{"id":1920,"date":"2023-10-02T07:21:39","date_gmt":"2023-10-02T07:21:39","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=1920"},"modified":"2023-10-02T07:28:26","modified_gmt":"2023-10-02T07:28:26","slug":"week-2-ss1-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss1-first-term-further-mathematics-notes\/","title":{"rendered":"Week 2 &#8211; SS1 First Term Further Mathematics Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK TWO<br \/>\n<\/strong><strong>TOPIC:  INDICIAL AND EXPONENTIAL EQUATIONS<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Exponential Equation of Linear Form<strong><br \/>\n\t\t\t\t<\/strong><\/li>\n<li>Exponential Equation of Quadratic Form<strong><br \/>\n\t\t\t\t<\/strong><\/li>\n<\/ul>\n<p><strong>Exponential Equation of Linear Form<br \/>\n<\/strong>Under exponential equation, if the base numbers of any equation are equal, then the power will be equal &amp; vice versa.<\/p>\n<p>\u00a0<strong>Examples<br \/>\n<\/strong>     Solve the following exponential equations<br \/>\n              a)  (1\/2) <sup>x<\/sup>  =  8   b)  (0.25) <sup>x+1<\/sup>  =  16    c)  3<sup>x<\/sup> = 1\/81\u00a0\u00a0\u00a0\u00a0   d) 10 <sup>x<\/sup> = 1\/0.001   e)  4\/2<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\t<strong>Solution<br \/>\n<\/strong>a)\u00a0\u00a0\u00a0\u00a0(1\/2) <sup>X<\/sup> = 8                                                     b)\u00a0\u00a0\u00a0\u00a0(0.25) <sup>x+1<\/sup> = 16<br \/>\n(2 <sup>-1<\/sup>) <sup>x<\/sup> = 2 <sup>3        <\/sup>                                                        (25\/100) <sup>x+1<\/sup> =  4<sup>2<\/sup><br \/>\n\t\t  2 <sup>\u2013x<\/sup> = 2 <sup>3<\/sup>                                                                 (1\/4) <sup>x+1<\/sup>  =  4<sup>2<\/sup><br \/>\n\t\t  -x = 3                                                                      (4<sup>-1<\/sup>)<sup> x + 1 <\/sup> = 4<sup>2<\/sup><br \/>\n\t\t   x = &#8211; 3                                                                     4 <sup>\u2013 x &#8211; 1   <\/sup>= 4<sup>2<\/sup><br \/>\n\t\t                                                                                                \u2013 x \u2013 1 = 2<br \/>\n\u00a0\u00a0\u00a0\u00a0                                                                                    &#8211; x = 2 + 1<br \/>\n                                                                                                 &#8211; x = 3<br \/>\n                                                                                                   X = &#8211; 3 <\/p>\n<p>\u00a0c)\u00a0\u00a0\u00a0\u00a03<sup>x<\/sup> = 1\/81                                                               d) 10 <sup>x<\/sup>  = 1\/0.001<br \/>\n            3<sup>x<\/sup> = 1\/3<sup>4<\/sup>                                                                     10 <sup>x<\/sup>  = 1000<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup>x<\/sup> = 3 <sup>-4<\/sup>                                                                       10 <sup>x<\/sup>  = 10 <sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0 x = -4                                                                         10 <sup>x<\/sup>  = 10 <sup>3<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0                                                                            x = 3<br \/>\ne)\u00a0\u00a0\u00a0\u00a04\/2<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a04\u00f72<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a02<sup>2<\/sup> \u00f72<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a02 <sup>2-x<\/sup> = (2 <sup>6<\/sup>) <sup>x<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a02 <sup>2-x<\/sup> = 2 <sup>6x<\/sup><br \/>\n\t\t2- x = 6x<br \/>\n2=6x+x<br \/>\n2 = 7x<br \/>\nDivide both sides by 7<br \/>\n2\/7 = 7x\/7<br \/>\nx = 2\/7<\/p>\n<p>\u00a0<strong>Evaluation<br \/>\n<\/strong>Solve the following exponential equations<br \/>\na)\u00a0\u00a0\u00a0\u00a02 <sup>x<\/sup> = 0.125 \u00a0\u00a0\u00a0\u00a0b) 25 <sup>(5x)<\/sup> = 625\u00a0\u00a0\u00a0\u00a0       c)\u00a0\u00a0\u00a0\u00a010 <sup>x<\/sup> = 1\/100000<\/p>\n<p>\u00a0<strong>Exponential Equation of Quadratic Form<br \/>\n<\/strong>Some exponential equation can be reduced to quadratic form as can be seen below.<br \/>\n<strong>Example<\/strong>:<br \/>\nSolve the following equations.<br \/>\na)\u00a0\u00a0\u00a0\u00a02<sup>2x<\/sup> \u2013 6 (2<sup>x<\/sup>) + 8 = 0<br \/>\nb)\u00a0\u00a0\u00a0\u00a05<sup>2x<\/sup> + 4 x 5 <sup>x+1<\/sup> \u2013 125 = 0<br \/>\nc)          3<sup>2x<\/sup> \u2013 9 = 0<\/p>\n<p>\u00a0<strong>Solution<br \/>\n<\/strong>      a)   2<sup>2x<\/sup> \u2013 6 (2<sup>x<\/sup>) + 8 = 0                                 When y = 4 then,         and \u00a0\u00a0\u00a0\u00a0When y = 2 then,<br \/>\n(2<sup>x<\/sup>)<sup>2<\/sup> \u2013 6 (2<sup>x<\/sup>) + 8 = 0                                          2 <sup>x<\/sup> = 4                                       2 <sup>x<\/sup> = 2<br \/>\nLet 2<sup>x<\/sup> = y                                                          2 <sup>x<\/sup> = 2 <sup>2                                                          <\/sup>2<sup> x<\/sup> = 2 <sup>1<br \/>\n<\/sup>Then y<sup>2<\/sup> \u2013 6y + 8 = 0                                           x = 2                                         x = 1\u00a0\u00a0\u00a0\u00a0<br \/>\nThen factorize                                                                     x = 1 and 2<br \/>\ny <sup>2<\/sup> \u2013 4y \u2013 2y + 8 = 0<br \/>\ny (y &#8211; 4) -2 (y -4) = 0<br \/>\n(y -2) (y &#8211; 4) = 0<br \/>\ny \u2013 2 = 0 or y \u2013 4 = 0<br \/>\ny = 2 or y= 4<br \/>\ny = 2, 4<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\nb)\u00a0\u00a0\u00a0\u00a05<sup>2x<\/sup> + 4 x 5<sup>x+1<\/sup> \u2013 125 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0(5<sup> x<\/sup>) <sup>2<\/sup> + 4 x (5<sup> x<\/sup> x 5<sup>1<\/sup>) \u2013 125 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0Let 5<sup> x<\/sup> = p<br \/>\n\u00a0\u00a0\u00a0\u00a0P <sup>2<\/sup> + 4 x (p x 5) \u2013 125 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0P<sup>2<\/sup> + 4 (5p) \u2013 125 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0P<sup>2<\/sup> + 20p \u2013 125 = 0<br \/>\nThen, Factorize p<sup>2<\/sup> + 25p \u2013 5p \u2013 125 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0                p (p + 25) &#8211; 5 (p + 25) = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0                   (p &#8211; 5) (p + 25) = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0                p \u2013 5 = 0 p + 25 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0                p = 5 or p = &#8211; 25<br \/>\n                           Since 5<sup>x <\/sup>= p, \u00a0\u00a0\u00a0\u00a0p = 5<br \/>\n                                     5<sup>x <\/sup> = 5 <sup>1<\/sup><br \/>\n\t\t                                       x = 1<br \/>\n                           