{"id":46942,"date":"2023-12-20T14:15:24","date_gmt":"2023-12-20T08:45:24","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=46942"},"modified":"2023-12-20T17:21:26","modified_gmt":"2023-12-20T11:51:26","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-9-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-9-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 9 Logarithms Chapter Test"},"content":{"rendered":"
Question 1.
\nExpand \\(\\log _{a} \\sqrt[3]{x^{7} y^{8} \\div \\sqrt[4]{z}}\\)
\nAnswer:
\n<\/p>\n
Question 2.
\nFind the value of \\(\\log _{\\sqrt{3}} 3 \\sqrt{3}-\\log _{5}(0 \\cdot 04)\\)
\nAnswer:
\n<\/p>\n
Question 3.
\nProve the following
\n
\nAnswer:
\n
\n= 2[log 11 – log 13] + [log 130 – log 77] – [log 55 – log 91]
\n= 2 [log – log 13] + [log 13 \u00d7 10 – log 11 \u00d7 7] – [log 11 – log 13 \u00d7 7]
\n= 2 [log 11 – log 13] + [(log 13 + log 10) – (log 11 + log 7)] – [(log 11 + log 5) – (log 13 + log 7)]
\n= 2 log 11 – 2 log 13 + log 10 – log 11 – log 7 – log 11 – log 5 + log 13 + log 7
\n= (2 log 11 – log 11 – log 11) + (-2 log 13 + log 13 + log 13) + log 10 – log 5 + (log 7 – log 7)
\n= 0 + 0 + log 10 – log 5 + 0 = log 10 – log 5
\n= log\\(\\left(\\frac{10}{5}\\right)\\) = 1og 2 = R.H.S
\nHence, Result is proved.<\/p>\n
Question 4.
\nIf log (m + n) = log m + log n, show that n = \\(\\frac{m}{m-1}\\).
\nAnswer:
\nGiven log (m + n) = log m + log n
\n\u21d2 log (m + n) = log mn \u21d2 m + n = mn
\n\u21d2 m = mn – n \u21d2 m = n (m -1)
\n\u21d2 n (m – 1) = m \u21d2 n = \\(\\frac{m}{m-1}\\)
\nHence, result is proved.<\/p>\n
Question 5. Question 6. Question 7. (vi) log (x2<\/sup> – 21) = 2 (vii) log6<\/sub> (x – 2) (x + 3) = 1 {\u2235 loga<\/sub> a= 1} (viii) log6<\/sub> (x – 2) + log6<\/sub> (x + 3) = 1 (ix) log (x +1) + log (x -1) = log 11 + 2 log 3 Question 8. Question 9. ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 9 Logarithms Chapter Test Question 1. Expand Answer: Question 2. Find the value of Answer: Question 3. Prove the following Answer: = 2[log 11 – log 13] + [log 130 – log 77] – [log 55 – log 91] = 2 [log – log 13] + […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[],"yoast_head":"\n
\nIf \\(\\log \\frac{x+y}{2}=\\frac{1}{2}(\\log x+\\log y)\\), prove that x = y.
\nAnswer:
\n
\nSquaring
\n\u21d2 (x + y)2<\/sup> = 4xy \u21d2 x2<\/sup> + y2<\/sup> + 2xy = 4xy
\n\u21d2 x2<\/sup> + y2<\/sup> + 2xy – 4xy = 0 \u21d2 x2<\/sup> + y2<\/sup> – 2xy = 0
\n\u21d2 (x – y)2<\/sup> = 0 \u21d2 x – y = 0
\n\u2234 x = y Hence proved.<\/p>\n
\nIf a, b are positive real numbers, a > b and a2<\/sup> + b2<\/sup> = 27 ab, prove that
\n
\nAnswer:
\na2<\/sup> + b2<\/sup> = 27ab
\n\u21d2 a2<\/sup> + b2<\/sup> – 2ab = 25ab
\n
\n
\nHence proved.
