=\"\"\/>4.\u00a0\u00a0\u00a0\u00a0  4 0.00083                     5.\u00a0\u00a0\u00a0\u00a03  0.06642<br \/>\n<strong>Solution<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a00.6735 x 0.928<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi32.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<strong>No.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log.___<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a00.6735\u00a0\u00a0\u00a0\u00a0            1.8283<br \/>\n\u00a0\u00a0\u00a0\u00a00.928\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01.9675<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi33.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi34.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<strong>0.6248\u00a0\u00a0\u00a0\u00a0             1.7958<br \/>\n<\/strong><br \/>\n\u00a0<strong>\u00a0\u00a0\u00a0\u00a0\\ 0.6735 x 0.928 = 0.6248<br \/>\n\t\t\t\t\t<\/strong><br \/>\n\u00a02.\u00a0\u00a0\u00a0\u00a00.005692 \u00b8  0.0943 <img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi35.png\" alt=\"\"\/><\/p>\n<p>\t\t\t<strong>No\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi36.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a00.005692\u00a0\u00a0\u00a0\u00a03.7553<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi37.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00b8 0.0943\u00a0\u00a0\u00a0\u00a02.9745<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi38.png\" alt=\"\"\/><br \/>\n\t\t\t<strong>0.06037\u00a0\u00a0\u00a0\u00a0             2.7808<br \/>\n<\/strong><br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi39.png\" alt=\"\"\/>3.\u00a0\u00a0\u00a0\u00a00.6104<sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0<strong>No\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log_____<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi40.png\" alt=\"\"\/><strong>\u00a0\u00a0\u00a0\u00a0<\/strong>0.6104<sup>3<\/sup>   \u00a0\u00a0\u00a0\u00a01.7856 x 3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi41.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a00.2274\u00a0\u00a0\u00a0\u00a0             1.3568<br \/>\n\u00a0\u00a0\u00a0\u00a0\\  0.6104<sup>3<\/sup>      =  <strong>0.2274<br \/>\n<\/strong><br \/>\n\u00a0<strong>\u00a0\u00a0\u00a0\u00a0<\/strong>\\  0.005692  \u00b8   0.943   =  <strong>0.6037<br \/>\n\t\t\t\t\t<\/strong><br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi42.png\" alt=\"\"\/>4.\u00a0\u00a0\u00a0\u00a04   0.00083<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi43.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<strong>No.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log._____<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi44.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi45.png\" alt=\"\"\/>4  0.00083\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04.9191  \u00b8 4<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi46.png\" alt=\"\"\/><strong>\u00a0\u00a0\u00a0\u00a0  <\/strong>0.1697\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0             1.2298<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi47.png\" alt=\"\"\/><\/p>\n<p>\t\t\t\\ 4  0.06642  =  <strong>0.1697<\/strong><\/p>\n<p>\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi48.png\" alt=\"\"\/>5.    3   0.6642<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi49.png\" alt=\"\"\/><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0<strong>No.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Log.____________<br \/>\n<\/strong><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi50.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi51.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<br \/>\n3  0.6642\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02.8223 \u00b8 3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi52.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02.1 + 1 + 0.8223 \u00b8 3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi53.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03 + 1 .8223 \u00b8 3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi54.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 + 0.6074<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi55.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<strong>0.405<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01.6074<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi56.png\" alt=\"\"\/><sup>    3<\/sup>    0.6642\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<strong>0.405<\/strong>\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0Note:  3 cannot divide 2 therefore subtract 1 from the negative integer and add 1 to the positive decimal fraction so as to have 3 which is divisible by 3 without remainder.<\/p>\n<p>\u00a0<strong>Evaluation:<br \/>\n<\/strong>Evaluate using logarithm tables:<br \/>\n(1)       \u221a12.3 x 0.0034<sup>3<\/sup><strong>\u00a0\u00a0\u00a0\u00a0<br \/>\n<\/strong><strong>            \u00a0\u00a0\u00a0\u00a0      <\/strong>132.5<br \/>\n(2)\u00a0\u00a0\u00a0\u00a023.97   x   0.7124<\/p>\n<ol>\n<li><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi57.png\" alt=\"\"\/>  x     52.18\n<\/li>\n<\/ol>\n<p>\u00a0<strong>General Evaluation<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0Solve the logarithmic equation: Log<sub>4<\/sub> (x<sup>2<\/sup> + 6x + 11) = \u00bd<br \/>\n2.         Log<sub>2 <\/sub>(x<sup>2<\/sup>&#8211; 2) =log<sub>2<\/sub>(x-1) + 1<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi58.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<br \/>\n3.\u00a0\u00a0\u00a0\u00a0Evaluate  5  (0.1684)<sup>3<\/sup><\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi59.png\" alt=\"\"\/>4.\u00a0\u00a0\u00a0\u00a06.28 x   304<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0             981<br \/>\n5.       16<sup>3\/2 <\/sup>x 8<sup>2\/<\/sup><sup>3<\/sup><\/p>\n<p>\t\t\t               32<sup>1\/5<br \/>\n<\/sup><br \/>\n\u00a0<strong>Reading Assignment<\/strong>: <em>Further Mathematics Project Book 1(New third edition).Chapter 2 pg.10- 16<\/em><\/p>\n<p>\u00a0<strong>Weekend Assignment<br \/>\n<\/strong>1.)\u00a0\u00a0\u00a0\u00a0If log81\/64 = x, find the value of x  (a) 2  (b) 1      (c) -3      (d) -4.<br \/>\n2.)\u00a0\u00a0\u00a0\u00a0Solve 9<sup>(1 &#8211; x)<\/sup> = (1\/27) <sup>x+1<\/sup>  (a) -5        (b) -1     (c) 1        (d) \u00bd<br \/>\nUse table to find the log of the following:<br \/>\n3.)\u00a0\u00a0\u00a0\u00a0900\u00a0\u00a0\u00a0\u00a0(a) 3.9542  (b) 1.9542         (c) 2.9542           (d) 0.9542<br \/>\n4.)\u00a0\u00a0\u00a0\u00a00.000197\u00a0\u00a0\u00a0\u00a0(a) 4.2945\u00a0\u00a0\u00a0\u00a0(b) 4.2945\u00a0\u00a0\u00a0\u00a0(c)   3.2945\u00a0\u00a0\u00a0\u00a0  (d) 3.2945<br \/>\n5.)\u00a0\u00a0\u00a0\u00a0Use antilog table to write down the number whose logarithms is 3.8226.<br \/>\n\u00a0\u00a0\u00a0\u00a0(a) 0.6646      (b) 0.06646     (c) 0.006646    (d) 66.46 <\/p>\n<p>\u00a0<strong>Theory<br \/>\n<\/strong><\/p>\n<ol>\n<li><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_0722_Week3SS1Fi60.png\" alt=\"\"\/>Find the value of x for which log<sub>10<\/sub> (4x<sup>2<\/sup> + 1) -2 log<sub>10<\/sub> x \u2013 log<sub>10<\/sub> 2 = 1 is valid.\n<\/li>\n<\/ol>\n<p>(2.) Evaluate using logarithm.    3   69.5<sup>2 <\/sup>\u2013 30.5<sup>2<br \/>\n<\/sup><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0 \u00a0WEEK THREE TOPIC: LOGARITHM &#8211; SOLVING PROBLEMS BASED ON LAWS OF LOGARITHM CONTENT Logarithm&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,173],"tags":[],"class_list":["post-1922","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss1-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1922","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=1922"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1922\/revisions"}],"predecessor-version":[{"id":1923,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1922\/revisions\/1923"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=1922"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=1922"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=1922"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}