{"id":3882,"date":"2023-10-06T06:17:14","date_gmt":"2023-10-06T06:17:14","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3882"},"modified":"2023-10-06T06:20:05","modified_gmt":"2023-10-06T06:20:05","slug":"week-2-ss3-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss3-first-term-mathematics-notes\/","title":{"rendered":"Week 2 &#8211; SS3 First Term Mathematics Notes"},"content":{"rendered":"<p><strong>  WEEK 2<br \/>\n<\/strong><strong>TOPIC: INDICIAL\/EXPONENTIAL EQUATION<br \/>\n<\/strong><strong>CONTENT:<br \/>\n<\/strong><\/p>\n<ul>\n<li>Exponential Equation of Linear Form<strong><br \/>\n\t\t\t<\/strong><\/li>\n<li>Exponential Equation of Quadratic Form<strong><br \/>\n\t\t\t<\/strong><\/li>\n<\/ul>\n<p>Under exponential equation, if the base numbers of any equation are equal, then the power will be equal&amp;vice versa.<br \/>\n<strong>Exponential Equation of Linear Form<br \/>\n<\/strong>Solve the following exponential equations<br \/>\na)\u00a0\u00a0\u00a0\u00a0(1\/2) <sup>x<\/sup> = 8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0b) (0.25) <sup>x+1<\/sup> = 16<br \/>\nc)\u00a0\u00a0\u00a0\u00a03<sup>x<\/sup> = 1\/81\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0d) 10 <sup>x<\/sup> = 1\/0.001<br \/>\ne)\u00a0\u00a0\u00a0\u00a04\/2<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\tSolution<br \/>\na)\u00a0\u00a0\u00a0\u00a0(1\/2) <sup>X<\/sup> = 8<br \/>\n(2 <sup>-1<\/sup>) <sup>x<\/sup> = 2 <sup>3<\/sup><br \/>\n\t2 <sup>\u2013x<\/sup> = 2 <sup>3<\/sup><br \/>\n\t-x = 3<br \/>\nX = &#8211; 3<br \/>\nb)\u00a0\u00a0\u00a0\u00a0(0.25) <sup>x+1<\/sup> = 16<br \/>\n\u00a0\u00a0\u00a0\u00a0(25\/100) <sup>x+1<\/sup> = 16<br \/>\n\u00a0\u00a0\u00a0\u00a0(1\/4) <sup>x+1<\/sup> = 4 <sup>2<\/sup><br \/>\n\t          (4 <sup>-1<\/sup>) <sup>x+1<\/sup> = 4 <sup>2<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0&#8211; x \u2013 1 = 2<br \/>\n\u00a0\u00a0\u00a0\u00a0&#8211; x = 2 + 1<br \/>\n\u00a0\u00a0\u00a0\u00a0&#8211; x = 3<br \/>\nX = &#8211; 3<br \/>\nc)\u00a0\u00a0\u00a0\u00a03<sup>x<\/sup> = 1\/81<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup>x<\/sup> = 1\/3<sup>4<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a03<sup>x<\/sup> = 3 <sup>-4<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0X = -4<br \/>\nd)\u00a0\u00a0\u00a0\u00a010<sup>x<\/sup> = 1\/0.001<br \/>\n\u00a0\u00a0\u00a0\u00a010 <sup>x<\/sup> = 1000<br \/>\n\u00a0\u00a0\u00a0\u00a010 <sup>x<\/sup> = 10 <sup>3<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a010<sup>x<\/sup> = 10 <sup>3<\/sup><br \/>\n\tX = 3<br \/>\ne)\u00a0\u00a0\u00a0\u00a04\/2<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a04\u00f72<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a02<sup>2<\/sup> \u00f72<sup>x<\/sup> = 64 <sup>x<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a02 <sup>2-x<\/sup> = (2 <sup>6<\/sup>) <sup>x<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a02 <sup>2-x<\/sup> = 2 <sup>6x<\/sup><br \/>\n\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<br \/>\n<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\u00a0c)\u00a0\u00a0\u00a0\u00a010 <sup>x<\/sup> = 1\/100000<br \/>\n<strong>Exponential Equation of Quadratic Form<br \/>\n<\/strong>Some exponential equation can be reduced to quadratic form as can be seen below.<br \/>\nExample :  Solve 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<br \/>\nSolution<br \/>\na)\u00a0\u00a0\u00a0\u00a02 <sup>2x<\/sup> \u2013 6 (2 <sup>x<\/sup>) + 8 = 0<br \/>\n(2 <sup>x<\/sup>)<sup>2<\/sup> \u2013 6 (2 <sup>x<\/sup>) + 8 = 0<br \/>\nLet 2<sup> x<\/sup> = y.<br \/>\nThen y<sup>2<\/sup> \u2013 6y + 8 = 0<br \/>\nThen factorize<br \/>\nY <sup>2<\/sup> \u2013 4 y \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 \/>\nSince 2 <sup>x<\/sup> = y, and y = 2<br \/>\n\u00a0\u00a0\u00a0\u00a02 <sup>x<\/sup> = 2<br \/>\n\u00a0\u00a0\u00a0\u00a02<sup> x<\/sup> = 2 <sup>1<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0x = 1<br \/>\nSince 2 <sup>x<\/sup> = y and y = 4<br \/>\n\u00a0\u00a0\u00a0\u00a02 <sup>x<\/sup> = 4<br \/>\n\u00a0\u00a0\u00a0\u00a02 <sup>x<\/sup> = 2 <sup>2<br \/>\n<\/sup>\u00a0\u00a0\u00a0\u00a0N = 2<br \/>\nX = 1 and 2<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 Factorise p<sup>2<\/sup> + 25p \u2013 5p \u2013 125 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0P (p + 25) -5 (p + 25) = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0(p &#8211; 5) (p + 25) = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0P \u2013 5 = 0 p + 25 = 0<br \/>\n\u00a0\u00a0\u00a0\u00a0P = 5 or p = &#8211; 25<br \/>\nSince 5<sup>x <\/sup>= p, \u00a0\u00a0\u00a0\u00a0p = 5<br \/>\n\u00a0\u00a0\u00a0\u00a05<sup> x <\/sup> = 5 <sup>1<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0X = 1<br \/>\n5x = -25 (Not simplified)\u00a0\u00a0\u00a0\u00a0<br \/>\n1)\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\u00a0a <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 \/>\nSince 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\u00a0\u00a0\u00a0\u00a0X = 1<br \/>\nSince 3<sup>x<\/sup> = a, \u00a0\u00a0\u00a0\u00a0when a = -3<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup> x <\/sup> = &#8211; 3\u00a0\u00a0\u00a0\u00a0(Not a solution)<br \/>\n<strong>EVALUATION<br \/>\n<\/strong>Solve the following exponential equations.<br \/>\na)\u00a0\u00a0\u00a0\u00a02 <sup>2x+ 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<strong>Reading Assignment<\/strong> : Further Maths Project Book 1(New third edition).Chapter 2 pg. 6- 10<\/p>\n<p>\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 (d) 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) It has no solution\n<\/li>\n<li>Solve for X : 10<sup>-X<\/sup> = 0.000001 (a) 4 (b) 6 (c) -6 (d) 5\n<\/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<strong><br \/>\n\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>WEEK 2 TOPIC: INDICIAL\/EXPONENTIAL EQUATION CONTENT: Exponential Equation of Linear Form Exponential Equation of Quadratic&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,302],"tags":[],"class_list":["post-3882","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3882","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=3882"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3882\/revisions"}],"predecessor-version":[{"id":3883,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3882\/revisions\/3883"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3882"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3882"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3882"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}