\nSolve the following equations for x<\/p>\n
\nSolve the following equations for x:
\n(i) logx<\/sub>\\(\\frac{1}{49}\\) = -2
\n(ii) logx<\/sub>\\(\\frac{1}{4 \\sqrt{2}}\\) = -5
\n(iii) logx<\/sub>\\(\\frac{1}{243}\\) = 10
\n(iv) logx<\/sub>32 = x – 4
\n(v) log7<\/sub> (2x2<\/sup> – 1) = 2
\n(vi) log (x2<\/sup> – 21) = 2
\n(vii) log6<\/sub> (x – 2) (x + 3) = 1
\n(viii) log6<\/sub> (x – 2) + log6<\/sub> (x + 3) = 1
\n(ix) log (x +1) + log (x -1) = log 11 + 2 log 3.
\nAnswer:
\n
\n
\n\u21d2 x = +5, -5<\/p>\n
\n(10)2<\/sup> = x2<\/sup> – 21 \u21d2 100 = x2<\/sup> – 21
\n\u21d2 x2<\/sup> – 21 = 100 \u21d2 x2<\/sup> = 100 + 21
\n\u21d2 x2<\/sup> = 121 \u21d2 x = \u00b1\\(\\sqrt{121}\\) \u21d2 x = \u00b111
\n\u2234 x = 11, -11<\/p>\n
\nComparing,
\n(x – 2)(x + 3) = 6
\n\u21d2 x2<\/sup> + 3x – 2x – 6 = 6
\n\u21d2 x2<\/sup> + x – 6 – 6 = 0
\n\u21d2 x2<\/sup> + x- 12 = 0
\n\u21d2 x2<\/sup> + 4x – 3x- 12 = 0
\n\u21d2 x (x + 4) – 3 (x + 4) = 0
\n\u21d2 (x + 4)(x – 3) = 0
\nEither x + 4 = 0, then x = -4
\nor x – 3 = 0, then x = 3
\nHence x = 3, -4<\/p>\n
\nlog6<\/sub> (x – 2) (x + 3) = 1 = log6<\/sub> 6 {\u2235 loga<\/sub> a = 1}
\nComparing,
\n(x – 2) (x + 3) = 6 => x2<\/sup> + 3x – 2x – 6 = 6
\n\u21d2 x2<\/sup> + x – 6 – 6 = 0 \u21d2 x2<\/sup> + x – 12 = 0
\n\u21d2 x2<\/sup> + 4x – 3x – 12 = 0
\n\u21d2 x (x + 4) – 3 (x + 4) = 0
\n\u21d2 (x + 4)(x – 3) = 0
\nEither x + 4 = 0, then x = -4
\nor x – 3 = 0, then x = 3
\n\u2234 x = 3, -4<\/p>\n
\n\u21d2 log [(x + 1) (x – 1)] = log 11 + log (3)2<\/sup>
\n\u21d2 log (x2<\/sup> – 1) = log 11 + log 9 [\u2235 a2<\/sup> – b2<\/sup> = (a + b)(a – b)]
\n\u21d2 log(x2<\/sup> – 1) = log(11 \u00d7 9) \u21d2 x2<\/sup> – 1 = 11 \u00d7 9
\n\u21d2 x2<\/sup> – 1 = 99 \u00d7 x2<\/sup> = 99 + 1 \u21d2 x2<\/sup> = 100
\n\u21d2 x2<\/sup> = (10)22<\/sup> \u21d2 x = 10<\/p>\n
\nSolve for x and y:
\n
\nAnswer:
\n
\n<\/p>\n
\nIf a = 1 + logx<\/sub>yz, b = 1 + logy<\/sub> zx and c = 1 + logz<\/sub>xy, then show that ab + bc + ca = abc.
\nAnswer:
\na = 1 + logx<\/sub>yz
\nb = 1 + logy<\/sub>zx
\nc = 1 + logz<\/sub>xy
\na = 1 + logx<\/sub>yz = logx<\/sub>x + logx<\/sub>yz
\n<\/p>\nML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"