5<sup>x<\/sup> = -25<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\nc)\u00a0\u00a0\u00a0\u00a03 <sup>2x<\/sup> \u2013 9 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0(3<sup> x<\/sup>) <sup>2<\/sup> &#8211; 9 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0Let 3<sup>x <\/sup> = a<br \/>\n\u00a0\u00a0\u00a0\u00a0 a<sup>2<\/sup> \u2013 9 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0a<sup>2<\/sup> = 9<br \/>\n\u00a0\u00a0\u00a0\u00a0a = \u00b1\u221a9<br \/>\n\u00a0\u00a0\u00a0\u00a0a = \u00b1 3<br \/>\n\u00a0\u00a0\u00a0\u00a0a = 3 or \u2013 3<br \/>\n           Since 3<sup>x <\/sup> = a, \u00a0\u00a0\u00a0\u00a0when a = 3<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup> x <\/sup> = 3<sup>1<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0x = 1<br \/>\n            Since 3<sup>x<\/sup> = a, \u00a0\u00a0\u00a0\u00a0when a = -3<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup> x <\/sup>= &#8211; 3<br \/>\n\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<strong>Evaluation:<\/strong><br \/>\n\t\tSolve: (a)   3(2<sup>2x + 3<\/sup>) &#8211; 5(2<sup>x+2<\/sup>) &#8211; 156 = 0         (b )\u00a0\u00a0\u00a0\u00a09<sup>2x+1 <\/sup>= (81 <sup>x-2<\/sup>\/3<sup>x<\/sup>)<\/p>\n<p>\u00a0<strong>General Evaluation<br \/>\n<\/strong>Solve the following exponential equations.<br \/>\na)\u00a0\u00a0\u00a0\u00a02<sup>2x<\/sup><br \/>\n\t\t\t<sup>+ 1<\/sup> \u2013 5 (2<sup>x<\/sup>) + 2 = 0<br \/>\nb)\u00a0\u00a0\u00a0\u00a03<sup>2x<\/sup> \u2013 4 (3<sup>x+1<\/sup>) + 27 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\n<strong>Reading Assignment<\/strong>: <em>Further Mathematics Project Book 1(New third edition).Chapter 2 pg. 6- 10<br \/>\n<\/em><br \/>\n\u00a0<strong>Weekend Assignment<br \/>\n<\/strong><\/p>\n<ol>\n<li>Solve for x : (0.25) <sup>X + 1<\/sup> = 16                                       (a) -3        (b) 3        (c) 4       (d) -4\n<\/li>\n<li>Solve for x : 3(3)<sup>X<\/sup> = 27                                               (a) 3         (b) 4        (c) 2       (d) 5\n<\/li>\n<li>Solve the exponential equation : 2<sup>2x<\/sup> + 2<sup>x+1<\/sup> \u2013 8 = 0   (a) 1         (b) 2        (c) 3       (d) 4\n<\/li>\n<li>The second value of x in question 3 is                      (a) -1        (b) 1        (c) 2       (d)  No solution\n<\/li>\n<li>\n<div>Solve for x : 10 <sup>-X<\/sup> = 0.000001                                   (a) 4         (b) 6        (c) -6      (d) 5\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>Theory<br \/>\n<\/strong>Solve the following exponential equations<br \/>\n         (1)  (3<sup>x<\/sup>)<sup>2<\/sup> + 2(3<sup>x<\/sup>)\u2013 3 = 0          (2)  5<sup>2x+1 <\/sup>&#8211; 26(5<sup>x<\/sup>) + 5 = 0<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK TWO TOPIC: INDICIAL AND EXPONENTIAL EQUATIONS CONTENT Exponential Equation of Linear Form Exponential Equation&#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-1920","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\/1920","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=1920"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1920\/revisions"}],"predecessor-version":[{"id":1921,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1920\/revisions\/1921"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=1920"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=1920"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=1920"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}